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Question

The standard reduction potential at STP (25C and 1 atm) of the following half cells are found as below.
Pb2+aq+2ePb(s) E0=0.121VPbO2(s)+4H+(aq)+SO24(aq)+2ePbSO4(s)+2H2O E0=1.71VPb4+(aq)+2ePb2+(aq) E0=1.689V
For the cell PbSO4(s)+2ePb(s)+SO24(aq),
The E0 is missing but it is found that in an aqueous solution of H2SO4 having a pH=1.7, when PbSO4(s) is put, then at saturated condition
Solubility of PbSO4=5×105Mol/L
(Take 2.303RTF=0.06 and log2=0.3)

List-I contains questions and List-II contains their answers.
List-IList-IIWhat is the E0(in V) for the half-cell(I) PbSO4(s)+2ePb(s)+SO24(aq)?(P) -0.168What is the E0cell(in V) of a lead storage battery? It has(II) the cell reaction asPb(s)+PbO2(s)+2H2SO4(aq)2PbSO4(s)+2H2O(l)(Q) 0.7What is the E0(in V) for the half cell(III)PbO2(s)+4H+(aq)+4ePb(s)+2H2O(I)(R) 2.0Ksp is the equilibrium constant for the reaction(IV) PbO2(s)+4H+(aq)Pb4+(aq)+2H2O(l)(S) -0.068What is the value of Eo (in V)?(T) -0.31(U) 2.02
Which of the following options has the correct combination considering the reactions in List-I and their answers in List-II?

A
(II), (R) and (IV), (S)
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B
(II), (U) and (IV), (S)
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C
(II), (R) and (IV), (P)
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D
(II), (U) and (IV), (P)
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Solution

The correct option is D (II), (U) and (IV), (P)
For PbSO4(s)+2ePb(s)+SO24(aq) E01
aqueous solution of H2SO4 having pH=1.7[H+]=0.02M
So, [SO24]=0.01M,and solubility =5×105mol/L
So Ksp=5×105×0.01=5×107
So, at standard condition when [SO24]=1M then [Pb2+]=Ksp=5×107M
So, from Pb2+(aq)+2ePb(s) E0=0.121 V
E01=0.1210.062log1Ksp=0.1210.062log15×107=0.31 V
Now combining cells
PbO2(s)+4H+(aq)+SO24(aq)+2ePbSO4(s)+2H2O E0=1.71 VAnd PbSO4(s)+2ePb(s)+SO24(aq) E01=0.31V to getPb(s)+PbO2(s)+2H2SO4(aq)2PbSO4(s)+2H2O(I)E0Cell=1.71(0.31)=2.02 V
Now subtracting Pb2+(aq)+2ePb(s) E0=0.121 VFrom PbO2(s)+4H+(aq)+4ePb(s)+2H2O(I) E0=0.7 V
We will get
PbO2(s)+4H+aq+2ePb2+(aq)+2H2O(I) E0=1.689 V
To form a complete cell PbO2(s)+4H+(aq)Pb4+(aq)+2H2O(I)
E0Cell=1.5211.689=0.168 V
at equilibrium E0Cell=0.168 V

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