{"id":10172,"date":"2019-12-06T11:54:41","date_gmt":"2019-12-06T06:24:41","guid":{"rendered":"https:\/\/kseebsolutions.guru\/?p=10172"},"modified":"2021-07-02T15:43:53","modified_gmt":"2021-07-02T10:13:53","slug":"2nd-puc-chemistry-question-bank-chapter-3","status":"publish","type":"post","link":"https:\/\/kseebsolutions.guru\/2nd-puc-chemistry-question-bank-chapter-3\/","title":{"rendered":"2nd PUC Chemistry Question Bank Chapter 3 Electrochemistry"},"content":{"rendered":"

You can Download Chapter 3 Electrochemistry Questions and Answers, Notes, 2nd PUC Chemistry\u00a0Question Bank with Answers<\/a> Karnataka State Board Solutions help you to revise complete Syllabus and score more marks in your examinations.<\/p>\n

Karnataka 2nd PUC Chemistry Question Bank Chapter 3 Electrochemistry<\/h2>\n
\n
\n
\n

2nd PUC Chemistry Electrochemistry NCERT Textbook Questions and Answers<\/h3>\n

Question 1.
\nArrange the following metals in the order in which they displace each other from the solution of their salts.
\nAl, Cu, Fe, Mg and Zn.
\nAnswer:
\nMg, Al, Zn, Fe, Cu.<\/p>\n

Question 2.
\nGiven the standard electrode potentials, K+<\/sup>\/K = -2.93V, Ag+<\/sup>\/Ag = 0.80V,
\nHg2+<\/sup>\/Hg = 0.79 V
\nMg2+<\/sup>\/Mg = -2,37 V, Cr3+<\/sup>\/Cr = – 0.74V
\nArrange these metals in their increasing order of reducing power.
\nAnswer:
\nThe lower the reduction potential, the higher is the reducing power. Hence, the reducing power of the given metals increases inthe following order.
\nAg < Hg < Cr < Mg < K.<\/p>\n

Question 3.
\nDepict the galvanic cell in which the reaction Zn(s)+2Ag+<\/sup>(aq) \u2014> Zn2+<\/sup>(aq)+2Ag(s) takes place. Further show:
\n(i) Which of the electrode is negatively charged?
\n(ii) The carriers of the current in the cell.
\n(iii) Individual reaction at each electrode.
\nAnswer:
\nThe galvanic cell in which the given reaction takes place is depicted as:
\nZn(s) | Zn2+<\/sup> (aq) || Ag+<\/sup> (aq) | Ag(s)
\n(i) Zn electrode (anode) is negatively charged
\n(ii) Tons are carriers of current in the cell and in the external circuit, current from silver to Zinc.
\n(iii) The reaction taking place at the anode is given by,
\nZn(s) -H \u2192 Zn2+<\/sup>(aq) + 2e–<\/sup>
\nThe reaction taking place at the cathode is given
\nAg+<\/sup>+ e–<\/sup> \u2192 Ag(s)<\/p>\n

\"KSEEB<\/p>\n

Question 4.
\nCalculate the standard cell potentials of galvanic cell in which the following reactions take place:
\n(i) 2Cr(s) + 3Cd2+<\/sup>(aq) \u2192 2Cr3+<\/sup>(aq) + 3Cd
\n(ii) Fe2+<\/sup>(aq) + Ag+<\/sup>(aq) \u2192 Fe3+<\/sup>(aq) + Ag(s)
\nCalculate the ArG9and equilibrium constant of the reactions.
\nAnswer:
\n(i) For the given reaction, the Nemst equation can be given as:
\n\"2nd
\n\u2234 E\u03b8<\/sup>\u00a0= 1.104V
\nWe know that,
\n\u0394r<\/sub>G\u03b8<\/sup>\u00a0= -nFE\u03b8<\/sup>
\n= -2 \u00d7 96487\u00d7 1.04
\n= – 213043.296J
\n= -213.04KJ<\/p>\n

