[Four-digit Integer] Calculate potential (in mV) of the cell: Cu — Electrochemistry Chemistry Question

💡 Solution & Explanation

\textbf{Setup:} The cell involves Cu and Mn²⁺/Mn. \textbf{Given:} $E^\circ(\text{Mn}^{2+}/\text{Mn}) = -1.185$ V; $E^\circ(\text{Cu}^{2+}/\text{Cu}) = +0.337$ V. \textbf{Step 1: Identify electrodes.} Cu has higher $E^\circ$ (+0.337 V) → cathode (reduction). Mn has lower $E^\circ$ (−1.185 V) → anode (oxidation). Net reaction: $\text{Mn} + \text{Cu}^{2+} \rightarrow \text{Mn}^{2+} + \text{Cu}$ \textbf{Step 2: Standard cell potential.} \[ E^\circ_{\text{cell}} = 0.337 - (-1.185) = 1.522\ \text{V} \] \textbf{Step 3: Apply Nernst equation.} For $n = 2$: \[ E_{\text{cell}} = E^\circ_{\text{cell}} - \frac{0.059}{2}\log\frac{[\text{Mn}^{2+}]}{[\text{Cu}^{2+}]} \] Substituting the specific concentrations from the question data gives the cell potential in mV. \textit{Note: The specific ion concentrations required for the numerical calculation were not recorded in the available question data. The result is obtained by substituting the given concentrations into the Nernst equation and converting to mV.}