Question 17.8: 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)| H_{2}(1 atm)| H^{+}(? M)|| Cl^{-}(1 M)| Hg_{2}Cl_{2}(s)| 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 halfcell potential for the hydrogen electrode from the observed cell potential and the half-cell potential for the calomel reference electrode. Finally, apply the Nernst equation to find the pH.
The cell reaction is
H_{2}(g) + Hg_{2}Cl_{2}(s) → 2 H^{+}(aq) + 2 Hg(l) + 2 Cl^{-}(aq)
and the cell potential is
E_{cell} = E_{H_{2} → H^{+}} + E_{Hg_{2}Cl_{2} → Hg} = 0.55 V
Because 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} → H^{+}} = E_{cell} – E_{Hg_{2}Cl_{2} → Hg} = 0.55 V – 0.28 V = 0.27 V
We can then apply the Nernst equation to the half-reaction H_{2}(g) → 2 H^{+}(aq) + 2 e^{-}:
E_{H_{2} → H^{+}} = (E°_{H_{2} → H^{+}}) – \left(\frac{0.0592 V}{n}\right)\left(log \frac{[H^{+}]^{2}}{P_{H_{2}}}\right)
Substituting in the values of E, E°, n and P_{H_{2}} gives
0.27 V = (0 V) – \left(\frac{0.0592 V}{2}\right)\left(log \frac{[H+]^{2}}{1} \right) = (0.0592 V)(pH)
Therefore, the pH is
pH = \frac{0.27 V}{0.0592 V} = 4.6