Q.Calculate the standard Gibbs energy change, $\Delta G^{\circ}$, for the Daniell cell reaction $\text{Zn}(s) + \text{Cu}^{2+}(aq) \to \text{Zn}^{2+}(aq) + \text{Cu}(s)$, given $E^{\circ}_{cell} = 1.10\ \text{V}$ and $n = 2$. (Take $F = 96500\ \text{C mol}^{-1}$.)

[!TLDR] $\Delta G^{\circ} = -nFE^{\circ}_{cell} = -2\times96500\times1.10$. [!ANSWER] $\Delta G^{\circ} = -212{,}300 …

The standard Gibbs energy change of a cell reaction is related to its standard EMF by $\Delta G^{\circ} = -nFE^{\circ}_{cell}$, where $n$ is the number of moles of electrons transferred per mole of reaction as written, and $F = 96500\ \text{C mol}^{-1}$ is the Faraday constant. For the Daniell cell, $n=2$ and $E^{\circ}_{cell}=1.10\ \text{V}$, so $\Delta G^{\circ} = -(2)(96500\ \text{C mol}^{-1})(1.10\ \text{V}) = -212{,}300\ \text{J mol}^{-1}$ (since $1\ \text{V}\times1\ \text{C} = 1\ \text{J}$), i.e. $-212.3\ \text{kJ mol}^{-1}$. The large negative …

Substitute $n$, $F$, and $E^{\circ}_{cell}$ directly into $\Delta G^{\circ}=-nFE^{\circ}_{cell …

Forgetting to convert joules to kilojoules, or using the wrong value of $n$ (must match the number of electrons in the specif …