{"id":9185,"date":"2019-12-02T09:47:11","date_gmt":"2019-12-02T04:17:11","guid":{"rendered":"https:\/\/kseebsolutions.guru\/?p=9185"},"modified":"2021-07-02T15:44:22","modified_gmt":"2021-07-02T10:14:22","slug":"1st-puc-chemistry-question-bank-chapter-8","status":"publish","type":"post","link":"https:\/\/kseebsolutions.guru\/1st-puc-chemistry-question-bank-chapter-8\/","title":{"rendered":"1st PUC Chemistry Question Bank Chapter 8 Redox Reactions"},"content":{"rendered":"

You can Download Chapter 8 Redox Reactions Questions and Answers, Notes, 1st PUC Chemistry Question Bank with Answers<\/a> Karnataka State Board Solutions help you to revise complete Syllabus and score more marks in your examinations.<\/p>\n

Karnataka 1st PUC Chemistry Question Bank Chapter 8 Redox Reactions<\/h2>\n

1st PUC Chemistry Redox Reactions One Mark Questions and Answers<\/h3>\n

Question 1.
\nDefine oxidation in terms of electronic concept.
\nAnswer:
\nOxidation is a process in which loss of electrons take place.<\/p>\n

Question 2.
\nDefine by reduction in terms of electronic concept.
\nAnswer:
\nReduction is a process in which gain of electrons takes place.<\/p>\n

Question 3.
\nWhat are redox reactions ? Give an example.
\nAnswer:
\nRedox reaction is reaction in which oxidation and reduction take place
\nsimultaneously, e.g. Zn(s)+ Cu2+<\/sup> (aq) \u2192 Zn2+<\/sup> (aq) + Cu(s):<\/p>\n

Question 4.
\nWhat is oxidation number of alkali metals in their compounds?
\nAnswer:
\n+1.<\/p>\n

Question 5.
\nZn(s)+Cu2+<\/sup> \u2192 Zn2+<\/sup> + Cu(s) is this reaction redox reaction ? If yes, name the
\n\u2234 oxidizing agent as well as reducing agent.
\nAnswer:
\nYes, Cu2+<\/sup> is oxidising agent whereas Zn is reducing agent.<\/p>\n

\"KSEEB<\/p>\n

Question 6.
\nWhich of the following does not conduct electric current and why?
\nMolten NaCl, Solid Pb, AgNO3<\/sub> solution and methanol.
\nAnswer:
\nMethanol will not conduct electric current because it does not ionise. .<\/p>\n

Question 7.
\nDefine cathode and anode.
\nAnswer:
\nCathode is electrode towards which cations are attracted. Anode is electrode which : attracts anions.<\/p>\n

Question 8.
\nWhat is oxidation state of Cr in CrO5<\/sup> and why?
\nAnswer:
\nCr has +6 oxidation state because it has 6 valence electrons, therefore can form 6 covalent bonds.
\n\"1st<\/p>\n

Question 9.
\n\"1st
\nwhich is the strongest oxidising agent out of them ?
\nAnswer:
\nFe3+<\/sup> is strongest oxidising agent because it has standard reducing potential. (i.e., +ve value) value.<\/p>\n

\"KSEEB<\/p>\n

Question 10.
\nWhat is meant by oxidation potential of an electrode ?
\nAnswer:
\nOxidation potential measures the tendency of an element or anion to lose the electrons.<\/p>\n

Question 11.
\nWhat is relationship between standard oxidation potential and standard reduction potential?
\nAnswer:
\nBoth are equal in magnitude but opposite in sign.<\/p>\n

Question 12.
\nIdentify the oxidant and reductant in the following reactions :
\n(i) Zn(s) + \\(\\frac { 1 }{ 2 }\\) O2<\/sub>(g) \u2192 ZnO(s)
\n(ii) Zn(s) + 2H+<\/sup> (aq) \u2192 Zn2+<\/sup>(aq) + H2<\/sub>(g)
\nAnswer:
\n(i) Zn is reducing agent (reductant) and O2<\/sub> is oxidizing agent (oxidant)
\n(ii) Zn is reducing agent (reductant) whereas H+<\/sup> is oxidizing agent (oxidant).<\/p>\n

Question 13.
\nHow will you define (i) Oxidant and (ii) Reductant in terms of oxidation no.?
\nAnswer:
\n(i) Oxidant: Which decreases its own oxidation no. and increases the ox. no. of the other in a reaction.
\n(ii) Reductant : Which increases its own ox. no. and decreases that of other in a chemical reaction.<\/p>\n

\"KSEEB<\/p>\n

Question 14.
\nWhat do you mean by disproportionation reaction ?
\nAnswer:
\nA disproportionation reaction is an oxidation-reduction process in which the same substance is oxidized and reduced e.g., 2Cu+<\/sup> \u2192 Cu + Cu2+<\/sup><\/p>\n

Question 15.
\nDetermine the oxidation number of C in the following : C2<\/sub>H6<\/sub>, C4<\/sub>H10<\/sub>, CO, CO2<\/sub> and HCO3<\/sub>–<\/sup>
\nAnswer:
\nO.N. of C = -3 in C2<\/sub>H6<\/sub> ; -2.5 in C4<\/sub>H10<\/sub> ; +2 in CO ; +4 in CO2<\/sub> and +4 in HCO3<\/sub>–<\/sup><\/p>\n

Question 16.
\nDetermine the oxidation number of O in the followiing : Na2<\/sub>O2 <\/sub>and CH3<\/sub>COOH.
\nAnswer:
\nO.N. of+1 in Na2<\/sub>O2<\/sub> and -2 in CH3<\/sub>COOH<\/p>\n

Question 17.
\nFind out the oxidation number of sulphur in the following species : and HSO4<\/sub>–<\/sup>.
\nAnswer:
\nO.N. of S = + 6 in HSO4<\/sub>–<\/sup>.<\/p>\n

Question 18.
\nDetermine the oxidation number of all the atoms in the following well known oxidants KCI04.
\nAnswer:
\nK = +1, Cl = +7, O = -2<\/p>\n

Question 19.
\nDetermine the change in the oxidation number of S in H2<\/sub>S and SO2<\/sub> in the following industrial reaction : 2H2<\/sub>S(g) + SO2<\/sub>(g) \u2192 3S(g) + 2H2<\/sub>O(g)
\nAnswer:
\nO.N. of S changes from -2 in H2<\/sub>S and +4 in SO2<\/sub> to zero in elemental sulphur.<\/p>\n

\"KSEEB<\/p>\n

Question 20.
\nWhat is oxidised and what is reduced in the reaction ?
\nH2<\/sub>S + Cl2<\/sub> \u2192 2HCl + S
\nAnswer:
\nH2<\/sub>S is oxidised to S and Cl2<\/sub> is reduced to HCl.<\/p>\n

Question 21.
\nIn the reaction SnCl2<\/sub> + 2HgCl2<\/sub> \u2192 SnCl4<\/sub> + Hg2<\/sub>Cl2<\/sub>
\nwhich is oxidant and which is reductant ?
\nAnswer:
\nSnCl2<\/sub> is a reductant and HgCl2<\/sub> is an oxidant.<\/p>\n

Question 22.
\nIdentify the strongest and weakest reducing agents from the following list: Zn, Cu, Ag, Na, Sn.
\nAnswer:
\nStrongest reducing agent in Zn and weakest reducing agent in Ag.<\/p>\n

Question 23.
\nCan oxidation number of an atom in a chemical species he fractional ? Illustrate by an example.
\nAnswer:
\nYes, it can be fractional. For example oxidation no. of Pb in Pb3<\/sub>O4<\/sub> = + \\(\\frac { 8 }{ 3 }\\)<\/p>\n

\"KSEEB<\/p>\n

1st PUC Chemistry Redox Reactions Two Marks Questions and Answers<\/h3>\n

Question 1.
\nCalculate the oxidation number of underlined atoms in the following compounds and ions: C<\/span>H4<\/sub>, Sb2<\/sub>O5<\/sub>, C<\/span>6<\/sub>H12<\/sub>O6<\/sub>
\nAnswer:
\nCH4<\/sub>
\nX + 4 = 0;
\nX = -4<\/p>\n

Sb2<\/sub>O5<\/sub>
\n2x – 10= 0;
\n2x = 10
\nx = +5<\/p>\n

C6<\/sub>H12<\/sub>O6<\/sub>
\n6x + 12 – 12 = 0
\nx = 0<\/p>\n

Question 2.
\nIdentify the oxidant and reductant in the following chemical reaction:
\n2I– <\/sup>(aq) + Cl2 <\/sub>(g) \u2192 2Cl– <\/sup>(aq) + I2<\/sub>(s)
\nAnswer:
\n2I– <\/sup>(aq) + Cl2 <\/sub>(g) \u2192 2Cl– <\/sup>(aq) + I2<\/sub>(s)
\nOxidant is Cl2 <\/sub>(g) reductant is I– <\/sup>(aq)<\/p>\n

Question 3.
\nThe standard reduction potentials of Al and Ni are -1.66 V and -0.28 V, respectively. Is Al a stronger or weaker reducing agent than Ni ? Explain.
\nAnswer:
\n‘Al’ is stronger reducing agent than Ni because it has lower reduction potential. Al3+<\/sup> is more stable than Ni2+<\/sup> ion.<\/p>\n

Question 4.
\nThe standard reduction potentials of Zn2+<\/sup>, Mg2+<\/sup> and Na+<\/sup> are -0.76 V, -2.37 V and -2.71 Y respectively, which of the following is the strongest oxidizing agent?
\nAnswer:
\nZn2+<\/sup> is strongest oxidising agent. It can gain electrons easily. It has highest standard reduction potential.<\/p>\n

Question 5.
\nWhich of the following is best reducing agent and why ? Li, Cu, Br2<\/sub>, F2<\/sub>, H2<\/sub>, K.
\nAnswer:
\nLithium (Li) is best reducing agent because it has lowest standard reduction potential, i.e., Li+<\/sup> is most stable.<\/p>\n

Question 6.
\nCalculate the oxidation number of (i) C in CH2<\/sub>Cl2<\/sub> and (iii) Pb in Pb3<\/sub>O4<\/sub>.
\nAnswer:
\n(i) C in CH2<\/sub>Cl2<\/sub>. Let the oxidation number of C in CH2<\/sub>Cl2<\/sub> be x. Writing the oxidation number of each atom above its symbol.
\n\"1st
\nx + 2(+1) + 2(-1) = 0 ,x = 0<\/p>\n

\"KSEEB<\/p>\n

Question 7.
\nBalance the equation by ion electron method.
\nFe(OH)2<\/sub>(s) + H2<\/sub>O2<\/sub> \u2192 Fe(OH)3 <\/sub>(s) + H2<\/sub>O is basic solution.
\nAnswer:
\n\"1st<\/p>\n

Question 8.
\nBalance the equation by ion electron method.
\nAl(s) + NO3<\/sub> \u2192 Al(OH)4<\/sub>–<\/sup> + NH3<\/sub> in basic solution.
\nAnswer:
\n\"1st<\/p>\n

Question 9.
\nBalance the equation by ion electron method.
\nPbO2<\/sub> + Cl–<\/sup> \u2192 ClO–<\/sup>+ [Pb(OH)3<\/sub>]–<\/sup> in basic solution.
\nAnswer:
\nHere 2e–<\/sup> + PbO2<\/sub> \u2192 [Pb(OH)3<\/sub>]–<\/sup>
\n2H2<\/sub>O + 2e–<\/sup>PbO22<\/sub> \u2192 [Pb(OH)3<\/sub>] + OH–<\/sup> …(I)
\n2OH + Cl–<\/sup> \u2192 ClO–<\/sup> + 2e–<\/sup>+H2<\/sub>O …(ii) Add(i)&(ii)
\nPbO2<\/sub> + OH–<\/sup> + Cl–<\/sup> + H2<\/sub>O \u2192 [Pb(OH)3<\/sub> ]–<\/sup> + ClO–<\/sup><\/p>\n

