Question

Cell Potential at Equilibrium For a single galvanic cell based on the (unbalanced) reaction: Ag+ (aq) + Zn(s) Zn2+ (aq) + Ag(s) What is the cell potential when the cell reaches equilibrium?

Please show all work step by step and final answer.

Answer 1

The given reaction is

$Ag^{+}_{(aq)} + Zn_{(s)} \rightleftharpoons Zn^{2+}_{(aq)} + Ag_{(s)}$

Hence, using Nernst's equation, we can write the cell potential of the reaction as

$E_{cell} = E^\circ_{(cell)} - \frac{RT}{nF} \ln Q$

Where Q is the reaction quotient.

Now, as the reaction reaches equilibrium, the reaction quotient Q become equal to equilibrium constant K.

$Q= K \\ \Rightarrow \ln Q = \ln K$

We know that the standard free energy change of the cell reaction $\Delta G^\circ = -nFE^\circ _{cell}$

The standard free energy change is also related to the equilibrium constant K as

$\Delta G^\circ = -RT \ln K$

Hence,

$-RT\ln K = -nFE_{cell}^\circ \\ \\ \Rightarrow \frac{RT}{nF}\ln K = E_{cell}^\circ$

Now, substituting the value above in the Nernst's equation

$E_{cell} = E^\circ_{(cell)} - \frac{RT}{nF} \ln Q \\ \\ \Rightarrow E_{Cell} = E^\circ _{cell} - \frac{RT}{nF} \ln K \\ \\ \Rightarrow E_{Cell} = E^\circ _{cell} - E^\circ _{cell} = 0$

Hence, the cell potential of the cell at equilibrium is 0 volts.