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InterviewSolution
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The reduction potential of the two half cell reaction (occuring in an electrochemical cell) are `PbSO_(4)(s)+2e^(-)rarrPb(s)+SO_(4)^(2-)(aq)(E^(@)=-0.31V)` `Ag^(+)(aq)+e^(-)rarrAg(s)(E^(@)=0.80V)` The fessible reaction will beA. `Pb(s)+SO_(4)^(2-)(aq)+2Ag^(+)+(aq)rarr2Ag(s)+PbSO_(4)(s)`B. `PbSO_(4)(s)+2Ag^(+)(aq)rarrPb(s)+SO_(4)^(2-)(aq)+2Ag(s)`C. `Pb(s)+SO_(4)^(2-)(s)+Ag(s)rarrAg^(+)(aq)+PbSO_(4)(s)`D. `PbSO_(4)(s)+Ag(s)rarrAg^(+)(aq)+Pb(s)+SO_(4)^(2-)(aq)` |
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Answer» Correct Answer - A Higher the reduction potential easier to reduce, Therefore two h alf cell reactions are: `Pb(s)+SO_(4)^(2-)(aq)rarrPbSO_(4)(s)+2e^(-)` `Ag^(+)(aq)+e^(-)rarrAg(s)xx2` `Pb (s)+SO_(4)^(2-) (aq)+2Ag^(+)(aq)rarrPbSO_(4)(S)+Ag(s)` |
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