Calculate the cell potential for the following reaction that takes place in an electrochemical cell at 25 °C: Cu(s) | Cu2+(aq, 3.2×10^(-3) mol L^-1) || Cu2+(aq, 4.48 mol L^-1) | Cu(s). Given E°(Cu2+/Cu) = +0.34 V, express your answer to two significant figures.

• A. +0.41 V
• B. +0.34 V
• C. +0.27 V
• D. +0.20 V

Answer: (C)

Verified step by step guidance

1. Identify the components of the electrochemical cell: The cell consists of two half-cells, both involving the Cu/Cu²⁺ redox couple. The concentrations of Cu²⁺ in the two half-cells are 3.2×10⁻³ mol L⁻¹ and 4.48 mol L⁻¹, respectively.
2. Use the Nernst equation to calculate the cell potential under non-standard conditions. The Nernst equation is: $E_{\mathrm{cell}}=E_{\mathrm{cell}}°-\frac{\mathrm{RT}}{\mathrm{nF}}\cdot \ln \left(Q\right)$, where $Q$ is the reaction quotient.
3. Calculate the reaction quotient $Q$ for the cell reaction. Since the reaction involves the transfer of electrons between Cu²⁺ ions, $Q$ is given by the ratio of the concentrations of Cu²⁺ in the two half-cells: $\frac{{\mathrm{Cu}}_2(\mathrm{aq},3.2\times {10}^{-})}{{\mathrm{Cu}}_2(\mathrm{aq},4.48)}$.
4. Substitute the values into the Nernst equation. Use the given standard cell potential $E_{\mathrm{cell}}°=0.34V$, the temperature $25°C$ (which is $298K$), and the number of electrons transferred $n=2$. The gas constant $R$ is $8.314J\cdot \mathrm{mol}-1\cdot K-1$, and the Faraday constant $F$ is $96485C\cdot \mathrm{mol}-1$.
5. Calculate the cell potential $E_{\mathrm{cell}}$ using the Nernst equation and express the answer to two significant figures.