Consider the cell $$ \mathrm{Pt}\left|\mathrm{H}_{2}\right| \mathrm{H}^{+} \| \mathrm{H}^{+}\left|\mathrm{H}_{2}\right| \mathrm{Pt} $$ In the anode half-cell, hydrogen gas at \(1.0 \mathrm{~atm}\) is bubbled over a platinum electrode dipping into a solution that has a \(\mathrm{pH}\) of \(7.0 .\) The other half-cell is identical to the first except that the solution around the platinum electrode has a pH of \(0.0\). What is the cell voltage?

The given cell can be written as: $$ \mathrm{Pt}\left|\mathrm{H}_{2}\right| \mathrm{H}^{+} \| \mathrm{H}^{+}\left|\mathrm{H}_{2}\right| \mathrm{Pt} $$ And the half reactions are: Anode: \(\mathrm{H}_2 \rightarrow 2\mathrm{H}^+ + 2e^-\) Cathode: \(\mathrm{H}_2 \rightarrow 2\mathrm{H}^+ + 2e^-\)

Since the half-cell reactions are the same, we can focus on the difference in hydrogen ion concentrations. The Nernst equation relates the cell potential difference to the concentrations of the ions involved in the redox reactions: $$E_{cell}=E^\circ_{cell}-\dfrac{RT}{nF}\ln Q $$ Here, \(E^\circ_{cell}\) is the standard cell potential, \(R\) is the gas constant, \(T\) is the temperature in Kelvin, \(n\) is the number of electrons transferred in the redox reaction, \(F\) is the Faraday constant, and \(Q\) is the reaction quotient. As both half-cell reactions are standard pHydrogen electrodes, \(E^\circ_{cell} =0V\). Given the pH values, the concentrations of \(\mathrm{H}^+\) ions in both half-cells are as follows: Anode: \([\mathrm{H}^{+}] = 10^{-7}\, \mathrm{M}\), as \(\mathrm{pH}=7\) Cathode: \([\mathrm{H}^{+}] = 10^{0} = 1.0\, \mathrm{M}\), as \(\mathrm{pH}=0\) Using these values, we can find the reaction quotient \(Q\): $$ Q = \dfrac{[\mathrm{H}^{+}]_\text{cathode}}{[\mathrm{H}^{+}]_\text{anode}} = \dfrac{1.0\mathrm{M}}{10^{-7}\,\mathrm{M}} = 1\times 10^7$$ With \(Q\), now we can find the cell potential difference: $$E_{cell} = 0 - \dfrac{(8.314\,\mathrm{J\, mol^{-1} K^{-1}}) (298\,\mathrm{K})}{(2)(96485\,\mathrm{C\, mol^{-1}})} \ln 1\times 10^7$$

Now we can plug values of the constants into the equation to calculate the cell voltage: $$E_{cell} = \dfrac{(8.314\,\mathrm{J\, mol^{-1} K^{-1}}) (298\,\mathrm{K})}{(2)(96485\,\mathrm{C\, mol^{-1}})} \ln 1\times 10^7$$ $$E_{cell} \approx 0.05916 \,\mathrm{V} \times \log(1\times 10^7)$$ $$E_{cell} \approx 0.05916 \,\mathrm{V} \times 7$$ $$E_{cell} \approx 0.4141\, \mathrm{V}$$ Therefore, the cell voltage is approximately \(0.414 \mathrm{V}\).