Chemistry Q71 – JEE-MAIN 2025

Home Chemistry ElectrochemistryQuestion #396

Consider the following electrochemical cell at standard condition.

Au(s) | {QH}_{2}, Q | {NH}_{4} X(0.01 M) | | {Ag}^{+} (1 M) | Ag(s) E_{cell} = + 0.4 V

The couple

{QH}_{2} / Q

represents quinhydrone electrode, the half cell reaction is given below :

Q + 2 e^{-} + 2 H^{+} \to {QH}_{2} E_{{Q/QH}_{2}}^{\circ} = + 0.7 V

Given :

E_{{Ag}^{+} / Ag}^{\circ} = + 0.8 V

and

\frac{2.303 RT}{F} = 0.06 V

The

{pK}_{b}

value of the ammonium halide salt (

{NH}_{4} X

) used here is _______. (nearest integer)

Enter Calculated Numerical Value:

Clue 1 of 3: Cell Potential and Nernst Equation

The overall cell potential is the difference between the cathode and anode potentials. The Nernst equation is needed for the quinhydrone electrode as its potential depends on pH.

🧠 Full Solution Path

● Video Walkthrough

Step 1: Calculate the pH of the anode compartment✦ Active

The cell potential is given by $E_{cell} = E_{cathode} - E_{anode}$ . For the cathode ( ${Ag}^{+} / Ag$ ), $E_{cathode} = E_{{Ag}^{+} / Ag}^{\circ} = + 0.8 V$ since $[{Ag}^{+}] = 1 M$ . For the anode (quinhydrone electrode), the reduction potential is $E_{anode} = E_{{Q/QH}_{2}}^{\circ} - \frac{2.303 RT}{F} pH$ . Substituting the given values:

0.4 = 0.8 - (0.7 - 0.06 pH)

0.4 = 0.1 + 0.06 pH

0.3 = 0.06 pH

pH = \frac{0.3}{0.06} = 5

Step 2: Calculate the

K_{a}

of the ammonium ion (

{NH}_{4}^{+}

)○ Expand

The anode compartment contains $0.01 M$ ${NH}_{4} X$ . The ${NH}_{4}^{+}$ ion hydrolyzes as ${NH}_{4}^{+} + H_{2} O ⇌ {NH}_{3} + H_{3} O^{+}$ . From $pH = 5$ , we have $[H_{3} O^{+}] = 10^{- 5} M$ . At equilibrium, $[{NH}_{3}] = [H_{3} O^{+}] = 10^{- 5} M$ and $[{NH}_{4}^{+}] \approx 0.01 M$ (since $10^{- 5} ≪ 0.01$ ). The acid dissociation constant for ${NH}_{4}^{+}$ is:

K_{a} ({NH}_{4}^{+}) = \frac{[{NH}_{3}] [H_{3} O^{+}]}{[{NH}_{4}^{+}]} = \frac{(10^{- 5}) (10^{- 5})}{0.01} = \frac{10^{- 10}}{10^{- 2}} = 10^{- 8}

💡 Teacher's Secret Hint

Remember that for a weak acid, the concentration of the conjugate base formed is equal to the concentration of $H^{+}$ ions produced.

Step 3: Calculate the

p K_{b}

of the ammonium halide salt○ Expand

The relationship between $K_{a}$ of a conjugate acid ( ${NH}_{4}^{+}$ ) and $K_{b}$ of its conjugate base ( ${NH}_{3}$ ) is $K_{a} \cdot K_{b} = K_{w}$ . Assuming standard temperature, $K_{w} = 10^{- 14}$ . Therefore:

10^{- 8} \cdot K_{b} = 10^{- 14}

K_{b} = \frac{10^{- 14}}{10^{- 8}} = 10^{- 6}

{pK}_{b} = - \log (K_{b}) = - \log (10^{- 6}) = 6

💡 Teacher's Secret Hint

Ensure you use the correct $K_{w}$ value for standard conditions (typically $10^{- 14}$ at 298 K).

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Chemistry Question 71 – JEE-MAIN 2025

Clue 1 of 3: Cell Potential and Nernst Equation

🧠 Full Solution Path