The correct option is C 0.315 V
From the cell representation,
Silver electrode will act as cathode and the M3+(aq)/M(s) couple will act as anode.
∴
The cell reaction will be,
M(s)+3Ag+(aq)→3Ag(s)+M3+(aq)
According to Nernst equation,
Ecell = E0cell − 0.05913log [M3+][Ag+]3
0.42 = E0cell − 0.05913log10(0.002)(0.01)3
0.42 = E0cell − 0.05913log 2×103
0.42 = E0cell − 0.05913×0.301−0.0591
0.42 = E0cell − 0.065
E0cell=0.42+0.065
E0cell=0.485 V
E0cell = Eocathode − Eoanode
E0anode =0.80 − 0.485
E0anode=0.315 V
Thus, the standard reduction potential of the half cell reaction is 0.315 V