22.4: Calculation of Cell Potentials from Electrode Potentials
- Page ID
- 333394
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\(\newcommand{\avec}{\mathbf a}\) \(\newcommand{\bvec}{\mathbf b}\) \(\newcommand{\cvec}{\mathbf c}\) \(\newcommand{\dvec}{\mathbf d}\) \(\newcommand{\dtil}{\widetilde{\mathbf d}}\) \(\newcommand{\evec}{\mathbf e}\) \(\newcommand{\fvec}{\mathbf f}\) \(\newcommand{\nvec}{\mathbf n}\) \(\newcommand{\pvec}{\mathbf p}\) \(\newcommand{\qvec}{\mathbf q}\) \(\newcommand{\svec}{\mathbf s}\) \(\newcommand{\tvec}{\mathbf t}\) \(\newcommand{\uvec}{\mathbf u}\) \(\newcommand{\vvec}{\mathbf v}\) \(\newcommand{\wvec}{\mathbf w}\) \(\newcommand{\xvec}{\mathbf x}\) \(\newcommand{\yvec}{\mathbf y}\) \(\newcommand{\zvec}{\mathbf z}\) \(\newcommand{\rvec}{\mathbf r}\) \(\newcommand{\mvec}{\mathbf m}\) \(\newcommand{\zerovec}{\mathbf 0}\) \(\newcommand{\onevec}{\mathbf 1}\) \(\newcommand{\real}{\mathbb R}\) \(\newcommand{\twovec}[2]{\left[\begin{array}{r}#1 \\ #2 \end{array}\right]}\) \(\newcommand{\ctwovec}[2]{\left[\begin{array}{c}#1 \\ #2 \end{array}\right]}\) \(\newcommand{\threevec}[3]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \end{array}\right]}\) \(\newcommand{\cthreevec}[3]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \end{array}\right]}\) \(\newcommand{\fourvec}[4]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \\ #4 \end{array}\right]}\) \(\newcommand{\cfourvec}[4]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \\ #4 \end{array}\right]}\) \(\newcommand{\fivevec}[5]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \\ #4 \\ #5 \\ \end{array}\right]}\) \(\newcommand{\cfivevec}[5]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \\ #4 \\ #5 \\ \end{array}\right]}\) \(\newcommand{\mattwo}[4]{\left[\begin{array}{rr}#1 \amp #2 \\ #3 \amp #4 \\ \end{array}\right]}\) \(\newcommand{\laspan}[1]{\text{Span}\{#1\}}\) \(\newcommand{\bcal}{\cal B}\) \(\newcommand{\ccal}{\cal C}\) \(\newcommand{\scal}{\cal S}\) \(\newcommand{\wcal}{\cal W}\) \(\newcommand{\ecal}{\cal E}\) \(\newcommand{\coords}[2]{\left\{#1\right\}_{#2}}\) \(\newcommand{\gray}[1]{\color{gray}{#1}}\) \(\newcommand{\lgray}[1]{\color{lightgray}{#1}}\) \(\newcommand{\rank}{\operatorname{rank}}\) \(\newcommand{\row}{\text{Row}}\) \(\newcommand{\col}{\text{Col}}\) \(\renewcommand{\row}{\text{Row}}\) \(\newcommand{\nul}{\text{Nul}}\) \(\newcommand{\var}{\text{Var}}\) \(\newcommand{\corr}{\text{corr}}\) \(\newcommand{\len}[1]{\left|#1\right|}\) \(\newcommand{\bbar}{\overline{\bvec}}\) \(\newcommand{\bhat}{\widehat{\bvec}}\) \(\newcommand{\bperp}{\bvec^\perp}\) \(\newcommand{\xhat}{\widehat{\xvec}}\) \(\newcommand{\vhat}{\widehat{\vvec}}\) \(\newcommand{\uhat}{\widehat{\uvec}}\) \(\newcommand{\what}{\widehat{\wvec}}\) \(\newcommand{\Sighat}{\widehat{\Sigma}}\) \(\newcommand{\lt}{<}\) \(\newcommand{\gt}{>}\) \(\newcommand{\amp}{&}\) \(\definecolor{fillinmathshade}{gray}{0.9}\)The potential of an electrochemical cell is the difference between the electrode potentials of the cathode and the anode
\[E_\text{cell} = E_\text{cathode} - E_\text{anode} \label{ecell} \]
where \(E_\text{cathode}\) and \(E_\text{anode}\) are both reduction potentials. Given a set of conditions, we can use the Nernst equation to calculate the cell potential, as shown by the following example.
Calculate (a) the standard state potential and (b) the potential when [Ag+] = 0.0020 M and [Cd2+] = 0.0050 M, for the following reaction at 25oC.
\[\mathrm{Cd}(s)+2 \mathrm{Ag}^{+}(a q)\rightleftharpoons2 \mathrm{Ag}(s)+\mathrm{Cd}^{2+}(a q) \nonumber \]
For part (b), calculate the potential twice, once using concentrations and once using activities assuming that the solution's ionic strength is 0.100.
Solution
(a) In this reaction Cd is oxidized at the anode and Ag+ is reduced at the cathode. Using standard state electrode potentials from Appendix 3, we find that the standard state potential is
\[E^{\circ} = E^{\circ}_{\text{Ag}^+/ \text{Ag}} - E^{\circ}_{\text{Cd}^{2+}/ \text{Cd}} = 0.7996 - (-0.4030) = 1.2026 \ \text{V} \nonumber \]
(b) To calculate the potential when [Ag+] is 0.0020 M and [Cd2+] is 0.0050 M, we use the appropriate relationship for the reaction quotient, Qr, when writing the Nernst equation
\[E = E^{\circ} - \frac{0.05916 \ \mathrm{V}}{n} \log \frac{\left[\mathrm{Cd}^{2+}\right]}{\left[\mathrm{Ag}^{+}\right]^{2}} \nonumber \]
\[E=1.2026 \ \mathrm{V}-\frac{0.05916 \ \mathrm{V}}{2} \log \frac{0.050}{(0.020)^{2}}=1.14 \ \mathrm{V} \nonumber \]
To calculate the potential using activities, we first calculate the activity coefficients for Cd2+ and Ag+. Following the approach outlined in the appendix in Chapter 35.7 gives
\[\log \gamma_{\ce{Cd^{2+}}} = \frac {-0.51 \times (+2)^2 \times \sqrt{0.100}} {1 + 3.3 \times 0.50 \times \sqrt{0.100}} = -0.2078 \nonumber \]
\[]log \gamma_{\ce{Ag^{+}}} = \frac {-0.51 \times (+1)^2 \times \sqrt{0.100}} {1 + 3.3 \times 0.25 \times \sqrt{0.100}} = -0.1279 \nonumber \]
\[a_{\ce{Cd^{2+}}} = \gamma_{\ce{Cd^{2+}}} \times [\ce{Cd}^{2+}] = 0.6197 \times 0.0050 = 0.003098 \nonumber \]
\[a_{\ce{Ag^{+}}} = \gamma_{\ce{Ag^{+}}} \times [\ce{Ag}^{+}] = 7449 \times 0.0020 = 0.00149 \nonumber \]
Finally, we substitute activities for concentrations in the Nernst equation to arrive at a potential of
\[E=1.2026 \ \mathrm{V}-\frac{0.05916 \ \mathrm{V}}{2} \log \frac{0.003098}{(0.00149)^{2}}=1.11 \ \mathrm{V} \nonumber \]