Question 10.
\nBalance the equation by ion electron method.
\nCrO4<\/sub>–<\/sup> + H2<\/sub>O2<\/sub> \u2192CrO4<\/sub>2-<\/sup> in basic solution.
\nAnswer:
\n[2OH–<\/sup> + CrO–<\/sup>3<\/sub> \u2192 CrO4<\/sub>2-<\/sup> + e–<\/sup> + H2<\/sub>O] x 2 …(i)
\n2e–<\/sup>+ H2<\/sub>O2<\/sub> \u2192 2OH–<\/sup> …(ii) (Add(i) & (ii))
\n2OH–<\/sup>+2CrO3<\/sub>–<\/sup> +H2<\/sub>O2<\/sub> \u2192 2CrO4<\/sub>2-<\/sup>+ 2H2<\/sub>O<\/p>\n

Question 11.
\nStarting with the correctly balanced half reactions write the overall net ionic reactions. Chloride ion is form oxidized to Cl2<\/sub> by MnO4<\/sub>–<\/sup> in acid
\nAnswer:
\n\"1st<\/p>\n

Question 12.
\nStarting with the correctly balanced half reactions write the overall net ionic reactions for HNO2<\/sub> reduces MnO4<\/sub>–<\/sup> in acid solution.
\nAnswer:
\n\"1st<\/p>\n

Question 13.
\nStarting with the correctly balanced half reactions write the overall net ionic reactions for HNO2<\/sub> oxidizes I–<\/sup> to I2<\/sub> in acid solution.
\nAnswer:
\n\"1st<\/p>\n

\"KSEEB<\/p>\n

Question 14.
\nStarting with the correctly balanced half reactions write the overall net ionic reactions. ClO3<\/sub>–<\/sup> oxidizes Mn2+<\/sup> to MnO2<\/sub>(s) in acid solution.
\nAnswer:
\n\"1st<\/p>\n

Question 15.
\nConsider the following cell notation :
\nAl(s)|Al3+<\/sup>(aq)|Ni2+<\/sup>(aq)|Ni(s), here
\nWhich substances act as anode ? Which of them acts as cathode ? Write the net ionic equation for the cell reaction.
\nAnswer:
\nAt acts as anode, Ni acts as cathode.
\n\"1st<\/p>\n

Question 16.
\nDescribe the standard half cell that is used in electrochemistry to measure standard potentials. Write an equation for its half reaction and show its standard potential.
\nAnswer:
\nStandard hydrogen elctrode is used to measure standard electrode potential.
\nIt consists of H2<\/sub>(g) filled at 1 atm, 298 K in a sealed tube having Pt rod for metallic contact, coated with platinum black which acts as catalyst. It is dipped in 1 M solution of HCl. The reaction 2H+<\/sup> (aq) + 2e–<\/sup> \u2192 H2<\/sub>(g) \\(\\mathrm{E}_{\\mathrm{H}^{+} \/ \\mathrm{H}_{2}}^{\\circ}=0\\)<\/p>\n

Question 17.
\nCalculate the E\u00b0cell<\/sub> using these electrodes whose half reactions are:
\nFe(OH)2<\/sub>(s) + 2e–<\/sup> \u2192 Fe(s) + 2OH–<\/sup>(aq)\u00a0 E\u00b0 = -0.88V
\nNiO2<\/sub>(s) + 2H2<\/sub>O(l) + 2e–<\/sup> \u2192 Ni(OH)2<\/sub>(s) + 2OH–<\/sup>(aq)\u00a0 E\u00b0 = +0.49V
\nAnswer:
\n\"1st<\/p>\n

Question 18.
\nCalculate E\u00b0cell<\/sub> for the reaction Cl2<\/sub>(g) + 2l–<\/sup> \u2192 I2<\/sub>(s) + 2Cl–<\/sup>(aq) with the help of these half cell reactions:
\nCl2<\/sub> (g) + 2e–<\/sup> \u2192 2Cl–<\/sup>(aq) ; E\u00b0 = +1.36V
\nI2<\/sub>(g) + 2e–<\/sup> \u2192 2I–<\/sup>(aq) ;E\u00b0 = +0.54V
\nAnswer:
\n\"1st<\/p>\n

Question 19.
\n(a) Give one use of heavy water in nuclear reactor.
\n(b) Write down balanced chemical equations of the reaction of cone. Nitric acid with (i) Copper and (ii) Iodine.
\nAnswer:
\n(a) Heavy water is used as moderate and coolant in nuclear reactor.
\n(b) (i) Cu(s) + 4HNO3<\/sub>(conc.) \u2192 Cu(NO3<\/sub>)2<\/sub> + 2NO2<\/sub> + 2H2<\/sub>O
\n\"1st<\/p>\n

Question 20.
\nBalance the following ionic reaction with the help of oxidation number method : MnO4<\/sub>–<\/sup> + I–<\/sup> \u2192 MnO2<\/sub> + IO3<\/sub>–<\/sup> (alkaline medium)
\nAnswer:
\nMnO4<\/sub>–<\/sup> + I–<\/sup> \u2192 MnO2<\/sub> + IO3<\/sub>–<\/sup> (alkaline medium)
\nH2<\/sub>O + 2MnO4<\/sub>–<\/sup> + I–<\/sup> \u2192 2MnO2<\/sub> + IO3<\/sub>–<\/sup> + 2OH–<\/sup> is balanced equation.<\/p>\n

\"KSEEB<\/p>\n

Question 21.
\nComplete the following equations :
\n(a) PbS(s) + H2<\/sub>O2<\/sub>(aq) \u2192 ………………………… (b) MnO4<\/sub>–<\/sup>(aq) + H2<\/sub>O2 <\/sub>\u2192………………….
\nAnswer:
\nPbS + 4H2<\/sub>O2<\/sub> \u2192 PbSO4<\/sub> +4H2<\/sub>O
\n2MnO4<\/sub>–<\/sup> + 3H2<\/sub>O2<\/sub> \u2192 2MnO2<\/sub> + 3O2<\/sub> + 2H2<\/sub>O + 2OH–<\/sup><\/p>\n

Question 22.
\nWrite the cell reactions for Zn|Zn2+<\/sup>(lM)||cd2+<\/sup>(lM)|Cd
\nE\u00b0Zn2+<\/sup> \/Zn = -0.76V, E\u00b0Cd2+<\/sup> \/Cd = 0.40V
\nAnswer:
\n\"1st<\/p>\n

Question 23.
\nZinc Silver oxide cell is used in heating aids or utensils and electric watches:
\nZn \u2192 Zn2+<\/sup> + 2e–<\/sup>; E\u00b0 = +0.76V
\nAg2<\/sub>O + H2<\/sub>O + 2e–<\/sup> \u2192 2Ag + 2OH–<\/sup>; E\u00b0 = +0.80V
\nWhich is oxidized and reduced?
\nAnswer:
\nZn is oxidised, Ag2<\/sub>O is reduced.<\/p>\n

Question 24.
\nThe standard emf value of some elements are listed below at 298 K. Which of the following two electrodes should be combined to form a cell having highest emf ? Identify the cathode and anode and write the cell reaction. Also mention the direction of flow of electrons in the external and internal circuit.
\n\"1st
\nAnswer:
\nFor maximum emf; anode should have minimum reduction potential and cathode should have maximum reduction potential. Hence A2+<\/sup> \/ A couple should be anode and C2+<\/sup>\/C couple acts as cathode.
\nCell reaction A + C2+<\/sup> \u2192 A2+<\/sup> + C
\nDirection of flow of electrons is from A to C in external and from C to A internal circuit. Direction of flow of current is opposite to the direction of flow of electrons.<\/p>\n

Question 25.
\nSilver jewellery turns greyish black due to the formation of Ag2<\/sub>S layer on it. Can this tarnish be removed by dipping tarnished jewllery in an aluminium vessel containing inert electrolytic solution. The standard electrode potential for the half reactions are given below :
\nAg2<\/sub>S + 2e–<\/sup> \u2192 2Ag + S2-<\/sup>; E\u00b0 =-0.71V ;and Al3+<\/sup> +3e–<\/sup> \u2192Al; E\u00b0 = -1.66V
\nAnswer:
\nTarnish (black colour) will be removed if the following reaction occur.
\nAl + Ag2<\/sub>S \u2192 A13+<\/sup> +2Ag + S2<\/sup> ; here E\u00b0 for this reaction = -0.71-(-1.66) = 0.95 V
\nEMF is positive. \u2206G will be negative this reaction will occur and tarnish will be removed.<\/p>\n

\"KSEEB<\/p>\n

Question 26.
\nJustify that the reaction : 2Na(s) + H2<\/sub>(g) \u2192 2NaH is a redox reaction.
\nAnswer:
\nIn this reaction, Na is oxidised to Na+<\/sup> and hydrogen is reduced from H2 and H- ion. Therefore it is a redox reaction.
\n2Na0<\/sup> + H20<\/sup> \u2192 2Na+<\/sup> H-1<\/sup> (s)<\/p>\n

Question 27.
\nOut of silver and aluminium vessel which one will be more suitable to store 1 M HCl solution and why ?
\nAnswer:
\nSince reduction potential of silver is more than that of hydrogen (E\u00b0H+<\/sup> \/H2<\/sub>Pt = 0), silver vessel will be suitable to store 1 M HCl. On the other hand E\u00b0 of Al3+<\/sup> \/Al is less than that of hydrogen (E\u00b0H+<\/sup> \/H2<\/sub>Pt) so that hydrogen will be liberated is stored in aluminium vessel.<\/p>\n

Question 28.
\nWhat is the oxidation number of Cr in (i) K2<\/sub>CrO4<\/sub> (ii) Ni and Ni(CO)4<\/sub>
\nAnswer:
\nLet the Ox. no. of Cr in K2<\/sub>CrO4<\/sub> be x.
\n2(+1) + x + 4(-2) = 0
\n+2 + X – 8 = 0
\nx = +6
\nOx. no. of Cr in K2<\/sub>CrO4<\/sub> = +6 A
\nOx. state of Ni in Ni(CO)4<\/sub> is x + 4(0) = 0;x – 0
\nOx. state of Fe in Fe(CO)5 <\/sub>is x + 5(0) = 0 or x = 0<\/p>\n

Question 29.
\nCan Fe3+<\/sup> oxidize Br– <\/sup>to Br2<\/sub> at 1 m concentrations ?
\nAnswer:
\nE\u00b0(Fe3+<\/sup>\/ Fe2+<\/sup>) is lower than that of E\u00b0(Br \/ B–<\/sup>). Therefore Fe2+<\/sup> reduce Br2<\/sub> but Br–<\/sup> cannot reduce Fe3+<\/sup> . Thus Fe3+<\/sup> cannot oxidize Br–<\/sup> to Br2<\/sub>.<\/p>\n

Question 30.
\nWhat is the oxidation state of S in (i) H2<\/sub>SO3<\/sub> (ii) P4<\/sub> (ii) (iii) PH3<\/sub> (iv) H3<\/sub>PO4<\/sub>
\nAnswer:
\n(i) H2<\/sub>SO3<\/sub> :2(+1) + x + 3(-2) = 0
\nx = +4
\nOx. state of is IV in H2<\/sub>SO3<\/sub>
\nAnswer:
\n(i) Ox. no. of P in P4<\/sub> = 0
\n(iii) Ox. no. of P in H3<\/sub>PO4<\/sub> .
\n3(+1) + x + 4(-2) = 0
\n+3 + x – 8 = 0
\nx = +5
\nOx. no. of P in H3<\/sub>PO4<\/sub> = +5<\/p>\n

