We are asked to calculate for the **equilibrium constant for the reaction**.

Recall that the** Nernst Equation** relates the concentrations of compounds and cell potential.

$\overline{){\mathbf{E}}{{\mathbf{\xb0}}}_{{\mathbf{cell}}}{\mathbf{=}}\frac{\mathbf{RT}}{\mathbf{nF}}{\mathbf{ln}}{\mathbf{}}{\mathbf{K}}}$

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're going to calculate for the equilibrium constant using the following steps:

*Step 1:** **Identify the anode and the cathode in the reaction.*** Step 2: **Calculate the cell potential of the reaction.

What is the equilibrium constant for the reaction taking place at room temperature (T = 25°C) in the battery Zn(s) I Zn^{2+}(aq) II Ce^{4+}(aq) I Ce^{3+}(aq) ? Assume that the number of electrons transferred in the reaction is n = 2.

Zn^{2+} + 2 e^{-} → Zn E°_{red} = -0.76 V

Ce^{4+} + e^{-} → Ce^{3+} E_{°red} = +1.61 V

1. 1:33 x 10^{80}

2. 2.37

3. 6:52 x 10^{79}

4. 1:84 x 10^{2}

5. 1:44 x 10^{2}

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