Question 19.11: Using the Nernst Equation in the Electrochemical Determinati...
Using the Nernst Equation in the Electrochemical Determination of ph
The following cell has a potential of 0.55 V at 25 °C:
Pt(s) \mid H_{2}(1 atm) \mid H^{+}(? M) \parallel Cl^{-}(1 M) \mid Hg_{2}Cl_{2}(s) \mid Hg(l)
What is the pH of the solution in the anode compartment?
STRATEGY
First, read the shorthand notation to obtain the cell reaction. Then, calculate the standard cell potential by looking up the standard reduction potentials in Appendix B. Finally, apply the Nernst equation to find the pH.
IDENTIFY
| Known | Unknown |
| E_{cell} = 0.55 V | pH |
| Redox reaction (from shorthand cell notation) | |
| Standard reduction potentials for half cells (Appendix D) | |
The cell reaction is
and the cell potential is
E_{cell} = E_{H_{2} \rightarrow H^{+}} + E_{Hg_{2}Cl_{2} \rightarrow Hg} = 0.55 VBecause the reference electrode is the standard calomel electrode, which has E = E° = 0.28 V (Appendix D), the half-cell potential for the hydrogen electrode is 0.27 V:
E_{H_{2} \rightarrow H^{+}} = E_{cell} – E_{Hg_{2}Cl_{2} \rightarrow Hg} = 0.55 V – 0.28 V = 0.27 VWe can then apply the Nernst equation to the half-reaction H_{2}(g) \rightarrow 2 H^{+}(aq) + 2 e^{-}:
E_{H_{2} \rightarrow H^{+}} = (E°_{H_{2} \rightarrow H^{+}}) -(\frac{0.0592 V}{n})(\log \frac{[H^{+}]^{2}}{P_{H_{2}}})Substituting in the values of E, E°, n, and P_{H_{2}} gives
0.27 V = (0 V) -(\frac{0.0592 V}{2})(\log \frac{[H^{+}]^{2}}{1})=(0.0592 V)(pH)Therefore, the pH is
pH =\frac{0.27 V}{0.0592 V}=4.6