Question 31.
\nAn iron rod is immersed in a solution containing NiSO4<\/sub> and ZnSO4<\/sub>. When the concentration of each salt is 1M, predict giving reasons which of the following reactions is likely to proceed ? (i) Iron reduces Zn2+<\/sup> ions, (ii) Iron reduces Ni2+<\/sup> ions.
\nAnswer:
\nGiven : E\u00b0(Zn2+<\/sup> \/Zn) = 0.76V, E\u00b0(Fe2+<\/sup> \/Fe) = 0.44V and E\u00b0(Ni2+<\/sup> \/Ni) = 0.25V
\nAnswer:
\n(i) The reduction potential of iron is more than that of zinc. Therefore, iron will be reduced. In other words Zn2+<\/sup> will not be reduced.
\n(ii) The reduction potential of Ni2+<\/sup> is more than that of Iron. Therefore, Ni2+<\/sup> will be reduced by iron.<\/p>\n

\"KSEEB<\/p>\n

Question 32.
\nWhich of the following redox reaction is oxidation and which is reduction?
\n(i) Zn \u2192 Zn2+<\/sup> + 2e–<\/sup>
\n(ii) Cl2 + 2e–<\/sup> \u2192 2Cl–<\/sup>
\n(iii) Fe \u2192 Fe2+<\/sup> + 2e–<\/sup>
\n(iv) Sn4+<\/sup> + 2e–<\/sup> \u2192 Sn2+<\/sup>
\nAnswer:
\n(i) Oxidation
\n(ii) Reduction
\n(iii) Oxidation
\n(iv) Reduction<\/p>\n

Question 33.
\nArrange the molecules, NH3<\/sub>, NO3<\/sub>, HN3<\/sub>, NO2<\/sub>–<\/sup>, N2<\/sub>O4<\/sub> and N2<\/sub>H4<\/sub> in the decreasing order of the oxidation states of nitrogen.
\nAnswer:
\n\"1st<\/p>\n

Question 34.
\nDetermine the change in oxidation number of S in H2<\/sub>S and SO2 <\/sub>in the following industrial process: 2H2<\/sub>S + SO2<\/sub> \u2192 3S + 2H2<\/sub>O
\nAnswer:
\nLet us write the oxidation numbers of atoms of all the reactants and products taking part in the reaction:
\n\"1st
\nO.N. of S in H2<\/sub>S increases by 2 O.N. of S in SO2<\/sub> decreases by 4<\/p>\n

Question 35.
\nCan the reaction Cr3<\/sub>O7<\/sub>2- <\/sup>+ H2<\/sub>O \u21cc CrO4<\/sub>2-<\/sup> +2H+<\/sup> be regarded as a redox reaction ?
\nAnswer:
\nOx. no. of Cr in Cr2<\/sub>O7<\/sub>2-<\/sup> \u21cc +6 Ox. no. of Cr in Cr2<\/sub>O7<\/sub>2-<\/sup>= +6
\nSince the ox. no. of Cr has neither increased nor decreased in the above reaction, therefore this reaction cannot be regarded as a redox reaction.<\/p>\n

Question 36.
\nBalance the oxidation reduction reaction FeS2<\/sub> + O2<\/sub> \u2192 Fe2<\/sub>O3<\/sub> + SO2<\/sub>
\nAnswer:
\n\"1st<\/p>\n

Question 37.
\nWhich of the two ClO2<\/sub>–<\/sup> or ClO4<\/sub>–<\/sup> show disproportionation reaction and why ?
\nAnswer:
\nThe oxidation state of Cl in ClO2<\/sub> is +3 and in ClO4<\/sub> is +7. Chlorine is present in highest oxidation state of +7 in ClO4<\/sub>–<\/sup> and it cannot increase its oxidation state. Hence ClO4<\/sub>–<\/sup> is not disproportionate.
\nThe disproportion reaction of ClO2<\/sub>–<\/sup> is
\n\"1st<\/p>\n

\"KSEEB<\/p>\n

Question 38.
\nWhich type of electrolytic are used in salt bridge ?
\nAnswer:
\nOnly those electrolytes for which cations and anions have nearly the same ionic mobilities (i.e. distance travelled by an ion per second under a potential gradient of one voltmeter) are used as electrolytes in the salt bridge. Thus KCl, KNO3<\/sub>, K4<\/sub>SO4<\/sub>, and NH4<\/sub>NO3<\/sub> are preferred over NaCl, NaNO3<\/sub> and Na2<\/sub>SO4<\/sub>.<\/p>\n

Question 39.
\nMention the cation and anion which have highest ionic mobility.
\nAnswer:
\nAmong cations, H+<\/sup> ion has the highest ionic mobility and among anions, OH–<\/sup> has the highest ionic mobility.<\/p>\n

Question 40.
\nThe standard electrode potential corresponding to the reaction Au3+<\/sup>(aq) + 3e–<\/sup> \u2192 Au(s) is 1.5 V. Predict if gold can be dissolved in 1M HCl solution and on passing hydrogen gas through gold salt solution, metallic gold will be precipitated or not.
\nAnswer:
\nConsider the half reaction, 2H+<\/sup>(aq) + 2e–<\/sup> \u2192 H2<\/sub>(g); E\u00b0 = 0.0V
\nAu3+<\/sup>(aq) + 3e–<\/sup> \u2192 Au(s); E\u00b0 = 1.50V
\nSince E\u00b0 (1.50V) for (Au3+<\/sup> \/ Au) is higher than that H+ \/ – \\(\\frac { 1 }{ 2 }\\)H2<\/sub> (0.0V), therefore, Au3+<\/sup>
\ncan be more easily reduced than H+<\/sup> ions Au3+<\/sup> can be reduced to Au metal by H2<\/sub> butr H+<\/sup> cannot oxidize metallic gold to Au3+<\/sup> ions. In other words, metallic gold does not dissolve in 1M HCl instead H2<\/sub> gas can reduce gold salt to metallic gold.<\/p>\n

Question 41.
\nWhat is the oxidation number of S in (i) Na2<\/sub>S4<\/sub>O6<\/sub>, (ii) S2<\/sub>Cl2<\/sub>?
\nAnswer:
\n(i)
\n\"1st
\nIn b & c, s has oxidation number is zero. In a & d s has oxidation number +5
\n(ii)
\n\"1st<\/p>\n

Question 42.
\nWhat is the name of the reaction ?
\n2CH3<\/sub>CH2<\/sub>CH2<\/sub>SH \u2192 cCH3<\/sub>CH2<\/sub>CH2<\/sub> – S – S – CH2<\/sub>CH2<\/sub>CH3<\/sub>
\nWhether condensation, oxidation, reduction or polymerization?
\nAnswer:
\nThis is an example of oxidation reaction since two H-atoms have been removed.<\/p>\n

\"KSEEB<\/p>\n

Question 43.
\nBoth Cr2<\/sub>O7<\/sub>2-<\/sup>(aq) and MnO4<\/sub>–<\/sup> (aq) can be used to titrate Fe2+<\/sup>(aq). If a given titration, 24.50 cm3 of 0.1 M Cr2<\/sub>O7<\/sub>2-<\/sup>\u00a0were used, then what volume of 0.1 M MnO4<\/sub>–<\/sup> solution would have been used for the same titration ?
\nAnswer:
\nSuppose V2<\/sub>cm3<\/sup> of M2<\/sub>Fe2+<\/sup> is titrated against 24.50 cm3 0.1M Cr2<\/sub>O7<\/sub>2-<\/sup>\u00a0\u00a0and V1<\/sub>cm3<\/sup> of 0.1M MnO4<\/sub>–<\/sup> solution, then
\n\"1st
\n\"1st<\/p>\n

1st PUC Chemistry Redox Reactions Three\/Four Marks Questions and Answers<\/h3>\n

Question 1.
\nBalance MnO2<\/sub> + HCl \u2192 MnCl2<\/sub> + Cl2<\/sub> + H2<\/sub>O
\nAnswer:
\nSolution x
\nStep – 1: Assign oxidation number:
\n\"1st<\/p>\n

Step – 2 : Write the oxidation number changes.
\nChange in oxidation number of Mn + 4 to +2 = 2
\nChange in oxidation number of Cl – 1 to 0 = +1<\/p>\n

Step – 3 : Cross multiply the numbers. Co-efficient 2 is multiplied to HCl and 1 is multiplied to MnO2<\/sub>.
\nMnO2<\/sub> + 2HCl \u2192 MnCl2<\/sub>+H2<\/sub>O + Cl2<\/sub><\/p>\n

Step – 4 : Check the oxidation number, and 2 molecules of HC1 on left hand side to balance the chlorine atoms of MnCl2<\/sub>. In order to balance oxygen and hydrogen atom 2 molecule of H2<\/sub>O has to be added on the right hand side.
\nMnO2<\/sub> + 4HCl \u2192 MnCl2<\/sub> + 2H2<\/sub>O + Cl2<\/sub><\/p>\n

Question 2.
\nBalance Fe2+<\/sup>+ Cr2<\/sub>O7<\/sub>2-<\/sup> \u2192 Fe3+<\/sup> + Cr3+<\/sup> is acid media.
\nAnswer:
\nSolution:
\nStep -1: (LEO = Loss of Electron Oxidation)
\n\"1st
\n(GER = Gain of Electron Reduction)<\/p>\n

Step-2: Fe2+<\/sup> \u2192 Fe3+<\/sup> … (1) (2 \u2192 3 = 1 unit ON) ,
\nCr2<\/sub>O7<\/sub>2-<\/sup> \u2192 2Cr3+<\/sup> … (2) (+6 \u2192+3 = 3 units 2 sets => 6 units ON)<\/p>\n

Step – 3 : Cross multiplication i.e., (1) x 6 and (2) x 1
\n(1) x 6 \u21d2 6Fe2+<\/sup> \u2192 6Fe3+<\/sup> … (3)
\n(2) x 1 \u21d2 Cr2<\/sub>O7<\/sub>-2<\/sup> \u2192 2Cr3+<\/sup> …(4)<\/p>\n

Step – 4 : Add (3) + (4) \u21d2 6Fe2+<\/sup> + Cr2<\/sub>O7<\/sub>2-<\/sup> \u2192 6Fe3+<\/sup> + 2Cr3+<\/sup><\/p>\n

Step – 5 : Balance Oxygen and then Hydrogen
\n6Fe2+<\/sup> + Cr2<\/sub>O7<\/sub>2-<\/sup> + 14H+<\/sup> \u2192 6Fe3+<\/sup> + 2Cr3+<\/sup>+ 7H2<\/sub>O<\/p>\n

\"KSEEB<\/p>\n

Question 3.
\nEvaluate the equivalent weight of I2 from the following equation.
\nI2<\/sub> + 2S2<\/sub>O3<\/sub> \u2192 S4<\/sub>O6<\/sub>2-<\/sup>+2I–<\/sup>
\nAnswer:
\n\"1st<\/p>\n

Question 4.
\nFind equivalent weight of KMnO4<\/sub> from
\n\"1st
\nAnswer:
\nBalance the given equation
\n2KMnO4<\/sub> + 3H2<\/sub>SO4<\/sub> \u2192 K2<\/sub>SO4 <\/sub>+ 2MnSO4 <\/sub>+ 3H2<\/sub>O + 5[O]<\/p>\n

Method – 1: Stoichiometry
\nWe know equivalent weight of Oxygen = 8
\nAtomic weight of Oxygen = 16
\n\u2234 2 equivalent of oxygen = 1 Atomic weight of oxygen
\nHere 5[O] = 10 equivalent of oxygen
\n\"1st<\/p>\n

Method – 2 : (Oxidation Number)
\n2 KMnO4<\/sub> + 3H2<\/sub>SO4<\/sub> \u2192 K2<\/sub>SO4<\/sub> + 2MnSO4<\/sub> + 3H2<\/sub>O + 5[O]
\n+7 \u2190 5 Unit GER Reduction \u2192 +3
\nKMnO4<\/sub> acts as oxidising agent
\n\"1st<\/p>\n

Question 5.
\nEvaluate the equivalent weight of oxidizing and reducing agents in the
\nfollowing reaction (4M) K2<\/sub>Cr2<\/sub>07<\/sub> +FeS04<\/sub> >Cr3+<\/sup> +Fe3+<\/sup>
\nAnswer:
\nAssigning oxidation number,
\n\"1st<\/p>\n

