Discussion Forum

The electrode potentials for Cu²⁺ (aq) + e^- →  Cu⁺ (aq) and Cu⁺ (aq) + e^- → Cu(s) are + 0. 15 V and + 0.50 V, respectively The value of E°_Cu²⁺/Cu will be: 

Answer:

Explanation:

Cu²⁺  + 1e^- →  Cu⁺    : ΔG^°₁ = -n₁E°₁F

Cu⁺  +  1e^- → Cu       :  ΔG₂^° = -n₂E°₂F

$Cu^{2+} + 2e^{-}\rightarrow Cu$ : $ΔG^{\circ} = ΔG_1^\circ + ΔG_2^\circ$

$-nE^{\circ}F = -1n_{1}$

$E_1^\circ F + (-1)n_{2}E_2^\circ F$

$ -nE^{\circ}F= -F(n_{1} E_1^\circ+n_{2}E_2^\circ F)$

$E^{\circ} =\frac{n_{1} E_1^\circ + n_{2}E_2^\circ}{n} $

= $\frac{0.15\times1 + 0.50\times1}{2}$ = 0.325V