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 + SO_(4)^(2-) + 2 Ag^(+) (aq) to 2 Ag (s) + PbSO_(4)`B. `PbSO_(4) + 2 Ag^(+) (aq) to Pb + SO_(4)^(2-) + 2 Ag(s)`C. `Pb + SO_(4)^(2-) + Ag (s) to Ag^(+) (aq) + PbSO_(4)`D. `PbSO_(4) + Ag (s) to Ag^(+) (aq) + Pb + SO_(4)^(2-)`

The reduction potential of the two half cell reaction (occuring in an

1.	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 + SO_(4)^(2-) + 2 Ag^(+) (aq) to 2 Ag (s) + PbSO_(4)`B. `PbSO_(4) + 2 Ag^(+) (aq) to Pb + SO_(4)^(2-) + 2 Ag(s)`C. `Pb + SO_(4)^(2-) + Ag (s) to Ag^(+) (aq) + PbSO_(4)`D. `PbSO_(4) + Ag (s) to Ag^(+) (aq) + Pb + SO_(4)^(2-)`
Answer» Correct Answer - A For EMF to be positive , oxidation should occur on lead electrode .

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