\"KSEEB<\/p>\n

Question 6.
\nThe half cell reaction with their oxidation potentials are
\nPb(s) \u2192 Pb2+<\/sup>(aq) + 2e–<\/sup>; Eo<\/sup>cell<\/sub> = +0.13V
\nAg(s) \u2192 Ag(aq) + e–<\/sup>; Eo<\/sup>cell<\/sub> = -0.80V
\nWrite the cell reaction and calculate its EMF.
\nAnswer:
\nRewrite the two equations in the reduction form thus.
\nPb2+<\/sup> (aq) + 2eE–<\/sup> \u2192 Pb(s) ; E\u00b0 = -0.13V .. (i)
\nAg+<\/sup>(aq) + e–<\/sup> \u2192 Ag(s) ; E\u00b0 = +0.80V …(ii)
\nTo obtain the equation for the cell reaction, multiply Eq. (ii) with 2 and subtract Eq. (i) from it, we have,
\nPb(s) + 2Ag+<\/sup>(aq) \u2192 Pb2+<\/sup>(aq) + 2Ag(s) ; Eo<\/sup>cell<\/sub> =+0.80-(-0.13) =+0.93V<\/p>\n

Question 7.
\nJustify that the reaction ; 2Cu2<\/sub>O(s) + Cu2<\/sub>S(s) \u2192 6Cu(s) + SO2<\/sub>(g) is a redox
\nreaction. Identify the species oxidized, reduced, which acts as oxidation and which acts as reductant.
\nAnswer:
\nWriting the oxidation number of each atom above its symbol we have,
\n\"1st
\nHere, in the reaction, the oxidation number of copper decreases from +1 in Cu2<\/sub>O or Cu2<\/sub>S to 0 in copper metal, therefore, copper is reduced and is .oxidizing agent or oxidants. Further, the oxidation number of S increases from -2 in Cu2<\/sub>S to +4 in SO2<\/sub>. Therefore, sulphur is oxidized. Thus, it is reducing agent of reductant.<\/p>\n

Question 8.
\nUsing stock notation, represent the following compounds : HAuCl2<\/sub> , TI2<\/sub>O, FeO, Fe2<\/sub>O3<\/sub>, Cul, CuO, MnO and MnO2<\/sub>.
\nAnswer:
\nBy applying the various rules for determining the oxidation number of different atoms in a compound, the oxidation number of each metallic element in the given compounds is as follows:
\nIn HAuCl2<\/sub> , Au has +3; in TI2<\/sub>0, TI has +1; in FeO, Fe has +2; in Fe2<\/sub>O2<\/sub> , Fe has +3 ; . in Cul, Cu has +1 : in CuO, Cu has +2 ; in MnO, Mn has +2 while in Mn2<\/sub>O , Mn has +4 oxidation state. Therefore, these compounds may be represent as :
\nHAu(II)Cl4<\/sub>, Tl2<\/sub>(1)O, Fe(II)O, Fe2<\/sub>(III)O3<\/sub>, Cu(I)I, Cu(II)O, Mn(II)O, and Mn(IV)O2<\/sub>.<\/p>\n

Question 9.
\nBalance the equation, Mg(aq) + HNO3<\/sub>(aq) \u2192 Mg(NO3<\/sub>)2<\/sub>(aq) + N2<\/sub>O(g) + H2<\/sub>O(1)
\nAnswer:
\nFind out the elements which undergo a change in oxidation number (O.N.)
\nO.N. increases by 2 per Mg atom
\n\"1st
\nHere, O.N. of Mg increases from 0 in Mg metal to +2 in Mg(NO3<\/sub>)2<\/sub> and that of N decreases from +5 in HNO3<\/sub> to +1 in N2<\/sub>O.<\/p>\n

Question 10.
\nSplit the following redox reactions into the oxidation and reduction half reactions.
\n(a) 2K(s) + Cl2<\/sub>(g) \u21922KCl(s) (b) 2Al(s) + 3Cu2+<\/sup>(aq) \u2192 2Al3+<\/sup>+(aq) + 3Cu(s)
\nAnswer:
\n\"1st<\/p>\n

\"KSEEB<\/p>\n

Question 11.
\nPredict the oxidation state of (a) C in C3<\/sub>O2<\/sub>, (b) Cr in Br3<\/sub>O3<\/sub>, (c) S in S3<\/sub>O3<\/sub>2-<\/sup>
\nAnswer:
\n(a) The structure of C3<\/sub>O2<\/sub>\u00a0(carbon -2 +2 0 +2 -2 suboxide) isO = C = C = C = 0 each of the two terminal oxygen atoms have an oxidation state of -2 and the carbons to which they are atteached have an oxidation state of +2. (Since when ever a covalent bond is formed between similar atoms, each of the atoms is given an oxidation state of zero), therefore, the oxidation state of central carbon atom in C3<\/sub>O2<\/sub>\u00a0<\/sup>is zero.<\/p>\n

Thus, the two terminal carbon atoms are present in +2 oxidation state each, whereas the third one is present in zero oxidation state. Therefore, the average oxidation state of carbon in
\n\"1st<\/p>\n

(b) Tribromooctaoxide (Br3<\/sub>O8<\/sub>)
\n\"1st<\/p>\n

We can easily find out that the terminal bromine atoms are present in +6 oxidation state while the middle one has an oxidation state of +4. However, the average oxidation number of Br in Br308 turns out to be
\n\"1st<\/p>\n

(c) Tetrathionate ion
\n\"1st
\nSince each of the two terminal sulphur atoms is connected to two oxygen atoms by a double bond and one oxygen atom by a single bond, therefore, the oxidation state of each of these terminal sulphur atoms is +5.<\/p>\n

Since two central sulphur atoms are linked to each other by a single bond and each sulphur is further attached to similar species on either side, the electron pair forming the S – S bond remain in the centre and hence each of the two central sulphur atoms has an oxidation state of zero.
\nHowever, the average oxidation state of the sulphur atoms is
\n\"1st<\/p>\n

Question 12.
\nMention which is reduced and oxidized in the following reaction :
\n(a) 4Na(s) + O2<\/sub>(g) \u2192 2Na2<\/sub>0(s)
\n(b) 2Na(s) + Cl2<\/sub>(g) \u2192 2NaCl(s) ,
\n(c) 2Na(s) + S(s) \u2192 Na2<\/sub>S(s)
\nAnswer:
\n\"1st<\/p>\n

Question 13.
\nIdentify the oxidant and reductant in the following reactions :
\n(a) 10H+<\/sup> (aq) + 4Zn(s) + NO3<\/sub>–<\/sup> (aq) \u2192 4Zn2+<\/sup>(aq) + NH+<\/sup>4<\/sub> +3H2<\/sub>O(l)
\n(b) I2<\/sub>(g) + H2<\/sub>S(g) \u2192 2IH(g) + S(s)
\nAns:
\n(a) Writing the O.N. of all the atoms above their symbols, we have
\n\"1st
\nO.N. of Zn changes from zero in Zn to +2 in Zn2+<\/sup> and, therefore, it is oxidized and hence Zn acts as a reductant.
\nThe O.N. of N decreases from +5 in NO3<\/sub>–<\/sup> to -3 in NO4<\/sub>+<\/sup> and, therefore, it is reduced and hence NO3<\/sub>–<\/sup>\u00a0acts as the oxidant.<\/sub><\/p>\n

(b) Writing the ON of all the atoms above their symbols, we have
\n\"1st
\nThe ON of I2<\/sub> decreases from zero in I2 to -1 in HI, therefore, I2<\/sub> is reduced and hence it acts as on oxidant.
\nThe O.N. of S increases from -2 in H2<\/sub>S to zero in S, therefore, H2<\/sub>S is oxidized and hence it acts as the reductant.<\/p>\n

\"KSEEB<\/p>\n

Question 14.
\n0.2g of a sample of H2<\/sub>O2<\/sub> reduced 20 ml of 0.1M KMnO4<\/sub> solution in acidic medium. What is the percentage purity of H2<\/sub>O2<\/sub> ?
\nAnswer:
\nNo of moles of KMnO4<\/sub> present in 20ml of 0.1M
\nKMnCh solution = \\(\\frac{20}{1000}\\) x 0.1 = 2 x 10-3<\/sup><\/p>\n

The balanced equation for the redox reactions is :
\n2KMnO4<\/sub> + 5H2<\/sub>O2<\/sub>\u00a0+ 3H2<\/sub>SO4<\/sub> \u2192 K2<\/sub>SO4<\/sub> + 2MnSO4<\/sub> + 8H2<\/sub>O + 5O2<\/sub> …(1)
\nFrom the equation, 2 moles of KMnO4<\/sub> = 5 moles of H2<\/sub>O2<\/sub>.
\n2 x 10-3<\/sup> moles of KMnO4<\/sub> will react with H2<\/sub>O2<\/sub> = \\(\\frac{5}{2}\\) x 2 x 10-3<\/sup> = 5 x 10-3<\/sup> moles
\nMolecular wt. of H2<\/sub>O2<\/sub> = 34
\nAmount of H2O2 actually present = 34 x 5 x 10-3<\/sup> = 0.17g .
\nPercentage purity of in 20 ml is = \u00a0\\(\\frac{0.17}{0.20}\\) x l00 = 85<\/p>\n

Question 15.
\n16.6g of pure potassium iodide was dissolved in water and the solution was . made upto one litre. V cm3<\/sup> of this solution was acidified with 20 cm3<\/sup> of 2M HCl. The resulting solution required 10 cm3 of decinormal KlO3<\/sub> for complete oxidation of I- ions to ICl. Find out the value of V.
\nAnswer:
\nThe chemical equation for the redox reactions is :
\n\"1st<\/p>\n

Question 16.
\nWhich of the following species, do not show disproportionation reaction and why ? CIO–<\/sup>, CIO2<\/sub>–<\/sup>, CIO3<\/sub>–<\/sup> and CIO4<\/sub>–<\/sup> Also write reactions for each of the species that disproportionates.
\nAnswer:
\nThe oxidation state of Cl in all the given species are :
\n\"1st
\nOut of above species, CIO4<\/sub>–<\/sup> does not undergo disproportionation since in this oxoanion chlorine is already present in the highest oxidation state of +7 and hence cannot be further oxidized.<\/p>\n

All the remaining oxoanions have oxidation state lower than the highest (i.e., +7) and higher than the lowest (i.e. -1) therefore, all these species undergo disproportionation reactions as shown below.
\n\"1st<\/p>\n

Question 17.
\nHow does CU2<\/sub>O act as both oxidant and reductant ? Explain with proper reactions showing the change of oxidation numbers in each example.
\nAnswer:
\nCu2<\/sub>O undergoes disproportionation to form Cu2+<\/sup> and Cu.
\n2Cu2+<\/sup>(aq) >Cu2+<\/sup>(aq) + Cu(s)
\nThus, Cu2<\/sub>O acts both as an oxidant as well as a reductant.
\n(i)
\n\"1st
\nand Cu2<\/sub>O acts as a reductant and reduces O2<\/sub> to O2-<\/sup><\/p>\n

(ii) When heated with Cu2<\/sub>S it oxidizes S2-<\/sup> to SO2<\/sub> and hence Cu2<\/sub>O acts as an oxidants
\n\"1st<\/p>\n

\"KSEEB<\/p>\n

Question 18.
\nDetermine the volume of \\(\\frac { M }{ 8 }\\) KMnO4<\/sub> solution required to react completely with 25.0cm3 \\(\\frac { M }{ 4 }M\/4\\) FeSO4<\/sub> solution in acidic medium.
\nAnswer:
\nThe balanced ionic equation for the reaction is
\nMnO4<\/sub>–<\/sup>+ 5Fe2+<\/sup> + 8H+<\/sup> \u2192 Mn2+<\/sup> + 5Fe3+<\/sup> + 4H2<\/sub>O
\nFrom the balanced equation, it is evident that 1 mole of KMnO4<\/sub> = 5 moles of FeSO4<\/sub>.
\nApplying molarity equation to the balanced redox equation,
\n\"1st
\nThus, the volume of \\(\\frac { M }{ 4 }M\/4\\) KMnO4<\/sub> solution required =10.0 ml.<\/p>\n

