The equilibrium constant of an electrochemical cell's redox reaction can be calculated using the Nernst equation and the relationship between standard cell potential and free energy. This example problem shows how to find the equilibrium constant of a cell's redox reaction.

## Problem

The following two half-reactions are used to form an electrochemical cell:

Oxidation:

SO_{2}(g) + 2 H_{2}0(ℓ) → SO_{4}^{-}(aq) + 4 H^{+}(aq) + 2 e^{-} E°_{ox} = -0.20 V

Reduction:

Cr_{2}O_{7}^{2-}(aq) + 14 H^{+}(aq) + 6 e^{-} → 2 Cr^{3+}(aq) + 7 H_{2}O(ℓ) E°_{red} = +1.33 V

What is the equilibrium constant of the combined cell reaction at 25 C?

Solution

**Step 1: Combine and balance the two half-reactions.**

The oxidation half-reaction produces 2 electrons and the reduction half-reaction needs 6 electrons. To balance the charge, the oxidation reaction must be multiplied by a factor of 3.

3 SO_{2}(g) + 6 H_{2}0(ℓ) → 3 SO_{4}^{-}(aq) + 12 H^{+}(aq) + 6 e^{-}

+ Cr_{2}O_{7}^{2-}(aq) + 14 H^{+}(aq) + 6 e^{-} → 2 Cr^{3+}(aq) + 7 H_{2}O(ℓ)

3 SO_{2}(g) + Cr_{2}O_{7}^{2-}(aq) + 2 H^{+}(aq) → 3 SO_{4}^{-}(aq) + 2 Cr^{3+}(aq) + H_{2}O(ℓ)

By balancing the equation, we now know the total number of electrons exchanged in the reaction. This reaction exchanged six electrons.

**Step 2: Calculate the cell potential.**

This electrochemical cell EMF example problem shows how to calculate cell potential of a cell from standard reduction potentials.**

E°_{cell} = E°_{ox} + E°_{red}

E°_{cell} = -0.20 V + 1.33 V

E°_{cell} = +1.13 V**Step 3: Find the equilibrium constant, K.**

When a reaction is at equilibrium, the change in free energy is equal to zero.

The change in free energy of an electrochemical cell is related to the cell potential of the equation:

ΔG = -nFE_{cell}

where

ΔG is the free energy of the reaction

n is the number of moles of electrons exchanged in the reaction

F is Faraday's constant (96484.56 C/mol)

E is the cell potential.

The** **cell potential and free energy example shows how to calculate free energy of a redox reaction.

If ΔG = 0:, solve for E_{cell}

0 = -nFE_{cell}

E_{cell} = 0 V

This means, at equilibrium, the potential of the cell is zero. The reaction progresses forward and backward at the same rate, meaning there is no net electron flow. With no electron flow, there is no current and the potential is equal to zero.

Now there is enough information known to use the Nernst equation to find the equilibrium constant.

The Nernst equation is:

E_{cell} = E°_{cell} - (RT/nF) x log_{10}Q

where

E_{cell} is the cell potential

E°_{cell} refers to standard cell potential

R is the gas constant (8.3145 J/mol·K)

T is the absolute temperature

n is the number of moles of electrons transferred by the cell's reaction

F is Faraday's constant (96484.56 C/mol)

Q is the reaction quotient

**The Nernst equation example problem shows how to use the Nernst equation to calculate cell potential of a non-standard cell.**

At equilibrium, the reaction quotient Q is the equilibrium constant, K. This makes the equation:

E_{cell} = E°_{cell} - (RT/nF) x log_{10}K

From above, we know the following:

E_{cell} = 0 V

E°_{cell} = +1.13 V

R = 8.3145 J/mol·K

T = 25 °C = 298.15 K

F = 96484.56 C/mol

n = 6 (six electrons are transferred in the reaction)

Solve for K:

0 = 1.13 V - [(8.3145 J/mol·K x 298.15 K)/(6 x 96484.56 C/mol)]log_{10}K

-1.13 V = - (0.004 V)log_{10}K

log_{10}K = 282.5

K = 10^{282.5}

K = 10^{282.5} = 10^{0.5} x 10^{282}

K = 3.16 x 10^{282}**Answer:**

The equilibrium constant of the cell's redox reaction is 3.16 x 10^{282}.