Question

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 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?

Short answer

In summary, the given hydrogen cell with hydrogen gas bubbled over two platinum electrodes dipped into a solution with pH values 7.0 and 0.0 has a cell voltage of approximately 0.118 V.

Step-by-step solution

Write the half-cell reactions

For each electrode, we have hydrogen gas H2 converting to aqueous H+ ions, which involves the transfer of electrons. We can write the half-cell reactions as follows: Anode (pH 7.0): $$H_2(g) \rightarrow 2H^+(aq) + 2e^-$$ Cathode (pH 0.0): $$H_2(g) \rightarrow 2H^+(aq) + 2e^-$$

Determine the standard reduction potential

Since we are given pH values and dealing with hydrogen half-cells, the standard reduction potential \(E^0\) for both half-cells is \(0.000 V\) (by definition, the hydrogen electrode is the reference electrode).

Use the Nernst equation to find the cell voltage

The Nernst equation relates the cell voltage, standard reduction potentials, and concentrations of reactants and products: $$E_{cell} = E^0_{cathode} - E^0_{anode} - \frac{RT}{nF} \times \ln Q$$ To find the cell voltage \(E_{cell}\), we can plug in the values. The standard reduction potentials \(E^0_{cathode}\) and \(E^0_{anode}\) are both \(0.000 V\), so their difference will also be zero. Since our half-cell reactions involve the transfer of 2 electrons (n = 2), we have: $$E_{cell} = -\frac{RT}{2F} \times \ln Q$$ In this case, R (universal gas constant) is \(8.314\,\text{J} / \text{molK}\), T (temperature) is \(298\,\text{K}\), and F (Faraday's constant) is \(96485\,\text{C/mol}\). Now we need to calculate the reaction quotient Q, which is the ratio of the product's concentration to the reactant's concentration: $$Q = \frac{[H^+]_{cathode}^2}{[H^+]_{anode}^2}$$ Since pH = \(-log[H^+]\), we can find the concentration of H+ ions in each half-cell: \([H^+]_{anode} = 10^{-7}\) and \([H^+]_{cathode} = 10^{0}\) Now we can calculate Q: $$Q = \frac{(10^{0})^2}{(10^{-7})^2} = 10^{14}$$ Finally, we can plug in all the values into the Nernst equation to find the cell voltage: $$E_{cell} = -\frac{(8.314\,\text{J} / \text{molK})(298\,\text K)}{2(96485\,\text{C/mol})}\times \ln\,(10^{14})$$ Evaluating this expression, we find: $$E_{cell} \approx 0.118\,\text{V}$$ So, the cell voltage for this hydrogen cell is approximately 0.118 V.