Calculate the cell potential for the following reaction that takes place in an electrochemical cell at 25 °C:
$\quad$ $\quad$ $\quad$ Al(s) | Al³⁺(aq,0.115 mol L^-1) $\|$ Al³⁺(aq,3.89 mol L^-1) | Al(s)
E°(Al³⁺/Al) = -1.66 V

Question

Calculate the cell potential for the following reaction that takes place in an electrochemical cell at 25 °C:

\quad

\quad

\quad

Al(s) | Al³⁺(aq,0.115 mol L^-1)

\|

Al³⁺(aq,3.89 mol L^-1) | Al(s)
E°(Al³⁺/Al) = -1.66 V

Quizplus · Accepted Answer

The cell potential can be calculated using the Nernst equation: $E = E° - \frac{RT}{nF} \ln Q$. Here, $E° = 0$ because both electrodes are the same. The reaction quotient $Q = \frac{[Al^{3+}]_{cathode}}{[Al^{3+}]_{anode}} = \frac{3.89}{0.115}$. At 25°C, $\frac{RT}{nF} \approx 0.0257 \, \text{V}$, and $n = 3$. Thus, $E = 0 - 0.0257/3 \ln(3.89/0.115) \approx +0.030 \, \text{V}$.

Calculate the Cell Potential for the Following Reaction That Takes \quad

Calculate the Cell Potential for the Following Reaction That Takes $\quad$