Question

For the cell reaction, $C{u}^{2+}\left(C_1,aq\right)+Zn(s)\rightleftharpoons Z{n}^{2+}\left(C_2,aq\right)+Cu(s)$ of an electrochemical cell, the change in free energy $(\Delta G)$ at a given temperature is a function of

• A. $In(C_1)$
• B. $In(C_2/C_1)$
• C. $In(C_2)$
• D. $In(C_1+C_2)$

Answer 1

Answer: (B)

$\Delta ;G=-nF{E}^0$
For concentration cell,
From Nernst equation,
$E=E_{cell}^0-\frac{RT}{nF}\text{In}\frac{C_1}{C_2}$
$E_{\text{cell }}^0=0.00V$
$E=\frac{-RT}{nF}\text{In}\frac{C_2}{C_1}$
$E=\frac{RT}{nF}\text{In}\frac{C_2}{C_1}$
$C_2>C_1$
$E=\frac{RT}{nF}\text{In}\frac{C_2}{C_1}$
( $R,T,n$ and $F$ are constant)
therefore, $E^0$ is based upon $\ln \frac{C_2}{C_1}$
$\Delta ;G=-nF{E}^0$
$=-nF\times \frac{RT}{nF}\text{In}\frac{C_2}{C_1}$
$=-RT\text{In}\frac{C_2}{C_1}$
Hence, at constant temperature Gibbs free energy $\Delta ;G$ depends upon $\ln \frac{C_2}{C_1}$.