The cell reaction of an electrochemical cell is $Cu^{2+}(C_1) + Zn \rightarrow Cu + Zn^{2+}(C_2)$. The change

Question

The cell reaction of an electrochemical cell is $Cu^{2+}(C_1) + Zn \rightarrow Cu + Zn^{2+}(C_2)$. The change in free energy will be the function of:

Accepted Answer

For the given cell reaction: $Zn(s) + Cu^{2+}(aq, C_1) \rightarrow Zn^{2+}(aq, C_2) + Cu(s)$
The change in Gibbs free energy ($\Delta G$) is related to the reaction quotient ($Q$) by the equation:
$\Delta G = \Delta G^{\circ} + RT \ln Q$
Where $Q$ is the ratio of concentration of products to reactants: $Q = \frac{[Zn^{2+}]}{[Cu^{2+}]} = \frac{C_2}{C_1}$
Substituting $Q$ in the equation, we get: $\Delta G = \Delta G^{\circ} + RT \ln\left(\frac{C_2}{C_1}\right)$.
Therefore, the change in free energy is a function of $\ln(C_2/C_1)$.

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