Question 5.
\nWrite the Nernst equation and em! of the
\nfollowing cells at 298 K:
\n(I) Mg(s)|Mg+2<\/sup>(O.OO1M) ||Cu+2<\/sup>(0.0001M)|Cu(s) .
\n(ii) Fe(s)|Fe+2<\/sup>(O.OO1 M)||H+<\/sup>(1M)H2<\/sub>(g) (1bar)|Pt(s)
\n(iii) Sn(s) |Sn2+<\/sup>(O.050 M)||H+<\/sup>(0.020M|H2(g) (1 bar)|Pt(s)
\n(iv) Pt(s)|Br2<\/sub>(l)|Br–<\/sup>(O.O1O M)||H+<\/sup>(O.030 M)| H2<\/sub>(g) (1 bar)|Pt(s).
\nAnswer:
\n(i) E\u03b8<\/sup>Cr3+<\/sup> \/Cr = 0.74V
\nE\u03b8<\/sup>Cd2+<\/sup> \/ Cd = – 0.40V
\nThe galvanic cell of the reaction IC depicted as :
\nCr(s)|Cr3+<\/sup> (aq) || (Cd2+<\/sup> (aq) Cd(s)
\nNow, the standard cell potential is
\nE\u03b8<\/sup>cell<\/sub> = E\u03b8<\/sup>R<\/sub>– E\u03b8<\/sup>L
\n= -40 – (-0.74)
\n= +0.34V
\n\u2206r<\/sub>G\u03b8<\/sup> = \u2014nFE\u03b8<\/sup>cell<\/sub>
\nIn the given equation,
\nn = 6
\nF = 96487 C mol-1<\/sup>
\nE\u03b8<\/sup>cell<\/sub> = + 0.34V
\nThen, \u2206r<\/sub>G\u03b8<\/sup> =-6 \u00d7 96487 mol-1<\/sup> \u00d7 0.34V
\n= -196833.48 CV mol-1<\/sup>
\n= -196833.48 J mol-1<\/sup>
\n= -196.83 KJ mol-1<\/sup>
\nAgain
\n\u2206r<\/sub>G\u03b8<\/sup>\u00a0= -RT In K
\n=>\u2206r<\/sub>G\u03b8<\/sup>\u00a0=-2.303RT log K
\n\"2nd<\/p>\n

(ii) E\u03b8<\/sub> Fe3+<\/sup> \/Fe2+<\/sup> = 0.77V
\nE\u03b8<\/sub> Ag–<\/sup>\/Ag = 0.80V
\n(ii) For the given reaction, the Nernst equation can be given as:
\n\"2nd<\/p>\n

(iii) Far the giv\u00ebn reaction, the Nernst equation can be given as:
\n\"2nd
\n\"2nd<\/p>\n

0.14 – 0.0295 \u00d7 log 125
\n= 0.14-0.062
\n= 0.078 V
\n= 0.08 V (approx)<\/p>\n

(iv) For the given reaction , the nernst equation can be given as:
\n\"2nd
\n= – 1.09 – 0.02955 \u00d7 log ( 1.11\u00d7 107<\/sup>)
\n= – 1.09 – 0.02955(0.0453 + 7)
\n= -1.09 – 0.208 =-1.298 V.<\/p>\n

\"KSEEB<\/p>\n

Question 6.
\nIn the button cells widely used in watches and other devices the following reaction takes place:
\nZn(s) + Ag2<\/sub>O(s) + H2<\/sub>O(l) \u2192 Zn2 <\/sup>+ (aq) + 2Ag(s) + 2OH-(aq)
\nDetermine \u2206r<\/sub> G\u03b8<\/sup> and E\u03b8<\/sup> for the reaction.
\nAnswer:
\nThe galvanic cell of the reaction is depicted as:
\nFe2+<\/sup> (aq) | Fe3+<\/sup> (aq) || Ag+<\/sup> (aq) | Ag(s)
\nNow, the standard cell potential is
\nE\u03b8<\/sup>cell<\/sub>\u00a0= E\u03b8<\/sup>R<\/sub> – El\u03b8<\/sup>
\n= 0.80 – 0.77 ‘
\n= 0.03 V
\nHere, n = 1
\nThen, \u2206r<\/sub> G\u03b8<\/sup> = – nFE\u03b8<\/sup>cell<\/sub>
\n= -1 \u00d7 96487 C mol-1<\/sup> \u00d7 0.03V
\n= – 2894.61 J mol-1<\/sup>
\n= – 2.89 KJ mol-1<\/sup>
\n\u2206r<\/sub> G\u03b8<\/sup> = 2.303 RT In K
\n\"2nd<\/p>\n

