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.
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)
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) -0.13
Mg2+(aq) + 2e− → Mg(s) -2.37
Br2(l) + 2e− → 2 Br−(aq) 1.09
Cl2(g) + 2e−→ 2 Cl−(aq) 1.36
MnO2(s) + 4 H+(aq) + 2e− → Mn2+(aq) + 2 H2O(l) 1.21
Cu2+(aq) + 2e− → Cu(s) 0.16
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