Question 19.
\nCalculate the concentration of hypo (Na2<\/sub>S2<\/sub>O3<\/sub>.5H2<\/sub>O) solution in gdm-3<\/sup> if 10.0ml of this solution decolourised 15mL of M\/40 iodine solution.
\nAnswer:
\nThe balanced equation for the redox reaction is 2S2<\/sub>O3<\/sub>2- <\/sup>+ I2<\/sub> \u2192 2I–<\/sup> + S4<\/sub>O6<\/sub> 2-<\/sup>
\nFrom the balanced equation, it is evident that 2 moles of Na2<\/sub>S2<\/sub>C>O3<\/sub> \u2261 1 mole of I2<\/sub>.
\n\"1st
\n\"1st<\/p>\n

Question 20.
\nCalculate the volume of 0.05M KMnO4<\/sub> solution required to oxidize completely 2.70g of oxalic acid (H2<\/sub>C2<\/sub>O4<\/sub>) in acidic medium.
\nAnswer:
\nBalanced equation for the redox reaction is :
\n2KMnO4<\/sub> + 5(COOH)2<\/sub> + 3H2<\/sub>SO4<\/sub> \u2192 K2<\/sub>SO4<\/sub> + 2MnSO4<\/sub> + 10CO2<\/sub> + 8H2<\/sub>O
\nNo of moles oxalic acid = 2.70\/90 = 0.03 mole
\nFrom the balanced equation, 5 moles of (COOH)2<\/sub> = 2 mole KMn04<\/sub> .
\nThen 0.03 mole (COOH)2<\/sub> = pppp\u2014 x 0.03 = 0.012 moles of KMn<\/sub>O4<\/sub>.
\nNow, 0.05 mole of KMnO4<\/sub> is present in solution given = 1000 cm3<\/sup>.
\n0.012 mole of KMnO4<\/sub> is present in solution = \\(\\frac { 2 }{ 5 }\\) = 240 cm3<\/sup><\/p>\n

Question 21.
\nWrite the rules for assigning oxidation number.
\nAnswer:<\/p>\n

    \n
  • All atoms in the elemental or molecular state have ‘zero’ oxidation state.<\/li>\n
  • Elements of group I and II in the periodic table always have ‘+1’ and ‘+2’ oxidation sstates respectively.<\/li>\n
  • Hydrogen ‘+1’ oxidation state in all its compounds except metal hydrides where it is ‘-1’<\/li>\n
  • Oxygen has been assigned an oxidation number of ‘-2’ in all its compounds except peroxides and oxygen fluoride. In peroxide it is ‘-1’ and in oxygen fluoride it is ‘+2’.<\/li>\n
  • Halogens generally have ‘-1’ oxidation state. Except fluorine, other halogens may have positive oxidation states in their oxides and inter halogen compounds (Fluorine always has ‘-1’ state).<\/li>\n
  • The algebraic sum of oxidation numbers of all the elements in a compound is zero and that in an ion is equal to the net charge on the ion.<\/li>\n<\/ul>\n

    \"KSEEB<\/p>\n

    Question 22.
    \nBalance MnO4<\/sub>–<\/sup> + Fe2+<\/sup> \u2192 Fe3+<\/sup> + Mn2+<\/sup> in acidic medium by ion electron method.
    \nAnswer:
    \nMnO4<\/sub>–<\/sup> + Fe2+<\/sup> \u2192 Fe3+<\/sup> +Mn2+<\/sup>
    \n[Fe2+<\/sup> \u2192 Fe3+<\/sup>] x 5 ……………(i)
    \n8H+<\/sup> + MnO4<\/sub>–<\/sup> >Mn2+<\/sup> + 4H2<\/sub>O …(ii)
    \n… (ii) Adding (i) and (ii), we get
    \n5Fe2+<\/sup> + 8H+<\/sup> + MnO4<\/sub>–<\/sup> \u2192 Mn2+<\/sup> + 4H2<\/sub>O + 5Fe3+<\/sup><\/p>\n

    Question 23.
    \nFind the oxidation state of sulphur in the following compounds :
    \nAnswer:
    \n\"1st<\/p>\n

    Question 24.
    \nFind the oxidation number of element underlined in the following species :
    \n\"1st
    \nAnswer:
    \n\"1st<\/p>\n

    Question 25.
    \nCalculate the oxidation number of (i) N in NO3<\/sub>–<\/sup> ; (ii) P in H3<\/sub>P2<\/sub>O7<\/sub>–<\/sup> CO3<\/sub>2-<\/sup>.
    \nAnswer:
    \n(i) N in NO3<\/sub>–<\/sup> , let the oxidation number of N in NO3<\/sub>–<\/sup> be x. Writing the oxidation number of each atom above its symbol. \\(\\begin{array}{cc}{\\mathbf{x}} & {-2} \\\\ {\\mathbf{N}} & {\\mathbf{O}_{3}}\\end{array}\\)
    \nx + 3(-2) = -1
    \nx – 6 = -1, x = +5
    \n(ii) P in H3<\/sub>P2<\/sub>O7<\/sub>–<\/sup> , let the oxidation number of P in H3<\/sub>P2<\/sub>O7<\/sub>–<\/sup> be x. Writing the oxidation
    \nnumber of each atom above its symbol.
    \n\\(\\begin{array}{ll}{\\text { th } x} & {-2} \\\\ {\\text { H }_{3} P_{2}} & {\\text { o_{ } }}\\end{array}\\)
    \nH3<\/sub>P2<\/sub>O7<\/sub>–<\/sup>= 3(+1) + 2 (x) = -1 +7(-2) or 2x – 11 = -1; x = + 5
    \nThus, the oxidation number of P in H3<\/sub>P2<\/sub>O7<\/sub>–<\/sup> is +5.<\/p>\n

    \"KSEEB<\/p>\n

    Question 26.
    \nWhat is the oxidation number of metals (i) [Fe(CN)6<\/sub> ]4-<\/sup> and (ii) MnO4<\/sub>–<\/sup> ?
    \nAnswer:
    \n(i) Fe in [Fe(CN)6<\/sub> ]4-<\/sup>. Let the oxidation number of Fe in [Fe(CN)6<\/sub> ]4-<\/sup> bexand CN be -1 i.e..,
    \n\"1st
    \n\u2234\u00a0 Sum of oxidation numbers of all the atoms in
    \n[Fe(CN)6<\/sub> ]4-<\/sup> =x + 6(-l) = -4 = x-6 = -4; x = +2
    \nThus, the oxidation number of Fe in [Fe(CN)6<\/sub> ]4-<\/sup> f is +2.<\/p>\n

    (ii) Mn in MnO4<\/sub>–<\/sup>. Let the oxidation number of Mn in MnO4<\/sub>–<\/sup> be x. Writing oxidation number of each atom above its symbol, we get.
    \n\"1st
    \nThus, the oxidation number of Mn in MnO4<\/sub>–<\/sup> is +7
    \nPredict the oxidation number of the element underlined NaCl, MgSO4<\/sub>, AlF3<\/sub>, SiCl4<\/sub>, PF3<\/sub>, P4<\/sub>O10<\/sub>, SF6<\/sub>, SO3<\/sub>, HClO4<\/sub>, Cl2<\/sub>O7<\/sub>.
    \nAnswer:
    \n\"1st<\/p>\n

    Question 27.
    \nWhat is the maximum and minimum oxidation numbers of N, S and Cl?
    \nAnswer:
    \n(i) The highest oxidation number (O.N) of N is +5 since it has five electrons in the valence shell (3s2<\/sup> 2p3<\/sup> ) and its minimum O.N. is -3 since it can be accept three more electrons to acquire the nearest inert gas (Ne) configuration.<\/p>\n

    (ii) Similarly, the highest O.N. of S is +6 since it has six electrons in the valence shell (3s2<\/sup> 2p4<\/sup>) and its minimum O.N. is -2 since it needs two more electrons to acquire the nearest inert gas (Ar) configuration.<\/p>\n

    (iii) Likewise the maximum O.N. of Cl is +7 since it has seven electrons in the valance shell (3s2<\/sup> 2p5<\/sup> ) and its minimum O.N. is -1 since it needs only one more electron to acquire the nearest (Ar) Noble gas configuration.<\/p>\n

    Question 28.
    \nNitric acid acts only as an oxidising agent while nitrous acid acts both as an oxidising as well as a reducing agent. Why ?
    \nAns:
    \n(i) Nitric Acid HNO3<\/sub> : Here Oxidation number of N is HNO3<\/sub> = +5 ; But
    \nMaximum oxidation number of N = +5 ; Minimum oxidation number of N = -3 Since the oxidation number of N in HNO3<\/sub> is maximum (+5), therefore it can only decrease by accepting electrons. Hence HNO3<\/sub> acts only as an oxidizing agent.<\/p>\n

    (ii) Nitrous Acid HNO2<\/sub> : Here Oxidation number of N in HNO2<\/sub> = +3
    \nMaximum oxidation number of N = +5 ; Minimum oxidation number of N = -3 Thus, the oxidation number of N can either increase by losing electrons or can decrease by accepting electrons. Therefore, HNO2<\/sub> acts both as an oxidizing as well as a reducing agent.<\/p>\n

    Question 29.
    \nWrite correctly the balanced equations for the following redox reactions using half reaction:
    \n(i) H2<\/sub>S + Fe3+<\/sup> \u2192 Fe2+<\/sup> + S + H+<\/sup>
    \n(ii) I–<\/sup>+IO3<\/sub>–<\/sup> + H+<\/sup> \u2192 I2<\/sub>+H2<\/sub>O
    \n(iii) I–<\/sup> + O2<\/sub> + H2<\/sub>O \u2192 I2<\/sub> + OH–<\/sup>
    \n(iv) Cu(s) + Au+<\/sup> \u2192 Au(s) + Cu2+<\/sup>
    \nAnswer:
    \n\"1st<\/p>\n

    \"KSEEB<\/p>\n

    Question 30.
    \nConsider the following reactions that produce electricity in the galvanic cell:
    \n(i) 2Fe3+<\/sup>(aq) + 2Cl–<\/sup>(aq) \u2192 2Fe2+<\/sup> (aq) + Cl2<\/sub>(g)
    \n(ii) 2Cr(s) + 3Cu2+<\/sup> \u2192 3Cu(s) + 2Cr3+<\/sup>
    \nWrite the anode and cathode reactions for the galvanic cell. Write cell in the usual notation.
    \nCathode
    \nAnswer:
    \n\"1st
    \n\"1st<\/p>\n

    Question 31.
    \nBalance the following reaction :
    \n(i) Cu + NO3<\/sub>–<\/sup> \u2192 NO2<\/sub> + Cu2+<\/sup>
    \nWrite correctly balanced half cell reactions and overall equations for the following equations by ion electron method:
    \nAnswer:
    \n\"1st
    \nQuestion 32.
    \nNO3<\/sub>–<\/sup> + Bi(s) \u2192 Bi3<\/sup> + NO2<\/sub> in acid solution by ion electron method.
    \nAnswer:
    \n\"1st<\/p>\n

    Question 33.
    \nCr2<\/sub>O7<\/sub>2-<\/sup> + C2<\/sub>H4<\/sub>O– <\/sup>\u2192 C2<\/sub>H4<\/sub>O2<\/sub> + Cr3+<\/sup> in acid solution by ion electron method.
    \nAnswer:
    \n\"1st<\/p>\n

