Question

Which of the changes below will increase the voltage of the following cell? \(\mathrm{Co}\left|\mathrm{Co}^{2+}(0.010 \mathrm{M}) \| \mathrm{H}^{+}(0.010 \mathrm{M})\right| \mathrm{H}_{2}(0.500 \mathrm{~atm}) \mid \mathrm{Pt}\) (a) Increase the volume of \(\mathrm{CoCl}_{2}\) solution from \(100 \mathrm{~mL}\) to \(300 \mathrm{~mL}\). (b) Increase \(\left[\mathrm{H}^{+}\right]\) from \(0.010 \mathrm{M}\) to \(0.500 \mathrm{M}\). (c) Increase the pressure of \(\mathrm{H}_{2}\) from 0.500 atm to \(1 \mathrm{~atm}\). (d) Increase the mass of the Co electrode from \(15 \mathrm{~g}\) to \(25 \mathrm{~g}\). (e) Increase \(\left[\mathrm{Co}^{2+}\right]\) from \(0.010 \mathrm{M}\) to \(0.500 \mathrm{M}\).

Short answer

Answer: (b) Increase \(\left[\mathrm{H}^{+}\right]\) from \(0.010 \mathrm{M}\) to \(0.500 \mathrm{M}\), and (c) Increase the pressure of \(\mathrm{H}_{2}\) from 0.500 atm to \(1 \mathrm{~atm}\).

Step-by-step solution

Write the half-reactions

Taking the cell \(\mathrm{Co}\left|\mathrm{Co}^{2+}(0.010 \mathrm{M}) || \mathrm{H}^{+}(0.010 \mathrm{M})\right| \mathrm{H}_{2}(0.500 \mathrm{~atm}) \mid \mathrm{Pt}\) into account, we have the following half-reactions: Oxidation: \(\mathrm{Co}(s)\to\mathrm{Co}^{2+}(aq)+2e^{-}\) Reduction: \(2\mathrm{H}^{+}(aq)+2e^{-}\to\mathrm{H}_{2}(g)\)

Write the Nernst Equation

As we have already written the Nernst Equation in the analysis, recall that it is: \(E_{cell}=E°_{cell}-\frac{RT}{nF}\ln(Q)\) Here, \(Q = \frac{[\mathrm{Co}^{2+}]}{P_{\mathrm{H}_{2}}\times [\mathrm{H}^{+}]^{2}}\)

Analyze each proposal's effect on the reaction quotient and hence the cell voltage

Now, examine how each proposal affects the reaction quotient and hence the cell voltage: (a) Since the volume change of \(\mathrm{CoCl}_{2}\) solution does not affect the concentrations or pressures involved in the cell, the reaction quotient (Q) remains unchanged and so does the cell voltage. (b) When the \(\left[\mathrm{H}^{+}\right]\) increases, the denominator of the reaction quotient (Q) increases. This causes the reaction quotient (Q) to decrease and hence the cell voltage (\(E_{cell}\)) to increase. (c) An increase in the pressure of \(\mathrm{H}_{2}\) causes the denominator of the reaction quotient(Q) to increase. This causes the reaction quotient(Q) to decrease and hence the cell voltage (\(E_{cell}\)) to increase. (d) The mass of the Co electrode does not affect the concentrations or pressures involved in the cell, so the reaction quotient (Q) remains unchanged and so does the cell voltage. (e) An increase in the \(\left[\mathrm{Co}^{2+}\right]\) concentration causes the numerator of the reaction quotient(Q) to increase. Consequently, the reaction quotient(Q) increases and the cell voltage (\(E_{cell}\)) decreases. From the analysis above, it is clear that both (b) and (c) increase the cell voltage. Therefore, the answer is: (b) Increase \(\left[\mathrm{H}^{+}\right]\) from \(0.010 \mathrm{M}\) to \(0.500 \mathrm{M}\), and (c) Increase the pressure of \(\mathrm{H}_{2}\) from 0.500 atm to \(1 \mathrm{~atm}\).