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)

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 = 0.149 V

R = 8.314 J·K⁻¹mol⁻¹

T = 25 °C = 298.15 K

n = 10

F = 96 485 C/mol

The value of the equilibrium constant is

1. they vary with a higher number because boiling is way hotter than melting