Question 7.
\nDefine conductivity and molar conductivity for the solution of an electrolyte.
\nDiscuss their variation with concentration.
\nAnswer:
\nConductivity of a solution is defined as the conductance of a solution 1 cm in length and area of cross section cm2.1 is represented by K.<\/p>\n

Conducti vity always decreases with a decrease in concentration both for weak and strong electrolytes. This is because the number of ions per unit volume that carry the current in a solution decreases with a decrease in concentration.<\/p>\n

Molar conductivity of a solution at a given concentration is the conductance of volume V of a solution containing 1 mole of the electrolyte kept between two electrodes with the area area of cross-section A and distance of unit length.<\/p>\n

Molar conductivity increases with a decrease in concentration. This is because the total volume of the solution containing one mole of the electrolyte increases on dilution.<\/p>\n

Question 8.
\nThe conductivity of 0.20 M solution of KCl at 298 K is 0.0248 S cm-1<\/sup>. Calculate its molar conductivity.
\nAnswer:
\n\"2nd<\/p>\n

Question 9.
\nThe resistance of a conductivity cell containing 0.001M KCl solution at 298 K is 1500\u2126. What is the cell constant if conductivity of 0.001M KCl solution at 298 K is 0.146 \u00d7 10-3<\/sup> S cm-1<\/sup>.
\nAnswer:
\nCell constant = conductivity \u00d7 Resistance
\n= 0.146 \u00d7 10-3<\/sup>S Cm-1<\/sup> \u00d7 1500 \u2126 = 0.219 cm-1<\/sup><\/p>\n

\"KSEEB<\/p>\n

Question 10.
\nThe conductivity of sodium chloride at 298 K has been determined at different concentrations and thfe results are given below:
\n\"2nd<\/p>\n

Calculate \u2206m<\/sub> for all concentrations and draw a plot between \u2206m<\/sub> and c1\/2<\/sup>. Find the value of \u22060<\/sup>m<\/sub>
\nAnswer:
\nK = 7.896 \u00d7 10-5<\/sup> S cm-1<\/sup>
\nM = 0.00241
\n\"2nd<\/p>\n

Question 11.
\nHow much charge is required for the following reductions?
\n(i) 1 mol of Al3+<\/sup> to Al
\n(ii) 1 mol of Cu2+<\/sup> to Cu
\n(iii) 1 mol of MnO4-<\/sup> to Mn2+<\/sup>
\nAnswer:
\nAl3+<\/sup> + 3e \u2192 A1
\ncharge required = 3F
\n(ii) Cu2+<\/sup> + 2e \u2192 Cu
\ncharge required = 2F
\n(iii) MnO4-<\/sup> + 8H+<\/sup> + Se–<\/sup> \u2192 Mn2+<\/sup> + H2<\/sub>O
\ncharge required = 5F<\/p>\n

Question 12.
\nHow much electricity in terms of Faraday \u00a1s required to produce
\n(j) 20.Ogat Ca from molten CaCl2<\/sub>
\n(ii) 40.0 g of Al from Alm\u00f3lten Al2<\/sub>O3<\/sub>
\nAnswer:
\n(i) Ca2+<\/sup>2 + 2e \u2192 Ca
\n2F can produce I mole (=40 g) Ca
\n\u2234 To produce 20 g Ca requires, \\(\\frac{2 \\mathrm{F} \\times 20}{40}\\) = 1F<\/p>\n

(ii) Al3+<\/sup> + 3e \u2192 Al
\n3F can produce 1 mole (= 27g) Al
\n\u2234 To produce 40 g A1 requires
\n\\(\\frac{3 \\mathrm{F} \\times 40}{27}\\) = 4.44F<\/p>\n

