Problem: Calculate the equilibrium constant for each of the reactions at 25°C.a. Pb2+(aq) + Mg(s) → Pb(s) + Mg2+(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

FREE Expert Solution

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=RTnFln 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 3Get the overall reaction by balancing the number of electrons transferred then adding the reduction half-reaction and oxidation half-reaction.
Step 4. Calculate E
Step 5Calculate K using the Nernst Equation.

Overall reaction: Pb2+(aq) + Mg(s) → Pb(s) + Mg2+(aq)

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Problem Details

Calculate the equilibrium constant for each of the reactions at 25°C.

a. Pb2+(aq) + Mg(s) → Pb(s) + Mg2+(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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