Answer:
![\boxed{\text{57.9}}](https://tex.z-dn.net/?f=%5Cboxed%7B%5Ctext%7B57.9%7D%7D)
Explanation:
Step 1. Determine the cell potential
<u> </u><u>E°/V</u><u> </u>
2MnO₄⁻ + 16H⁺ + 10e⁻ → 2Mn²⁺+ 8H₂O 1.507
<u>10Cl⁻ ⟶ 5Cl₂ + 10e⁻ </u> <u>-1.358 27
</u>
2MnO₄⁻ + 10Cl⁻ + 16H⁺⟶ 2Mn²⁺ + 5Cl₂ + 8H₂O 0.149
Step 2. Calculate K
The formula relating K and E is
![E = \dfrac{RT}{nF} \ln K](https://tex.z-dn.net/?f=E%20%3D%20%5Cdfrac%7BRT%7D%7BnF%7D%20%5Cln%20K)
E = 0.149 V
R = 8.314 J·K⁻¹mol⁻¹
T = 25 °C = 298.15 K
n = 10
F = 96 485 C/mol
![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}](https://tex.z-dn.net/?f=0.149%20%3D%20%5Cdfrac%7B8.314%20%5Ctimes%20298.15%7D%7B10%5Ctimes%2096%20485%7D%20%5Cln%20K%5C%5C%5C%5C0.149%20%3D%200.002%20570%20%5Cln%20K%5C%5C%5C%5C%5Cln%20K%20%3D%20%5Cdfrac%7B0.149%7D%7B0.002570%7D%20%3D%20%5Ctextbf%7B57.9%7D)
The value of the equilibrium constant is ![\boxed{\textbf{57.9}}](https://tex.z-dn.net/?f=%5Cboxed%7B%5Ctextbf%7B57.9%7D%7D)