# Problem: A disproportionation reaction involves a substance that acts as both an oxidizing and a reducing agent, producing higher and lower oxidation states of the same element in the products. Which of the following disproportionation reactions are spontaneous under standard conditions? Calculate ΔG° and K at 25°C for those reactions that are spontaneous under standard conditions.a. 2Cu+(aq) → Cu 2+(aq) + Cu(s)b. 3Fe2+(aq) → 2Fe 3+(aq) + Fe(s)c. HClO2(aq) → ClO3 -(aq) + HClO(aq) (unbalanced)Use the half-reactions:ClO3- + 3H +  + 2e- → HClO2 + H 2O         ε° = 1.21 VHClO2 + 2H + + 2e- → HClO + H 2O         ε° = 1.65 V

###### FREE Expert Solution

ΔG° can be calculated from the cell potential using the following equation:

$\overline{){\mathbf{∆}}{\mathbf{G}}{\mathbf{=}}{\mathbf{-}}{\mathbf{nFE}}{{\mathbf{°}}}_{{\mathbf{cell}}}}$

ΔG° = Gibbs Free Energy, J
n = # of e- transferred
F = Faraday’s constant = 96485 C/(mol e-)
cell = standard cell potential, V

Recall: 1 V = 1 J/C

a.) 2Cu+(aq) → Cu 2+(aq) + Cu(s)

The reaction taking place at cathode is:

Cu+(aq) + e- → Cu(s) E° = 0.52 V

The reaction taking place at anode is:

Cu+(aq) → Cu2+(aq) + e- E° = -0.16 V

86% (357 ratings) ###### Problem Details

A disproportionation reaction involves a substance that acts as both an oxidizing and a reducing agent, producing higher and lower oxidation states of the same element in the products. Which of the following disproportionation reactions are spontaneous under standard conditions? Calculate ΔG° and K at 25°C for those reactions that are spontaneous under standard conditions.

a. 2Cu+(aq) → Cu 2+(aq) + Cu(s)

b. 3Fe2+(aq) → 2Fe 3+(aq) + Fe(s)

c. HClO2(aq) → ClO3 -(aq) + HClO(aq) (unbalanced)

Use the half-reactions:

ClO3- + 3H +  + 2e→ HClO2 + H 2O         ε° = 1.21 V

HClO2 + 2H + + 2e→ HClO + H 2O         ε° = 1.65 V