Answer:
-0.303 V
Step-by-step explanation:
From Nernst equation
[In3+] =0.10 M
[Zn2+]=Zn in 0.0050 M
n= 6
E= 0.47V
Thus
E= E°- 0.0592/n log [Red]/[Ox]
0.47= E° - 0.0592/6 log 0.10/0.0050
0.47= E° - 0.0128
E°= 0.47- 0.0128
E°= 0.457 V
But
E°cell= E° cathode- E°anode
E°cathode= E°cell + E°anode
Hence
E°cathode= 0.457 + ( -0.76)
E°cathode= -0.303 V