The e.m.f of a cell involving the cell reaction
`H_(2)+2Ag^(+) to 2H^(+) +2Ag ` is 0.80 V. The standard reduction potential of `Ag^(+)|Ag` electron is

Options

A. 0.20 V
B. 0.40 V
C. 0.60 V
D. 0.80 V

Answer

D

Explanation

`E_("cell")^(0)=underset(Ag)(E_(R)^(0))-underset(H_(2))(E_(R)^(0))`
`therefore 0.8= E_(Ag)^(0)-0 therefore E_(Ag)^(0) =0.8V`