    Question 34.
    \nMnO4<\/sub> + H2<\/sub>C2<\/sub>O4<\/sub> \u2192 Mn2+<\/sup> + CO2<\/sub> in acid solution by ion electron method.
    \nAnswer:
    \n\"1st<\/p>\n

    \"KSEEB<\/p>\n

    Question 35.
    \nA cell is prepared by dipping a zinc rod in 1M ZnSO4<\/sub> solution and a lead rod in 1 M Pb(NO3<\/sub>)2<\/sub> solution. The standard electrode a potentials for Pb+2<\/sup>\/Pb and Zn+2<\/sup> \/Zn electrode are -0.126 V and -0.763 V respectively. _
    \n(a) How will you represent the cell ?
    \n(b) Write the half cell reactions and the overall cell reaction.
    \nAnswer:
    \n\"1st<\/p>\n

    Question 36.
    \nCopper dissolves in dilute HNO3 <\/sub>but not in dilute HCl. Why ? Explain.
    \n\"1st
    \nAnswer:
    \nElectrode potential of Cu2+<\/sup>\/Cu is higher than H+<\/sup>\/H2<\/sub> and so H+<\/sup> ions cannot oxidize Cu to Cu2+<\/sup> ions and therefore copper does not dissolve in dilute HCl. On the other hand, the electrode potential of NO3<\/sub>–<\/sup> ion i.e., NO3<\/sub>–<\/sup>\/NO electrode is higher than that of copper electrode and hence it can oxidize Cu to Cu2+<\/sup> ion; so copper dissolve in dilute HNO3<\/sub>.
    \nThus copper dissolved in dilute HNO3<\/sub> due to oxidation of Cu by NO3<\/sub>–<\/sup> ions and not by H+<\/sup>ions.<\/p>\n

    1st PUC Chemistry Redox Reactions Five Marks Questions and Answers<\/h3>\n

    Question 1.
    \nWrite the following redox reactions using half equations.
    \n(i) Zn(s) + PbCl2<\/sub> (aq) \u2192 Pb(s) + ZnCl2<\/sub>(aq)
    \n(ii) 2Fe3+<\/sup> (aq) + 2e–<\/sup> (aq) \u2192\u00a0 I2<\/sub>(s) + 2Fe2+<\/sup> (aq)
    \n(iii) 2Na(s) + Cl2<\/sub> (g) \u2192\u00a0 2NaCl(s)
    \n(iv) Mg(s) + Cl2<\/sub>(s) \u2192 MgCl2<\/sub>(s)
    \n(v) Zn(s) + 2H+<\/sup> (aq) \u2192\u00a0 Zn2+<\/sup> (aq) + H2<\/sub> (g)
    \nIn each of the reactions given above, mention
    \n(i) Which reactant is oxidized ? To what ?
    \n(ii) Which reactant is the oxidizer ?
    \n(iii) Which reactant is reduced ? To what ?
    \n(iv) Which reactant is the reducer ?
    \nAnswer:
    \n(i) Zn \u2192 Zn2+<\/sup> + 2e–<\/sup> (oxidation), Pb2+<\/sup> \u2192 2e–<\/sup> +Pb (reduction) Zn is oxidized to Zn2+<\/sup>, Pb2+<\/sup> is reduced to Pb; Pb2+<\/sup> is the oxidizer and Zn is the reducer.<\/p>\n

    (ii) 2Fe3+<\/sup> + 2e–<\/sup> \u2192 2Fe2+<\/sup> (reduction), 2l–<\/sup> \u2192 I2<\/sub> +2e–<\/sup> (oxidation)
    \nFe3+<\/sup> is reduced to Fe2+<\/sup>, I–<\/sup> is oxidized to I2<\/sub>; I–<\/sup> is the reducer and Fe3+<\/sup> is the oxidizer.<\/p>\n

    (iii) 3Na \u2192 2Na+<\/sup> +2e–<\/sup> (oxidation), Cl2 <\/sub>+ 2e–<\/sup> > 2Cl–<\/sup> (reduction)
    \nNa is oxidized to Na+<\/sup> and Cl2<\/sub> is reduced to Cl–<\/sup>; Na is the reducer and Cl2<\/sub> is the oxidizer.<\/p>\n

    (iv) Mg \u2192 Mg2+<\/sup> + 2e–<\/sup> (oxidation), Cl2<\/sub> + 2e–<\/sup> \u2192 2Cl (reduction)
    \nMg is oxidized to Mg while Cl2 is reduced to Cl ; Mg is the reducer and Cl2 is the oxidizer.<\/p>\n

    (v) Zn \u2192 Zn2+<\/sup> +2e–<\/sup> (oxidation), 2H+<\/sup> + 2e–<\/sup> >H2<\/sub> (reduction)
    \nZn is oxidized to Zn2+<\/sup> while H+<\/sup> is reduced to H2<\/sub>;Zn is the reducer and H+<\/sup> is the oxidizer.<\/p>\n

    \"KSEEB<\/p>\n

    Question 2.
    \nWrite the reaction at anode and cathode and the net cell reaction in the following cells. Which electrode would be +ve terminal in each cell ?
    \n\"1st
    \nAnswer:
    \n\"1st<\/p>\n

    Question 3.
    \nMention which element undergoes disproportionate reaction in alkaline media.
    \n\"1st
    \nAnswer:
    \n\"1st
    \n\"1st<\/p>\n

    Question 4.
    \nCalculate emf of the following cell:
    \n\"1st
    \nAnswer:
    \n\"1st<\/p>\n

    Question 5.
    \nIn the reaction given below^ identify the species undergoing oxidation and reduction:
    \n(i) CH4<\/sub>(g) + 2O2<\/sub>(g) \u2192 CO2<\/sub>(g) + 2H2<\/sub>O(l)
    \n(ii) CH2<\/sub> = CH2<\/sub>(g) + H2<\/sub>(g) \u2192 H3<\/sub>C – CH3<\/sub>(g)
    \n(iii) H2<\/sub>S(g) + O2<\/sub>(g) \u2192 2S(s) + 2H2<\/sub>O(1)
    \n\"1st
    \nAnswer:
    \n(i) CH4<\/sub> is oxidized to CO2<\/sub> while O2<\/sub> is reduced to H2<\/sub>O.
    \n(ii) H2<\/sub>S is oxidised to S while O2<\/sub> is reduced to H2<\/sub>O.
    \n(iii) CH2<\/sub> = CH2<\/sub> is reduced to CH3<\/sub> – CH3<\/sub> while H2<\/sub> is oxidized to CH3<\/sub> – CH3<\/sub>.
    \n(iv) Hg2+<\/sup> is reduced to Hg while O2-<\/sup> is oxidized to O2<\/sub>.<\/p>\n

    \"KSEEB<\/p>\n

    Question 6.
    \nUsing electron-transfer concept, identify the oxidation and reduction in the following redox reactions.
    \n(a) Zn(s) + 2H+<\/sup>(aq) \u2192 Zn2+<\/sup>(aq) + H2<\/sub>(g)
    \n(b) 2[Fe(CN)6<\/sub>]4-<\/sup> (aq) + H2<\/sub>O2<\/sub>(aq) + 2H+<\/sup>(aq) \u2192 2[Fe(CN)6<\/sub>]3-<\/sup> (aq) + 2H2<\/sub>O(aq)
    \n(c) 2[Fe(CN)6<\/sub>]3-<\/sup> (aq) + 2OH–<\/sup>(aq) + H2<\/sub>O2<\/sub>(aq) \u2192 2[Fe(CN)6<\/sub>]4-<\/sup>(aq) + 2H2<\/sub>O(l)
    \n(d) BrO3<\/sub>–<\/sup>(aq) + F2<\/sub>(g) + 2OH–<\/sup>(aq) \u2192 BrO4<\/sub>–<\/sup>(aq) + 2F–<\/sup>(aq) + H2<\/sub>O(l)
    \n(e) 2NaClO3<\/sub>(aq) + l2<\/sub>(aq) \u2192 2NaIO2<\/sub>(aq) + Cl2<\/sub>(g)
    \nAnswer:
    \nOxidants : (a) H+<\/sup> (b) H2<\/sub>O2<\/sub>, (c) [Fe(CN)6<\/sub>]3-<\/sup>, (d) Fe, (e) l2<\/sub>.
    \nReductants : (a) Zn, (b) [Fe(CN)6<\/sub>]4-<\/sup> , (c) H2<\/sub>O2, (d) BrO3<\/sub>–<\/sup> , (e)NaClO3<\/sub><\/p>\n

    Question 7.
    \nWhy do the following reactions proceed differently ?
    \nPb3<\/sub>O4<\/sub> + 8HCl \u2192 3PbCl2<\/sub> +Cl2<\/sub> + 4H2<\/sub>O and
    \nPb3<\/sub>O4<\/sub> + 4HNO3<\/sub> \u00a0 \u2192 2Pb(NO3<\/sub>)2<\/sub> + PbO2<\/sub> + 4H2<\/sub>O
    \nAnswer:
    \nPb3<\/sub>O4<\/sub> is actually a stoichiometric mixture of 2 moles of PbO and one mole of PbO2<\/sub>, i.e., 2PbO. PbO2<\/sub>. In PbO2<\/sub>, lead is present in + 4 oxidation state, whereas in PbO, lead is present in +2 oxidation state. Since +2 oxidation state or Pb is more stable, therefore, PbO2<\/sub> acts as an oxidant (oxidizing agent) and hence oxidises OF ions of HCl into Cl2<\/sub>. Further more, PbO is a basic oxide and hence reacts with HCl to form PbCl2<\/sub> and H2<\/sub>O. Thus the reaction of Pb2<\/sub>O4<\/sub> with HCl can be split into two , reactions, namely, acid-base and redox reaction as shown below :
    \n\"1st<\/p>\n

    Since HNO3<\/sub> is an oxidizing agent, it does not react with PbO2<\/sub> which is also an oxidizing agent. Therefore, no redox reaction occurs between PbO2<\/sub> and HNO3<\/sub>. However, acid-base reaction between PbO and HNO3<\/sub> occurs as follows :<\/p>\n

    2PbO + 4HNO3<\/sub> \u2192 2Pb(NO3<\/sub> )2<\/sub> + 2H2<\/sub>O<\/p>\n

    The overall reaction of PbO3<\/sub> with HNO3<\/sub> can then be written as
    \n2PbO + PbO2<\/sub> + 4HNO3<\/sub> \u2192 2Pb(NO3<\/sub> )2<\/sub> + PbO2<\/sub> + 2H2<\/sub>O or
    \nPb3<\/sub>O4<\/sub>+ 4HNO3<\/sub> \u2192 2Pb(NO3<\/sub> )2<\/sub> + PbO2<\/sub> + 2H2<\/sub>O
    \nThus, it is the passive nature of PbO2<\/sub> against HNO3<\/sub>that makes the reaction of Pb3<\/sub>O4<\/sub> with HNO3<\/sub> different from that with HCl.<\/p>\n

    Question 8.
    \nExplain applications of Redox Reactions.
    \nAnswer:
    \n1. Extraction of metals : By using a suitable reducing agent, metal oxides can be ‘ reduced to metals. For example, Fe2<\/sub>O3<\/sub> is reduced to iron in the blast furnace using coke as the reducing agent.<\/p>\n

    Fe2<\/sub>O3<\/sub>(s) + 3C(S) \u2192 2Fe(s) + 3CO(g)<\/p>\n

    Simialrly, Al2<\/sub>O3<\/sub> is reduced to aluminium by cathodic reduction in an electrolytic cell. Other metals such as lithium, sodium, potassium, magnesium, calcium, etc., are also obtained commercially by electrolytic methods.<\/p>\n

    2. Electrochemical cells or batteries : Electrochemical cells or batteries based on redox reactions are widely used in our day today life to run a number of small and big gadgets and equipments. For example storage cells are used to supply all the electrical needs of our cars, trucks, buses, trains aeroplanes, etc. Similarly, electrical energy needed in the space capsule is obtained by the reaction of hydrogen and oxygen in fuel cells which are electro chemical cells using oxygen and hydrogen electrodes.<\/p>\n

