Q.A cell is constructed for the reaction $\text{Cu}(s) + \text{Zn}^{2+}(aq) \to \text{Cu}^{2+}(aq) + \text{Zn}(s)$ (the reverse of the Daniell reaction), for which $E^{\circ}_{cell} = -1.10\ \text{V}$. State, with reasons, whether this reaction is spontaneous as written, and what sign $\Delta G^{\circ}$ must have.

[!TLDR] A negative $E^{\circ}_{cell}$ means the reaction is non-spontaneous as written, so $\Delta G^{\circ}$ must be positive. [!AN …

The sign of $E^{\circ}_{cell}$ and the sign of $\Delta G^{\circ}$ are linked by $\Delta G^{\circ} = -nFE^{\circ}_{cell}$, and they are always opposite in sign (since $n$ and $F$ are both positive). Here $E^{\circ}_{cell} = -1.10\ \text{V}$ is negative, so $\Delta G^{\circ} = -nF\times(-1.10) = +nF(1.10)$ is positive. A positive $\Delta G^{\circ}$ means the forward reaction, $\text{Cu}(s) + \text{Zn}^{2+}(aq) \to \text{Cu}^{2+}(aq) + \text{Zn}(s)$, is non-spontaneous under standard conditions — it will not proceed on its own. This makes physical sense: this reaction is simply …

Use the sign rule directly: $E^{\circ}_{cell}<0 \Leftrightarrow \Delta G^{\circ}>0 \Leftrightarrow$ non-sponta …

Believing a 'cell EMF' of any sign still implies a working, spontaneous cell — a negative $E^{\circ}_{cell}$ means that reaction direction simply will not o …