The reduction potential of the two half cell reactions (occuring in an electrochemical cell) are
`PbSO_(4)+ 2e^(-)rarrPb+SO_(4)^(2-) (E^(@)=-0.31V)`
`Ag^(+)(aq)+e^(-)rarrAg(s)(E^(@)=+0.80V)`
The fessible reaction will be

Options

A. `Pb^(2+)+2Agrarr2Ag^(+)+Pb`
B. `Pb^(2+)+H_(2)rarr2H^(+)+Pb`
C. `2H^(+)+2Agrarr2Ag^(+)+H_(2)`
D. `2Ag^(+)+PbrarrPb^(2+)+2Ag`

Answer

D

Explanation

For the cell reaction to be spontaneous `E_("cell")` should be positive. Therefore, two half cell reaction are
`Ag^(+)+e^(-)rarrAg (E^(@)=+0.80V)`
`PbrarrPb^(2+)+2e^(-)(E^(@)=+0.13V)`
and net reaction is
`2Ag^(+)+Pbrarr2Ag+Pb^(2+)`