Question 13.
\nHow much electricity is required in coulomb for the oxidation of
\n(i) 1 mol of H2<\/sub>O to O2<\/sub>
\n(ii) 1 mol of FeO to Fe2<\/sub>O3<\/sub>.
\nAnswer:
\n(i) 2H2<\/sub>O \u2192 4H+<\/sup> + O2<\/sub> + 4e
\n2F of electricity is required for oxidation of 1 mole of H2<\/sub>O
\n(ii) Fe2+<\/sup> \u2192 Fe3+<\/sup> + e
\nIF of electricity is required for oxidation of 1 mole FeO<\/p>\n

Question 14.
\nA solution of Ni(NO3<\/sub>)2<\/sub> is electrolysed between platinum electrodes using a current of 5 amperes for 20 minutes. What mass of Ni is deposited at the cathode?
\nAnswer:
\nNi2+<\/sup> + 2e–<\/sup> \u2192 Ni
\n2F (2 \u00d7 96500 C) can produce 58.7 g of Ni
\nQ = It = 5 \u00d7 20 \u00d7 60 =6000C
\n\"2nd<\/p>\n

Question 15.
\nThree electrolytic cells A,B,C containing = 0.439 g of Zn solutions of ZnSO4<\/sub>, AgNO3<\/sub> and CuSO4<\/sub>, respectively are connected in series. A steady current of 1.5 amperes was passed through them until 1.45 g of silver deposited at the cathode of cell B. How long did the current flow? What mass of copper and zinc were deposited?
\nAnswer:
\n\"2nd
\ni.e. 108 g of Ag is deposited by 96487 C
\nTherefore, 1 .45g of Ag is deposited by
\n\"2nd
\nGiven,
\nCurrent = 1.5A
\n\\(\\frac{1295.43}{1.5}\\) S
\n\u2234 Time = 863.6S
\n= 864 S
\n= 14.40 min
\nAgain,
\n\"2nd
\ni.e. 2 \u00d7 96487 C of charge deposit = 63.5 g of Cu
\n\"2nd
\nTherefore, 1295.43 C of charge will deposit = 0.426g of Cu
\n\"2nd
\ni.e. 2 \u00d7 96487 C of charge deposit = 65.4 g of Zn
\n\"2nd
\nTherefore, 1295.43 C of charge will deposit = 0.439 g of Zn<\/p>\n

\"KSEEB<\/p>\n

Question 16.
\nPredict the products of electrolysis in
\neach of the following:
\n(1) An aqueous solution of AgNO3<\/sub> with silver electrodes.
\n(ii) An aqueous solution of AgNO3<\/sub> with platinum electrodes.
\n(iii) A dilute solution of H2<\/sub>SO4<\/sub> with platinum electrodes.
\n(iv) An aqueous solution of CuCl2<\/sub> with platinum electrodes.
\nAnswer:
\n(i) At cathode:
\nThe following reduction reactions compete to take place at the cathode
\nAg+<\/sup>(aq) + e–<\/sup> \u2192 Ag(s); E\u03b8<\/sup> = 0.80 V
\nH+<\/sup> (aq) + e–<\/sup> \u2192 \\(\\frac { 1 }{ 2 }\\) H2<\/sub> (g); E\u03b8<\/sup> = 0.00V
\nThe reaction with a higher value of E\u03b8<\/sub> takes place of the cathode. Therefore, deposition of silver will take place at the cathode.<\/p>\n

At anode:
\nThe Ag anode is attacked by NO3<\/sub>–<\/sup> ions. Therefore, the silver electrode at the anode dissolves in the solution to from Ag+<\/sup>.<\/p>\n

(ii) At cathode: Same as above
\nAt anode: Anode is not attackable and hence OH–<\/sup> ions have lower discharge potential than NO3<\/sub>–<\/sup> ions and OH–<\/sup> ions react to give O2<\/sub>
\nOH–<\/sup> \u2192 OH + e–<\/sup>
\n4OH \u2192 2H2<\/sub>O + O2<\/sub> (g)
\n(iii) H2<\/sub>SO4<\/sub> \u2192 2H+<\/sup> + SO2-<\/sup>4<\/sub>
\nHO2<\/sub> \u21cc H+<\/sup> + OH–<\/sup><\/p>\n

