Cd2+(aq) + 2e- -----> Cd(s) E° = -0.403 V
Sn2+(aq) + 2e- -------> Sn(s) E° = - 0.136 V
Sn2+ is more positive so it will reduced at cathode and Cd2+ oxidized at anode.
E°cell = E°cathode - E°anode
E°cell = -0.136 V - (-0.403V)
E°cell = +0.267V
~ +0.27V
Hence, the correct option is (d) 0.27
A voltaic cell is based on the following two half-reactions: Cd2 (ag) +2e-> Cd (s) Sn2(aq)+...
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