Answer:
[tex]\boxed{\text{57.9}}[/tex]
Explanation:
Step 1. Determine the cell potential
E°/V
2MnO₄⁻ + 16H⁺ + 10e⁻ → 2Mn²⁺+ 8H₂O 1.507
10Cl⁻ ⟶ 5Cl₂ + 10e⁻ -1.358 27
2MnO₄⁻ + 10Cl⁻ + 16H⁺⟶ 2Mn²⁺ + 5Cl₂ + 8H₂O 0.149
Step 2. Calculate K
The formula relating K and E is
[tex]E = \dfrac{RT}{nF} \ln K[/tex]
E = 0.149 V
R = 8.314 J·K⁻¹mol⁻¹
T = 25 °C = 298.15 K
n = 10
F = 96 485 C/mol
[tex]0.149 = \dfrac{8.314 \times 298.15}{10\times 96 485} \ln K\\\\0.149 = 0.002 570 \ln K\\\\\ln K = \dfrac{0.149}{0.002570} = \textbf{57.9}[/tex]
The value of the equilibrium constant is [tex]\boxed{\textbf{57.9}}[/tex]