At cathode:
\n2H+<\/sup>+ 2e–<\/sup> \u2192 H2<\/sub>
\nAt anode: 4OH–<\/sup> \u2192 2H2<\/sub>O + O2<\/sub> + 4e–<\/sup>
\ni. e., H2<\/sub> will be liberated at cathode and O2<\/sub> at anode.<\/p>\n

(iv) CuCl2<\/sub> \u2192 Cu2+<\/sup>+2Cl–<\/sup>
\n\"2nd
\nAt Cathode: Cu2+<\/sup> ions will be reduced in preference to H+<\/sup> ions
\nCu2+<\/sup> + 2e \u2192 Cu
\nAt anode: Cl’ ions will be oxidised in preference to OH–<\/sup> ions.
\n2Cl–<\/sup> \u2192 Cl2<\/sub> + 2e–<\/sup>
\ni.e., Cu will be deposited on the cathode and
\nCl2<\/sub> will be liherated at the anode.<\/p>\n

2nd PUC Chemistry Electrochemistry Additional Questions and Answers<\/h3>\n

Question 1.
\nA solution of sodium chloride is a better
\nconductor of electricity at a temperature of 50\u00b0C than at room temperature. Why?
\nAnswer:
\nA solution of NaCl shows greater conduction of electricity at a temperature of 50\u00b0C than at room temperature because the ionic mobility of a strong electrolyte such as NaCl increases with an increase in temperature.<\/p>\n

Question 2.
\nGive the relationship between molar conductivity and specific conductivity.
\nAnswer:
\nMolar conductivity and specific conductivity are related to each other by the given equation.
\n\"2nd
\nWhere,
\n\u0394m<\/sub> = Molar conductivity
\nK = Specific conductivity
\nC = Molar concentration<\/p>\n

Question 3.
\nWhy is it not possible to measure single electrode potential?
\nAnswer:
\nThe process of oxidation or reduction cannot take place alone. However, electrode potential is a relative tendency and can be measured with respect to a reference electrode such as standard hydrogen electrode.<\/p>\n

Question 4.
\nWhy is the rusting of iron faster in saline water than in pure water?
\nAnswer:
\nStrong electrolytes such as sodium chloride are present in saline water. The ions produced from NaCl help in the reduction of oxygen to form water. Hence, the rusting of iron is faster in saline water than in pure water.<\/p>\n

Question 5.
\nWhat happens \u03940<\/sup>m<\/sub> for weak electrolytes obtained by using Kohlrausch law if the migration of ions is increased to three
\nAnswer:
\n\u03940<\/sup>m<\/sub> for weak electrolytes obtained by using Kohlrausch law is independent of the migration of ions.<\/p>\n

\"KSEEB<\/p>\n

Question 6.
\nDefine molar conductivity of a solution.
\nAnswer:
\nThe molar conductivity of a solution at a given concentration is the conductance of volume \u2018V\u2019 of a solution containing 1 mole of _ the electrolytic kept between two electrodes with cross sectional area \u2018A\u2019 and distance of unit length.
\nOr,
\n\u0394m<\/sub> = \\(\\frac{\\Delta}{1}\\) K
\nNow,
\n1 = 1 and \u0394 = V (volume containing 1 mole of the electrolyte)
\n\u2234 \u0394m<\/sub> = KV<\/p>\n

Question 7.
\nWhat are the factors that affect the conductivity of an ionic (electrolytic) solution?
\nAnswer:
\nThe conductivity of an ionic (electrolytic) solution depends upon the following factors.<\/p>\n

    \n
  • Temperature<\/li>\n
  • Concentration of electrolyte<\/li>\n
  • Nature of the electrolyte added<\/li>\n
  • Nature of solvent and its viscosity<\/li>\n
  • Size of the ions produced and their solvation.<\/li>\n<\/ul>\n

    Question 8.
    \nThe electrolysis of a salt solution of a metal was carried out by passing a current of 4A for 45 minutes. This resulted in the deposition of 2.977g of the metal. If the atomic mass of the metal is 106.4 g mol-1<\/sup>, then calculate the charge present in the metal cation.
    \nAnswer:
    \nLet the charge on the metal cation be h i.e. the metal cation is M+<\/sup>
    \nAccordingly,
    \nMh+<\/sup> + 4e–<\/sup> \u2192 M
    \nTherefore, a current of h \u00d7 96500 coulomb will deposit 106.4 g of metal quantity of charge passed = 10800 coulombs
    \nNow, 10800 coulombs will deposit
    \n\"2nd
    \nHence, the charge present on the metal cation is + 4.<\/p>\n

