Calculate the cell potential for the following reaction that takes place in an electrochemical cell at 25 °:
$\quad$ $\quad$ Cu(s) | Cu²⁺(aq,0.0032 mol L^-1) $\|$ Cu²⁺(aq,4.48 mol L^-1) | Cu(s)
E°(Cu²⁺/Cu) = +0.34 V

Question

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

\quad

\quad

Cu(s) | Cu²⁺(aq,0.0032 mol L^-1)

\|

Cu²⁺(aq,4.48 mol L^-1) | Cu(s)
E°(Cu²⁺/Cu) = +0.34 V

Quizplus · Accepted Answer

The Nernst equation is used to calculate the cell potential of an electrochemical cell:$$E = E° - \frac{RT}{nF} \ln Q$$where:* $E$ is the cell potential in volts (V)* $E°$ is the standard cell potential in volts (V)* $R$ is the ideal gas constant (8.314 J/mol·K)* $T$ is the temperature in Kelvin (K)* $n$ is the number of moles of electrons transferred in the balanced chemical equation* $F$ is the Faraday constant (96,485 C/mol)* $Q$ is the reaction quotientIn this case, the balanced chemical equation is:$$Cu^{2+}(aq) + Cu(s) \rightarrow Cu(s) + Cu^{2+}(aq)$$and $n = 2$. The reaction quotient is:$$Q = \frac{[Cu^{2+}]_{initial}}{[Cu^{2+}]_{final}} = \frac{0.0032 mol/L}{4.48 mol/L} = 7.14 \times 10^{-4}$$Substituting these values into the Nernst equation, we get:$$E = 0.34 V - \frac{8.314 J/mol·K \times 298 K}{2 \times 96,485 C/mol} \ln (7.14 \times 10^{-4}) = 0.093 V$$Therefore, the correct answer is B) +0.093 V.

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

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