For the reduction of silver ions with copper metal, the standard cell potential was found to be $+0.46 ext{

Question

For the reduction of silver ions with copper metal, the standard cell potential was found to be $+0.46 	ext{ V}$ at $25^\circ	ext{C}$. The value of standard Gibbs energy, $\Delta G^\circ$ will be: ($F = 96500 	ext{ C mol}^{-1}$)

Accepted Answer

1.  **Identify the Cell Reaction:** The reduction of silver ions ($Ag^+$) by copper metal ($Cu$) is represented by the balanced chemical equation:
    $$Cu(s) + 2Ag^+(aq) ightarrow Cu^{2+}(aq) + 2Ag(s)$$
    In this reaction, Copper loses 2 electrons and two Silver ions gain 1 electron each. Therefore, the number of electrons transferred, $n = 2$ .
2.  **Formula:** The relationship between standard Gibbs energy change ($\Delta_r G^\circ$) and standard cell potential ($E^\circ_{cell}$) is given by:
    $$\Delta_r G^\circ = -nFE^\circ_{cell}$$
    Where $F$ is the Faraday constant .
3.  **Calculation:**
    *   $n = 2$
    *   $F = 96500 	ext{ C mol}^{-1}$
    *   $E^\circ_{cell} = +0.46 	ext{ V}$
    $$\Delta_r G^\circ = -2 	imes 96500 	ext{ C mol}^{-1} 	imes 0.46 	ext{ V}$$
    $$\Delta_r G^\circ = -193000 	imes 0.46 	ext{ J mol}^{-1}$$
    $$\Delta_r G^\circ = -88780 	ext{ J mol}^{-1}$$
    Converting to kJ:
    $$\Delta_r G^\circ \approx -88.8 	ext{ kJ}$$
    Rounding to the nearest option gives $-89.0 	ext{ kJ}$.

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