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 equation for this reaction is: \[ E = E^\circ - \frac{0.06}{2} \log [H^+]^2 \] \[ E = E^\circ - 0.06 \log [H^+] \] Now, using the given data and solving for pH: \[ \text{pH} = - \log [H^+] = \frac{E - E^\circ}{0.06} = \frac{0.4 - 0.1}{0.06} = 5 \] Now, we consider the ammonium halide salt (NH₄X) with the relation: \[ \text{pH} + \text{NH}_4X = 7 - \frac{1}{2} pK_a - \frac{1}{2} \log C \] Substituting values: \[ 5 = 7 - \frac{1}{2} pK_a - \frac{1}{2} \log (10^{-3}) \] \[ 5 = 7 - \frac{1}{2} pK_a + \frac{1}{2} \times 3 \] \[ 5 = 7 - \frac{1}{2} pK_a + 1.5 \] \[ \Rightarrow pK_a = 6 \]

Answer 2

We are asked to find the pK\( _b \) value of the weak base corresponding to the ammonium halide salt (NH\( _4 \)X) used in the anode compartment of the given electrochemical cell. The overall cell potential and standard potentials for the half-cells are provided.

Concept Used:

The solution involves several key concepts from electrochemistry and ionic equilibrium:

1. Electrochemical Cell Potential: The potential of an electrochemical cell is the difference between the reduction potential of the cathode and the reduction potential of the anode. \[ E_{\text{cell}} = E_{\text{cathode}} - E_{\text{anode}} \]
2. Nernst Equation: This equation relates the reduction potential of a half-cell to its standard reduction potential and the concentrations of the species involved. For a general reduction reaction \( aA + n e^- \rightarrow bB \), the equation is: \[ E = E^\circ - \frac{2.303RT}{nF} \log \frac{[B]^b}{[A]^a} \] For the quinhydrone electrode (Q + 2H\( ^+ \) + 2e\( ^- \) \(\rightleftharpoons\) QH\( _2 \)), where [Q] = [QH\( _2 \)], the potential is directly related to the pH of the solution: \[ E_{\text{Q/QH}_2} = E^\circ_{\text{Q/QH}_2} - 0.0591 \times \text{pH} \quad (\text{using } \frac{2.303RT}{F} \approx 0.0591) \] The problem gives this value as 0.06.
3. Hydrolysis of Salts: The salt of a weak base and a strong acid (like NH\( _4 \)X) undergoes cationic hydrolysis, producing an acidic solution. \[ \text{NH}_4^+ + \text{H}_2\text{O} \rightleftharpoons \text{NH}_3 + \text{H}_3\text{O}^+ \]
4. Relationship between K\( _a \), K\( _b \), and K\( _w \): For a conjugate acid-base pair, the product of their dissociation constants is equal to the ionic product of water, K\( _w \). \[ K_a(\text{NH}_4^+) \times K_b(\text{NH}_3) = K_w = 1 \times 10^{-14} \]

Step-by-Step Solution:

Step 1: Identify the anode and cathode and calculate the potential of the cathode.

The cell is represented as: Au(s) | QH\( _2 \), Q | NH\( _4 \)X(0.01M) || Ag\( ^+ \)(1M) | Ag(s).

• The left half-cell is the anode (oxidation).
• The right half-cell is the cathode (reduction).

The cathode reaction is: Ag\( ^+ \)(aq) + e\( ^- \) \(\rightarrow\) Ag(s).

Using the Nernst equation for the cathode:

\[ E_{\text{cathode}} = E^\circ_{\text{Ag}^+/\text{Ag}} - \frac{0.06}{1} \log \frac{1}{[\text{Ag}^+]} \] \[ E_{\text{cathode}} = 0.8 \, \text{V} - 0.06 \log \frac{1}{1} = 0.8 \, \text{V} \]

Step 2: Calculate the potential of the anode.

Using the overall cell potential formula:

\[ E_{\text{cell}} = E_{\text{cathode}} - E_{\text{anode}} \] \[ 0.4 \, \text{V} = 0.8 \, \text{V} - E_{\text{anode}} \] \[ E_{\text{anode}} = 0.8 \, \text{V} - 0.4 \, \text{V} = 0.4 \, \text{V} \]

Step 3: Use the anode potential to find the pH of the solution.

The anode is the quinhydrone electrode. Its reduction potential is given by the Nernst equation for the reaction: Q + 2H\( ^+ \) + 2e\( ^- \) \(\rightarrow\) QH\( _2 \).

In a quinhydrone electrode, it is assumed that [Q] = [QH\( _2 \)].

\[ E_{\text{anode}} = E^\circ_{\text{Q/QH}_2} - \frac{0.06}{2} \log \frac{[\text{QH}_2]}{[\text{Q}][\text{H}^+]^2} \] \[ E_{\text{anode}} = 0.7 - 0.03 \log \frac{1}{[\text{H}^+]^2} = 0.7 + 0.03 \log([\text{H}^+]^2) \] \[ E_{\text{anode}} = 0.7 + 0.06 \log[\text{H}^+] = 0.7 - 0.06 \times \text{pH} \]

Now, substitute the value of \( E_{\text{anode}} \) from Step 2:

\[ 0.4 = 0.7 - 0.06 \times \text{pH} \] \[ 0.06 \times \text{pH} = 0.7 - 0.4 = 0.3 \] \[ \text{pH} = \frac{0.3}{0.06} = 5 \]

Step 4: Calculate the acid dissociation constant (K\( _a \)) for the NH\( _4 \)\( ^+ \) ion.

From the pH, we can find the hydrogen ion concentration:

\[ [\text{H}^+] = 10^{-\text{pH}} = 10^{-5} \, \text{M} \]

The H\( ^+ \) ions are produced by the hydrolysis of the NH\( _4 \)\( ^+ \) ions from the salt NH\( _4 \)X (concentration C = 0.01 M).

\[ \text{NH}_4^+ \rightleftharpoons \text{NH}_3 + \text{H}^+ \]

The expression for the acid dissociation constant is:

\[ K_a = \frac{[\text{NH}_3][\text{H}^+]}{[\text{NH}_4^+]} \]

Assuming the degree of hydrolysis is small, [NH\( _3 \)] \(\approx\) [H\( ^+ \)] and [NH\( _4 \)\( ^+ \)] \(\approx\) C.

\[ K_a = \frac{[\text{H}^+]^2}{C} = \frac{(10^{-5})^2}{0.01} = \frac{10^{-10}}{10^{-2}} = 10^{-8} \]

Step 5: Calculate the base dissociation constant (K\( _b \)) and pK\( _b \).

Using the relationship for a conjugate acid-base pair:

\[ K_a \times K_b = K_w \] \[ 10^{-8} \times K_b = 10^{-14} \] \[ K_b = \frac{10^{-14}}{10^{-8}} = 10^{-6} \]

Finally, we calculate pK\( _b \):

\[ \text{pK}_b = -\log_{10}(K_b) = -\log_{10}(10^{-6}) = 6 \]

The pK\( _b \) value of the ammonium halide salt is 6.