Question

Consider the following electrochemical cell at standard condition. $$ \text{Au(s) | QH}_2\text{ | QH}_X(0.01 M) \, \text{| Ag(1M) | Ag(s) } \, E_{\text{cell}} = +0.4V $$ The couple QH/Q represents quinhydrone electrode, the half cell reaction is given below: $$ \text{QH}_2 \rightarrow \text{Q} + 2e^- + 2H^+ \, E^\circ_{\text{QH}/\text{Q}} = +0.7V $$

Hint:

When dealing with electrochemical cells, use the Nernst equation to calculate potential at non-standard conditions. Additionally, in pH calculations, use the relationship between pH, pKa, and concentration to determine the equilibrium.

Answer 1

Correct Answer: 6

The cell reaction is:\[\text{QH}_2 + 2\text{Ag}^+ \rightarrow 2\text{Ag} + \text{Q} + 2H^+\]The corresponding equation is:\[E = E^\circ - \frac{0.06}{2} \log [H^+]^2\]This simplifies to:\[E = E^\circ - 0.06 \log [H^+]\]Using the provided data to solve for pH:\[\text{pH} = - \log [H^+] = \frac{E - E^\circ}{0.06} = \frac{0.4 - 0.1}{0.06} = 5\]Next, we consider the ammonium halide salt (NH₄X) with the following relation:\[\text{pH} + \text{NH}_4X = 7 - \frac{1}{2} pK_a - \frac{1}{2} \log C\]Substituting the values:\[5 = 7 - \frac{1}{2} pK_a - \frac{1}{2} \log (10^{-3})\]\[5 = 7 - \frac{1}{2} pK_a + \frac{1}{2} \times 3\]\[5 = 7 - \frac{1}{2} pK_a + 1.5\]Which leads to:\[\Rightarrow pK_a = 6\]