    3. Photosynthesis : Green plants convert carbon dioxide and water into carbohydrates in presence of sunlight. This reaction is called Photosynthesis and is sensitized by chlorophyll.
    \n\"1st
    \nDuring this reaction, CO2<\/sub> is reduced to carbohydrates while water is oxidized to oxygen. The energy needed for the reaction is provided by sunlight.<\/p>\n

    This reaction is a source of food for plants and animals. It also maintain a constant supply of 21% of O2<\/sub> by volume in the atmosphere needed for combustion of fuels and breathing of a living creatures in the world.<\/p>\n

    4. Supply of energy : The energy required for our daily needs is obtained by oxidation of fuels. For example, oxidation of fuels such as wood, gas, kerosene, petrol, etc., produces a large amount of energy which we need for various purposes in our daily life.
    \nFuels (wood, petrol, kerosene, gas) + O2<\/sub> \u2192 CO2<\/sub> + H2<\/sub>O + other products + Energy<\/p>\n

    Human body also needs energy for proper functioning. This is obtained by the oxidation of glucose in our body to CO2<\/sub> and water.
    \n\"1st<\/p>\n

    5. Production of chemicals : Many chemicals of our daily needs such as caustic soda, chlorine, fluorine, etc, are produced by electrolysis which is based on redox reaction.<\/p>\n

    6. Quantitative analysis : Redox reactions are very useful in quantitative analysis by redox titrations. These titrations involve the reactions between oxidizing and reducing agents and help in estimating the amount of unknown substances in solutions.<\/p>\n

    \"KSEEB<\/p>\n

    Question 9.
    \nBalance the following equations in acidic medium by oxidation number
    \n\"1st
    \nAnswer:
    \n\"1st<\/p>\n

    Question 10.
    \nBalance the following equation by ION-Electron method :
    \nZn(s) + NO4<\/sub>+<\/sup> \u2192 Zn2+<\/sup> (aq) + NH4<\/sub>+<\/sup> (aq) + H2<\/sub>O(l)
    \nAnswer:
    \n\"1st<\/p>\n

    Question 11.
    \nSplit the following redox reaction in the oxidation and reduction half reactions:
    \n(a) 2K(s) + Cl2<\/sub>(g) >2KCl(s)
    \n(b) 2Al(s) + 3Cu2+<\/sup>(aq) >2Al3+<\/sup>(aq) + 3Cu(s)
    \nAnswer:
    \n(a) Let us write the oxidation numbers of the atoms of all the reactants and products taking part in reaction.
    \n\"1st<\/p>\n

    (b) Let us write the oxidation numbers of the atoms of all the reactants and products taking part in the reaction.
    \nIncrease in Oxidation Number
    \n\"1st
    \nQuestion 12.
    \nBalance the following reaction by ION-ELECTRON method Cl2<\/sub>(g) + OH–<\/sup> \u2192 Cl–<\/sup> (aq) + CIO3<\/sub>–<\/sup> (aq) + H2<\/sub>O (in basic medium)
    \nAnswer:
    \nCl2<\/sub>(g) + OH–<\/sup> \u2192 Cl–<\/sup> (aq) + CIO3<\/sub>–<\/sup> (aq) + H2<\/sub>O
    \n\"1st
    \nCl2<\/sub>(g) + 6H2<\/sub>O(l) \u2192 2ClO3<\/sub>–<\/sup>(aq) (… Go for balancing oxygen is a abasic media odd off on RHS )
    \nCl2<\/sub>(g) + 6H2<\/sub>O(l) + 12OH–<\/sup> \u2192 2ClO3<\/sub>–<\/sup>(aq) + 12H2<\/sub>O(l)
    \nCl2<\/sub>(g) + 12OH–<\/sup> (aq) \u2192 2ClO3<\/sub>–<\/sup>(aq) + 6H2<\/sub>O(l) (now balance e-)
    \nCl2<\/sub>(g) I +12OH–<\/sup> (aq) \u2192 2ClO3<\/sub>–<\/sup>(aq) + 6H2<\/sub>O(l) + 10e–<\/sup> . ..(1)
    \nReduction half-reaction
    \nCl2<\/sub>(g) \u2192 Cl–<\/sup>(aq)
    \nCl2<\/sub>(g) \u2192 2Cl–<\/sup>(aq)
    \nCl2<\/sub>(g) + 3e \u2192 2Cl–<\/sup>(aq) …(2)
    \nMultiply equation no. (2) by 5 to equalize electrons
    \nAdd both the half-reactions
    \n\"1st<\/p>\n

    \"KSEEB<\/p>\n

    Question 13.
    \nWrite the over all net ionic reaction for the following change if the half reactions are correctly balanced: Cl–<\/sup> ion is oxidized to Cl2<\/sub> by MnO2<\/sub> in acidic solution: Chloride ion is oxidized to Cl2<\/sub> by MnO4<\/sub>–<\/sup>(in acid solution)
    \nAnswer:
    \nThe skeletal equation is
    \n\"1st
    \nWhere ox. no. of chlorine changes from -1 in Cl ion to O in Cl2<\/sub>
    \n\"1st
    \n\"1st<\/p>\n

    Question 12.
    \nA cell is prepared by dipping a chromium rod in 1 M Cr2<\/sub>(SO4<\/sub>)3<\/sub> solution and an ion rod in 1 M FeSO4<\/sub> solution. The standard reduction potentials of chromium and iron electrodes are -0.75V and -0.45 V respectively.
    \n(a) What will be the cell reaction ?
    \n(b) What will be the standard EMF of the cell ?
    \n(c) Which electrode will act as anode ?
    \n(d) Which electrode will act as cathode ?
    \nAnswer:
    \nThe two half cell reduction equations me :<\/p>\n

    Fe2+<\/sup>(aq) + 2e–<\/sup> \u2192 Fe(s);E\u00b0 =-0.45 V …(i)
    \nCr3+<\/sup>(aq) + 3e–<\/sup> \u2192 Cr(s);E\u00b0 = -0.75V …(ii)<\/p>\n

    Since Cr3+<\/sup> \/ Cr electrode has lower reduction potential therefore it acts as the anode while Fe2+<\/sup> \/Fe electrode with higher electrode potential acts as the cathode.<\/p>\n

    To equalize the number of electrons, multiply Eq. (i) by 3 and Eq. (ii) by 2. But do not multiply their E\u00b0 values. Thus,<\/p>\n

    3Fe2+<\/sup> (aq) + 6e–<\/sup> \u2192 3Fe(s); E\u00b0 = -0.45V .. .(iii)
    \n2Cr3+<\/sup>(aq) + 6e–<\/sup> \u2192 2Cr(s) ; E\u00b0 =-0.75V …(iv)<\/p>\n

    To obtain equation for the cell reaction, subtract Eq. (iv) from Eq. (iii), we have,
    \n2Cr(s) + 3Fe2+<\/sup>(aq) \u2192 2Cr3+<\/sup>(aq) + 3Fe(s)
    \nE\u00b0cell<\/sub> = -0.45-(-0.75V) = +0.30V Thus, the EMF of the cell = + 0.30V<\/p>\n

    Question 13.
    \nWrite the net ionic equation for the reaction of potassium dichromate (VI), K2<\/sub>Cr2<\/sub>O7<\/sub> with sodium sulphate, Na2<\/sub>SO3<\/sub>, in acid solution to give chromium (III) ion and sulphate ion.
    \nAnswer:
    \nStep-1: Write the skeleton equation for the given reaction.
    \n\"1st
    \nStep – 2 : Find out the elements which undergo a change in oxidation number (O.N.) O.N. decreases by 3 per Cr. atom
    \n\"1st
    \nHere, O.N. of Cr decreases from +6 in Cr2<\/sub>O7<\/sub>2-<\/sup> to +3 in Cr3+<\/sup> while that of S increases from +4 SO3<\/sub>-2<\/sup> to +6 in SO4<\/sub>2-<\/sup>
    \nStep – 3 : Balance increase \/ decrease in O.N. since the total increase in O.N. is 2 and decrease is 6, therefore, multiply SO3<\/sub>2-<\/sup> on L.H.S. and SO4<\/sub>2-<\/sup> on R.H.S. of Eq. (ii) by 3. Combining steps 2 and 3, we have,<\/p>\n

    \"1st<\/p>\n

    Step – 7 : Balance O atoms by adding H2<\/sub>O molecules.<\/p>\n

    \"1st<\/p>\n

    Step – 7 : Balance H atoms by adding H+<\/sup><\/p>\n

    \"1st<\/p>\n

    Thus, Eq. (v) represents the correct balanced equation.<\/p>\n

    \"KSEEB<\/p>\n

    Question 14.
    \nPermanganate ion react with bromide ion in basic medium to give manganese dioxide and bromate ion. Write the balance chemical equation for the reaction.
    \nAnswer:
    \nStep – 1: Write the skeletal equation. The skeletal equation for the given reaction is :
    \n\"1st<\/p>\n

    Step – 2 :
    \nFind out the elements which undergo a change in oxidation number (O.N.) O.N. increases by 6 per Br atom
    \n\"1st<\/p>\n

    Here, O.N. of Br inceases from – 1 Br–<\/sup> to +5 in BrO3<\/sub>–<\/sup>, therefore, Br–<\/sup> acts as reductant. Further, O.N. of Mn decreases from +7 in MnO4<\/sub>–<\/sup> to +4 in MnO2<\/sub>, therefore, MnO4<\/sub>–<\/sup> acts as oxidant.<\/p>\n

    Step – 3 : Balance O atoms by adding H2<\/sub>O molecules.
    \n\"1st<\/p>\n

    Step – 4 s Balance H atoms by adding H2<\/sub>O and OH–<\/sup> ions since the reaction occurs in basic medium, therefore, add 2H2<\/sub>O\u00a0to L.H.S. and 2OH–<\/sup> to R.H.S. of Eq. (iv), we have,
    \n\"1st<\/p>\n

    Question 15.
    \nWrite the half reactions for the following redox reactions.
    \n(a) Fe2+<\/sup>(aq) + 2I–<\/sup>(aq) \u2192 2Fe2+<\/sup>(aq) + I2<\/sub>(aq)
    \n(b) Zn(s) + 2H+<\/sup>(aq) \u2192 Zn2+<\/sup>(aq) + H2<\/sub>(g)
    \n(c) Al(s) + 3Ag+<\/sup>(aq) \u2192 Al3+<\/sup>(aq) + 3Ag(s)
    \nAnswer:
    \n\"1st<\/p>\n

    Question 16.
    \nBalance the equation, by ion electron method.
    \nAS2<\/sub>S3<\/sub>(S) + NO3<\/sub>–<\/sup>(aq) + H+<\/sup> (aq) \u2192 AsO4<\/sub>3-<\/sup>(aq) + S(s) + NO(g) + H2<\/sub>(l)
    \nAnswer:
    \nStep 1 : To identify the atoms whose oxidation numbers have undergone a change. Writing the oxidation number of each atom above its symbol, we have
    \n\"1st<\/p>\n

    Step 2:
    \n\"1st<\/p>\n

    Step 3 : To balance the oxidation half Eq. (i)
    \n(a) Balance all the atoms other than H and O. Multiply AsO4<\/sub>3-<\/sup> by 2 and S by 3 on RHS of Eq. (i) we have,
    \n\"1st<\/p>\n

    (c) Balance charge by adding H+ ions. The total charge on RHS of Eq. (iv) is -16 and zero the LHS therefore, add 16H+ to RHS of Eq. (iv) we have,
    \n\"1st
    \n(d) Balance O atoms by adding H2<\/sub>O molecules.
    \n\"1st
    \nThe H-atoms get automatically balanced. Thus Eq. (vi) represents the balanced oxidation half equation.<\/p>\n

    Step 4 : To balance the reduction half Eq. (ii)
    \n(a) Balance oxidation number by adding electrons.
    \n\"1st<\/p>\n

