Calculating the Cell emf for Nonstandard Conditions What is the emf of the following voltaic cell at 25°C? [latex]\mathrm{Zn}(s)\left|\mathrm{Zn}^{2+}\left(1.00 \times 10^{-5} M\right) \| \mathrm{Cu}^{2+}(0.100 M)\right| \mathrm{Cu}(s)[/latex] PROBLEM STRATEGY From the cell reaction, deduce the value of [latex]n[/latex]; then calculate [latex]Q[/latex]. Now substitute [latex]n[/latex], [latex]Q[/latex], and [latex]E_{\text {cell }}^{\circ}[/latex] into the Nernst equation to obtain [latex]E_{\text {cell }}[/latex].

The cell reaction is [latex]\mathrm{Zn}(s)+\mathrm{Cu}^{2+}(a q) \rightleftharpoons \mathrm{Zn}^{2+}(a q)+\mathrm{Cu}(s)[/latex] The number of moles of electrons transferred is two; hence, [latex]n=2[/latex]. The reaction quotient is [latex]Q=\frac{\left[\mathrm{Zn}^{2+}\right]}{\left[\mathrm{Cu}^{2+}\right]}=\frac{1.00 \times 10^{-5}}{0.100}=1.00 \times 10^{-4}[/latex] The standard emf is [latex]1.10 \mathrm{~V}[/latex], so the Nernst equation becomes [latex] \begin{aligned} E_{\text {cell }} & =E_{\text {cell }}^{\circ}-\frac{0.0592}{n} \log Q \\ & =1.10-\frac{0.0592}{2} \log \left(1.0 \times 10^{-4}\right) \\ & =1.10-(-0.12)=1.22 \mathrm{~V} \end{aligned} [/latex] The cell emf is [latex]\mathbf{1 . 2 2} \mathbf{V}[/latex].

Question 20.12: Calculating the Cell emf for Nonstandard Conditions What is ...

General Chemistry

Calculating the Cell emf for Nonstandard Conditions

What is the emf of the following voltaic cell at 25°C?

$\mathrm{Zn}(s)\left|\mathrm{Zn}^{2+}\left(1.00 \times 10^{-5} M\right) \| \mathrm{Cu}^{2+}(0.100 M)\right| \mathrm{Cu}(s)$

PROBLEM STRATEGY

From the cell reaction, deduce the value of $n$ ; then calculate $Q$ . Now substitute $n$ , $Q$ , and $E_{\text {cell }}^{\circ}$ into the Nernst equation to obtain $E_{\text {cell }}$ .

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The cell reaction is

$\mathrm{Zn}(s)+\mathrm{Cu}^{2+}(a q) \rightleftharpoons \mathrm{Zn}^{2+}(a q)+\mathrm{Cu}(s)$

The number of moles of electrons transferred is two; hence, $n=2$ . The reaction quotient is

$Q=\frac{\left[\mathrm{Zn}^{2+}\right]}{\left[\mathrm{Cu}^{2+}\right]}=\frac{1.00 \times 10^{-5}}{0.100}=1.00 \times 10^{-4}$

The standard emf is $1.10 \mathrm{~V}$ , so the Nernst equation becomes

$\begin{aligned} E_{\text {cell }} & =E_{\text {cell }}^{\circ}-\frac{0.0592}{n} \log Q \\ & =1.10-\frac{0.0592}{2} \log \left(1.0 \times 10^{-4}\right) \\ & =1.10-(-0.12)=1.22 \mathrm{~V} \end{aligned}$

The cell emf is $\mathbf{1 . 2 2} \mathbf{V}$ .