For the disproportionation of copper 2Cu+ Cu2+ + Cu, E is (Given E for Cu2+/Cu is 0.34 V and E for Cu2+/Cu+ is 0.15 V.)
- 0.49 V
- -0.19 V
- 0.38 V
- -0.38 V
Answer
(3) 0.38 V
Solution
Electrode potentials are not additive, so combine the half-reactions through G = -nFE :
Cu2+ + 2e- Cu; G 1 = -2F(0.34)
Cu2+ + e- Cu+; G 2 = -1F(0.15)
Subtracting: Cu+ + e- Cu; G 3 = G 1 - G 2
⇒ -1F E Cu+/Cu = -0.68F + 0.15F
⇒ E Cu+/Cu = 0.53 V
In 2Cu+ Cu2+ + Cu, Cu+ is reduced to Cu (cathode) and oxidised to Cu2+ (anode).
∴ E cell = E Cu+/Cu - E Cu2+/Cu+
= 0.53 - 0.15
= 0.38 V
Answer: (3)