Problem: Calculate the equilibrium constant for each of the reactions at 25°C.c. MnO2(s) + 4 H+(aq) + Cu(s) → Mn2+(aq) + 2 H2O(l) + Cu2+(aq)Standard Electrode Potentials at 25 ∘C:Reduction Half-Reaction                                                E°(V)Pb2+(aq) + 2 e− → Pb(s)                                                  -0.13Mg2+(aq) + 2e− → Mg(s)                                                 -2.37Br2(l) + 2e− → 2 Br−(aq)                                                   1.09Cl2(g) + 2e−→ 2 Cl−(aq)                                                   1.36MnO2(s) + 4 H+(aq) + 2e− → Mn2+(aq) + 2 H2O(l)            1.21Cu2+(aq) + 2e− → Cu(s)                                                     0.16

We are asked to calculate the equilibrium constant for the given reaction at 25°C. We will use the Nernst Equation to calculate the equilibrium constant. The Nernst Equation relates the concentrations of compounds and cell potential.

$\boxed{\mathbf{E}{\mathbf{°}}_{\mathbf{cell}}\mathbf{=}\frac{\mathbf{RT}}{\mathbf{nF}}\mathbf{ln}\mathbf{ }\mathbf{K}}$

E°_cell = cell potential, V
R = gas constant = 8.314 J/(mol·K)
T = temperature, K
n = mole e^- transferred
F = Faraday’s constant, 96485 C/mol e^- 
K = equilibrium constant

We will use the following steps:

Step 1. Write the two half-cell reactions and determine the half-cell potentials
Step 2. Identify the reduction half-reaction (cathode) and the anode half-reaction (anode)
Step 3. Get the overall reaction by balancing the number of electrons transferred then adding the reduction half-reaction and oxidation half-reaction.
Step 4. Calculate E°_cell.
Step 5. Calculate K using the Nernst Equation.

Overall reaction: MnO₂(s) + 4 H⁺(aq) + Cu(s) → Mn²⁺(aq) + 2 H₂O(l) + Cu²⁺(aq)