    Question 9.
    \n(a) At infinite dilution, the ionic conductance of Ba2+<\/sup> and Cl is 121 and 76 ohm1 cm respectively. What will be the equivalent
    \nconductance of BaCl2<\/sub> (in ohm-1<\/sup> cm2<\/sup>) at infinite dilution?
    \n(b) What effect does concentration have on the molar conductivity of a strong electrolyte?
    \nAnswer:
    \n(a) The molar conductivity of barium chloride is given by the following equation:
    \n\u03940<\/sup>m<\/sub>(BaCl2<\/sub>) = \u03940<\/sup>Ba2+<\/sup> + 2 \u03940<\/sup>Cl-<\/sub>
    \n= 127 + 2 \u00d7 76 = 279\u2126-1<\/sup> cm2<\/sup> mol-1<\/sup>
    \n\u03940<\/sup> eq = \\(\\frac{279}{2}\\) \u2126-1<\/sup>cm2<\/sup>Eq-1<\/sup>
    \n= 139.5 \u2126-1<\/sup>cm2<\/sup> Eq-1<\/sup>
    \n[\u2234 Eq.wt. of BaCl2 = \\(\\frac{1}{2}\\) \u00d7 mol.wt]
    \nHence the equivalent conductance of BaCl2<\/sub> at infinite dilution is 139.5 \u2126-1cm2<\/sup> eq-1<\/sup><\/sup><\/sup><\/p>\n

    (b) The molar conductivity of a strong electrolyte decreases with the square root of concentration in linear fashion, as shown below.
    \n\"2nd<\/p>\n

    Question 10.
    \n(a) The standard reduction potentials of Fe3+<\/sup> | Fe2+<\/sup> and I–<\/sup>3<\/sup> | I–<\/sup>are 0.77 V and 0.54 V respectively for the reaction 2Fe3+<\/sup> + 3I–<\/sup> \u21cc 2Fe2+<\/sup> + I–<\/sup>3<\/sub>. Calculate the value of equilibrium constant.
    \n(b) How much charge is required 1 mole of Cu2+<\/sup> to Cu
    \nAnswer:
    \n(a) The cell may be represented as
    \n\"2nd
    \nHence, the equilibrium constant for the given cell is 6.025 \u00d7 107<\/sup> (b) When copper is reduced, the following reaction takes place Cu2+<\/sup> + 2e– <\/sup>\u2192 Cu
    \nHence, the quantity of charge required for the reduct ion of 1 mole of Cu2+<\/sup>= 2F
    \n= 2 \u00d7 96500 C = 193000C<\/p>\n

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

    You can Download Chapter 3 Electrochemistry Questions and Answers, Notes, 2nd PUC Chemistry\u00a0Question Bank with Answers Karnataka State Board Solutions help you to revise complete Syllabus and score more marks in your examinations. Karnataka 2nd PUC Chemistry Question Bank Chapter 3 Electrochemistry 2nd PUC Chemistry Electrochemistry NCERT Textbook Questions and Answers Question 1. Arrange the …<\/p>\n","protected":false},"author":5,"featured_media":0,"comment_status":"open","ping_status":"open","sticky":false,"template":"","format":"standard","meta":{"footnotes":""},"categories":[82],"tags":[],"jetpack_sharing_enabled":true,"jetpack_featured_media_url":"","_links":{"self":[{"href":"https:\/\/kseebsolutions.guru\/wp-json\/wp\/v2\/posts\/10172"}],"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=10172"}],"version-history":[{"count":0,"href":"https:\/\/kseebsolutions.guru\/wp-json\/wp\/v2\/posts\/10172\/revisions"}],"wp:attachment":[{"href":"https:\/\/kseebsolutions.guru\/wp-json\/wp\/v2\/media?parent=10172"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/kseebsolutions.guru\/wp-json\/wp\/v2\/categories?post=10172"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/kseebsolutions.guru\/wp-json\/wp\/v2\/tags?post=10172"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}