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 be

Options

A. `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

A

Explanation

For EMF to be positive , oxidation should occur on lead electrode .