    (b) Balance charge by adding H+ ions.
    \n\"1st
    \n(c) Balance H+ by adding H20
    \n\"1st
    \nThus, Eq. (ix) represents the balanced reduction half equation Eq. (ix) by 10 and Eq,
    \n(vi) by 3 and (ix) 10 + (vi) x 3<\/p>\n

    Step 5 :
    \n\"1st<\/p>\n

    \"KSEEB<\/p>\n

    Question 17.
    \nPermanganate (VII) ion, in basic solution oxidizes iodide ion 1- to produce molecular iodine (I2<\/sub>) and manganese (IV) oxide (MnO2<\/sub>). Write a balanced ionic equation to represent this redox reaction.
    \nAnswer:
    \nStep 1 s Write pie skeletal equation for the given reaction.
    \n\"1st
    \nStep 2 : Write the O.N. of all the elements above their respective symbols.
    \n\"1st
    \n\"1st<\/p>\n

    Step 4 : To balance oxidation half Eq. … (ii)
    \n(a) Balance all atoms 2I–<\/sup>(aq) \u2192 I2<\/sub>(s) … (iv)
    \n(b) Balance O.N. by adding electrons.
    \n2I–<\/sup>(aq) \u2192 I2<\/sub>(s) + 2e–<\/sup> …(v)
    \nCharge on either side of Eq(v) is balanced. Thus, Eq. (v) represents the balanced oxidation half equation.<\/p>\n

    Step 5 : Balance the reduction half equation (iii)
    \nBalance O.N. by adding electrons.
    \n\"1st<\/p>\n

    Step 6: To balance the electrons lost in Eq. (v) and gained in Eq. (viii) Equation (v) x 3 + Equation (viii) x 2 we have,
    \n\"1st
    \nThis represents are final balanced redox equation.<\/p>\n

    Question 18.
    \nIn passing chlorine gas through a concentrated solution of alkali, we get chloride and chlorate ions. Obtain balanced chemical equation for this reaction by ion -electron method.
    \nAnswer:
    \nStep 1: Write the skeletal equation for the given reaction.
    \n\"1st<\/p>\n

    Step 2 : Write the ON of all the elements above their respective symbols O.N. of Cl increases by 5 per Cl atom.
    \n\"1st
    \nO.N. of Cl decreases by 1 per Cl atom Total increase = 2 x 5 = 10 Total decrease = 2 x -1 = -2<\/p>\n

    Step 3 : Find out the oxidant and the reductant and split the skeletal Eq (i) into two half reactions.<\/p>\n

    Step 4 : To balance the reduction half equation (ii). (a) Balance all atoms.
    \nCl2<\/sub>(aq) \u21922Cl–<\/sup>(aq) … (iv)
    \n(b) Balance oxidation number by adding electrons.
    \nCl2<\/sub>(g) + 2e–<\/sup> \u21922Cl–<\/sup>(aq) … (v)
    \nrepresents the balanced reduction half reactions.<\/p>\n

    Step 5 : To balance the oxidation half equation (iii)
    \n(a) Balance all atoms Cl2<\/sub>(g) \u00bb2ClO3<\/sub>–<\/sup>
    \n(b) Balance ON by adding electrons.
    \nCl2<\/sub>(g) \u2192 2ClO3<\/sub>–<\/sup>(aq) + 10e–<\/sup> … (vi)
    \n(c) Balance charge by adding OH” ions
    \nCl2<\/sub>(g) + 120H–<\/sup>(aq) \u2192 2ClO3<\/sub>–<\/sup>(aq) + 10e–<\/sup> … (ix)
    \n(d) Balance O atoms, the RHS of Eq. (viii) contains six O atoms but on the LHS there are 12. Therefore, add 5H2<\/sub>O to the RHS, we have
    \n\"1st
    \nEq (ix) represents the balanced oxidation half equation.<\/p>\n

    Step 6 : (v) x 5 + (ix) we have
    \n\"1st
    \nThis represents the final balanced redox equation.<\/p>\n

    Question 19.
    \nSuggest a scheme of classification of the following redox reactions.
    \n(a) N2<\/sub>(g) + O2<\/sub>(g) \u21922N(Xg)
    \nOb) 2Pb(NO3<\/sub>)2<\/sub>(s) \u2192 2PbO(s) + 2NO2<\/sub>(g) + 1\/2 O2<\/sub>(g)
    \n(c) NaH(s) + H2<\/sub>O(1)\u2192\u00a0 4NaOH(aq) + H2<\/sub>(g)
    \n(d) 2NO2<\/sub>(g) + 20H–<\/sup>(aq) \u2192 NO2<\/sub>–<\/sup>(aq) + NO3<\/sub>–<\/sup> (aq) + H2<\/sub>O(l)
    \nAnswer:
    \n(a)
    \n\"1st In this reaction, the compound nitric oxide is formed by combination of elemental substances like nitrogen and oxygen. Since the oxidation number of nitrogen increases from 0 to +2 and of oxygen decreases from 0 to -2 therefore, it is a combination redox reaction.<\/p>\n

    (b)
    \n\"1st<\/p>\n

    In this reaction, lead nitrate decomposes to form three products, viz, lead oxide, nitrogen dioxide and oxygen. Since the oxidation number of nitrogen decrease from +5 in lead nitrate to +4 in NO2 in O2, therefore, it is a decomposition redox reaction.<\/p>\n

    (c)
    \n\"1st
    \nIn this reaction, hydrogen of water has been displaced by hydride ion to form dihydrogen gas. Therefore, it is a displacement reaction. Since in this reaction, the oxidation number of hydrogen increases from -1 in hydride ion to zero in dihydrogen gas and that of hydrogen decreases from +1 in water to zero in dihydrogen, therefore, it is a displacement redox reaction.<\/p>\n

    (d)
    \n\"1st
    \nThis is a disproportionation reaction since here the oxidation state of nitrogen decreases from +4 in NO2<\/sub> to +3 in NO2<\/sub>–<\/sup> ion, as well as increases from +4 in NO2<\/sub> to +5 in NO3<\/sub> ion.<\/p>\n

    Question 20.
    \nCalculate the oxidation number of the element underline in each of the following cases,
    \n(a) Al in Al2<\/sub>O3<\/sub>
    \n(b) P in P2<\/sub>O5<\/sub>
    \n(c) S in Na2<\/sub>S2<\/sub>O3<\/sub>
    \n(d) Cl in NaClO3 <\/sub>
    \n(e) Mn in KMn<\/sub>O4<\/sub>
    \n(f) HA4<\/sub>Cl4<\/sub>
    \n(g) Na2<\/sub>PdCl4<\/sub>
    \n(h) K4<\/sub>MO2<\/sub>Cl8<\/sub>
    \n(i) ClO– <\/sup>
    \n(j) Si03<\/sub>2– <\/sup>
    \n(k) H2<\/sub>PO4<\/sub>– <\/sup>
    \n(1) Cr2<\/sub>O7<\/sub>2– <\/sup>
    \n(m) H2<\/sub>PO4<\/sub>– <\/sup>
    \nAnswer:
    \nSolution : The rule used here is that the algebraic sum of the oxidation numbers of all the atoms a molecule is zero.
    \n\"1st<\/p>\n

    b) P2<\/sub>O5<\/sub> : 2 x (oxidation no. of P) + 5(-2) = 0
    \noxidation of P = pp = +5<\/p>\n

    c) Na2<\/sub>S2<\/sub>O3<\/sub> : 2(+l) + 2(oxidation no. of S) + 3(-2) = 0
    \n2(oxidation no. of S) = 4<\/p>\n

    d) NaClO3<\/sub> : 1(+1) + (oxidation no. of Cl) +3 (-2) = 0
    \noxidation no. of Cl = +6 – 1 = +5<\/p>\n

    e) KMnO4<\/sub> : 1(+1) + (oxidation no. of Mn) + 4(-2) = 0
    \n+1 + (oxidation no. of Mn) – 8 =0
    \n\u2234 oxidation no. of Mn = + x7
    \nAnswer:
    \nSolution : The rule used here is that the algebraic sum of the oxidation numbers of all the atoms in an ion is equal to the net charge on the ion.<\/p>\n

    f) HA4<\/sub>Cl4<\/sub> 1(ON of H) + 1C (ON of A4) + (ON of Cl) = 0
    \n(+1) + x + 4(-l) = 0 => l + x-4 = 0 => x = +3<\/p>\n

    g) Na2<\/sub>PdCl2<\/sub> = 2(+l) + x + 4(-l) = 0 => 2+x-5=0 => x = +2<\/p>\n

    h) x be ON to be find
    \nK4<\/sub>MO2<\/sub>Cl8<\/sub> = 4(+l) + (x)2 + 8(-l) = 0
    \n4 + 2x – 8 = 0 => 2x – 4 = 0 => x = +2
    \n\"1st<\/p>\n

    \"KSEEB<\/p>\n

    Question 21.
    \nWhat is oxidation number of sulphur in the following molecules \/ ions (n) sinH2<\/sub>S sin SO4<\/sub>2-<\/sup> sinH2<\/sub>SO4<\/sub>
    \nAnswer:
    \n(i) The oxidation number of Hydrogen is +1, and s = x than
    \n\"1st<\/p>\n

    (ii) SO4<\/sub>2-<\/sup> let x be ON of s ; Oxidation Number of oxygen is -2
    \n\"1st<\/p>\n

    (iii) H2<\/sub>SO4<\/sub> let x, +1, -2 be on of S, H and O
    \n\"1st<\/p>\n

    Question 22.
    \nFind the oxidation number of the underlined atoms.
    \n(a) CCl4<\/sub>
    \n(b) CH4<\/sub>
    \n(C) ClO3<\/sub>–<\/sup>
    \n(d) BrF3<\/sub>
    \n(e) Na2<\/sub> B4<\/sub>O7<\/sub>
    \n(f) HClO3<\/sub>
    \n(g) HPO2<\/sub> –<\/sup>
    \n(h) Fe3<\/sub> O4<\/sub>
    \nAnswer:
    \n\"1st
    \n\"1st<\/p>\n

    \"KSEEB<\/p>\n","protected":false},"excerpt":{"rendered":"

    You can Download Chapter 8 Redox Reactions Questions and Answers, Notes, 1st PUC Chemistry Question Bank with Answers Karnataka State Board Solutions help you to revise complete Syllabus and score more marks in your examinations. Karnataka 1st PUC Chemistry Question Bank Chapter 8 Redox Reactions 1st PUC Chemistry Redox Reactions One Mark Questions and Answers …<\/p>\n","protected":false},"author":5,"featured_media":0,"comment_status":"open","ping_status":"open","sticky":false,"template":"","format":"standard","meta":{"footnotes":""},"categories":[81],"tags":[],"jetpack_sharing_enabled":true,"jetpack_featured_media_url":"","_links":{"self":[{"href":"https:\/\/kseebsolutions.guru\/wp-json\/wp\/v2\/posts\/9185"}],"collection":[{"href":"https:\/\/kseebsolutions.guru\/wp-json\/wp\/v2\/posts"}],"about":[{"href":"https:\/\/kseebsolutions.guru\/wp-json\/wp\/v2\/types\/post"}],"author":[{"embeddable":true,"href":"https:\/\/kseebsolutions.guru\/wp-json\/wp\/v2\/users\/5"}],"replies":[{"embeddable":true,"href":"https:\/\/kseebsolutions.guru\/wp-json\/wp\/v2\/comments?post=9185"}],"version-history":[{"count":0,"href":"https:\/\/kseebsolutions.guru\/wp-json\/wp\/v2\/posts\/9185\/revisions"}],"wp:attachment":[{"href":"https:\/\/kseebsolutions.guru\/wp-json\/wp\/v2\/media?parent=9185"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/kseebsolutions.guru\/wp-json\/wp\/v2\/categories?post=9185"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/kseebsolutions.guru\/wp-json\/wp\/v2\/tags?post=9185"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}