For an electrochemical cell represented as , the standard electrode reduction potentials are and . F — Electrochemistry Chemistry Question

💡 Solution & Explanation

The cell reaction as written is $M \rightarrow M^+ + e^-$ (anode) and $X + e^- \rightarrow X^-$ (cathode). $E^\circ_{cell} = E^\circ_{cathode} - E^\circ_{anode} = E^\circ_{X/X^-} - E^\circ_{M^+/M} = 0.33 - 0.44 = -0.11\text{ V}$. Since $E^\circ_{cell} < 0$, the forward reaction is non-spontaneous. The reverse reaction ($M^+ + X^- \rightarrow M + X$) has $E^\circ_{cell} = +0.11\text{ V}$ and is spontaneous. Because $M^+$ is reduced to $M$, $X^-$ must be the stronger reducing agent.