Question

M⁺ is not stable and undergoes disproportionation to form M and M²⁺. Calculate E° for M⁺ disproportionation E°$_{M^{2+}/M}$ = +0.153V, E°$_{M^+/M}$ = 0.53V

• A. +0.683V
• B. -0.367 V
• C. 0.754 V
• D. +0.3415V

Answer 1

Answer: (C)

0.754 V

Answer 2

The disproportionation reaction of M⁺ is given by:

2M⁺ $\rightarrow$ M + M²⁺

This reaction can be split into two half-reactions:

1. Reduction: M⁺ + e⁻ $\rightarrow$ M
2. Oxidation: M⁺ $\rightarrow$ M²⁺ + e⁻

We are given the standard reduction potentials:

E°$_{M^{2+}/M}$ = +0.153 V for M²⁺ + 2e⁻ $\rightarrow$ M E°$_{M^+/M}$ = 0.53 V for M⁺ + e⁻ $\rightarrow$ M

Let's denote the potentials as:

Reaction (1): M²⁺ + 2e⁻ $\rightarrow$ M, E°$_1 = 0.153$ V, n$_1 = 2$ Reaction (2): M⁺ + e⁻ $\rightarrow$ M, E°$_2 = 0.53$ V, n$_2 = 1$

The reduction half-reaction for the disproportionation is M⁺ + e⁻ $\rightarrow$ M, which is Reaction (2). Its potential is E°$_{red}$ = E°$_2 = 0.53$ V.

The oxidation half-reaction for the disproportionation is M⁺ $\rightarrow$ M²⁺ + e⁻. Let's call the standard potential for this oxidation E°$_{ox}$. This is the reverse of the reduction half-reaction M²⁺ + e⁻ $\rightarrow$ M⁺. Let's find the standard reduction potential for M²⁺ + e⁻ $\rightarrow$ M⁺, denoted as E°$_{M^{2+}/M^+}$.

We can relate the standard potentials using the relationship between standard Gibbs free energy ($\Delta G°$) and standard potential (E°): $\Delta G° = -nFE°$. The $\Delta G°$ values are additive for sequential reactions. Consider the reactions:

(A) M²⁺ + e⁻ $\rightarrow$ M⁺, $\Delta G°_A = -1F E°_{M^{2+}/M^+}$ (B) M⁺ + e⁻ $\rightarrow$ M, $\Delta G°_B = -1F E°_{M^+/M} = -1F(0.53)$

Adding (A) and (B) gives:

M²⁺ + 2e⁻ $\rightarrow$ M, $\Delta G°_{(A)+(B)} = \Delta G°_A + \Delta G°_B$. This overall reaction is Reaction (1). So, $\Delta G°_{(A)+(B)} = \Delta G°_1$. $\Delta G°_1 = -2F E°_1 = -2F(0.153)$.

Thus, $\Delta G°_1 = \Delta G°_A + \Delta G°_B$ $-2F(0.153) = -1F E°_{M^{2+}/M^+} + -1F(0.53)$ Dividing by -F: $2(0.153) = E°_{M^{2+}/M^+} + 0.53$ $0.306 = E°_{M^{2+}/M^+} + 0.53$ $E°_{M^{2+}/M^+} = 0.306 - 0.53 = -0.224$ V.

Now we have the standard reduction potential for M²⁺ + e⁻ $\rightarrow$ M⁺. The oxidation half-reaction in the disproportionation is M⁺ $\rightarrow$ M²⁺ + e⁻. This is the reverse of M²⁺ + e⁻ $\rightarrow$ M⁺. The standard potential for the oxidation half-reaction is E°$_{ox}$ = -E°$_{M^{2+}/M^+} = -(-0.224$ V) = +0.224 V.

The standard potential for the overall disproportionation reaction 2M⁺ $\rightarrow$ M + M²⁺ is the sum of the standard potentials of the reduction and oxidation half-reactions:

E°$_{disproportionation}$ = E°$_{red}$ + E°$_{ox}$ E°$_{disproportionation}$ = E°$_{M^+/M}$ + E°$_{M^+/M^{2+}}$ E°$_{disproportionation}$ = 0.53 V + 0.224 V = 0.754 V.

Alternatively, using E°$_{cell}$ = E°$_{cathode}$ - E°$_{anode}$:

In the disproportionation 2M⁺ $\rightarrow$ M + M²⁺, one M⁺ is reduced to M (cathode) and another M⁺ is oxidized to M²⁺ (anode). Reduction half-reaction: M⁺ + e⁻ $\rightarrow$ M, E°$_{cathode}$ = E°$_{M^+/M}$ = 0.53 V. Oxidation half-reaction: M⁺ $\rightarrow$ M²⁺ + e⁻. The potential for this oxidation is the negative of the reduction potential for M²⁺ + e⁻ $\rightarrow$ M⁺. E°$_{anode}$ = E°$_{M^{2+}/M^+}$ = -0.224 V (calculated above). E°$_{disproportionation}$ = E°$_{cathode}$ - E°$_{anode}$ = E°$_{M^+/M}$ - E°$_{M^{2+}/M^+}$ E°$_{disproportionation}$ = 0.53 V - (-0.224 V) = 0.53 + 0.224 = 0.754 V.

The standard potential for M⁺ disproportionation is +0.754 V. A positive E° for the disproportionation reaction indicates that the reaction is spontaneous under standard conditions, which is consistent with the statement that M⁺ is not stable and undergoes disproportionation.