Question

From standard reduction potentials, calculate the equilibrium constant at 25 ∘c for the reaction 2mno−4(aq)+10cl−(aq)+16h+(aq)→2mn2+(aq)+5cl2(g)+8h2o(l)

Answer 1

Answer:

$\boxed{\text{57.9}}$

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

$E = \dfrac{RT}{nF} \ln K$

E = 0.149 V

R = 8.314 J·K⁻¹mol⁻¹

T = 25 °C = 298.15 K

n = 10

F = 96 485 C/mol

$0.149 = \dfrac{8.314 \times 298.15}{10\times 96 485} \ln K\\\\0.149 = 0.002 570 \ln K\\\\\ln K = \dfrac{0.149}{0.002570} = \textbf{57.9}$

The value of the equilibrium constant is $\boxed{\textbf{57.9}}$