Question

Consider the following galvanic cell at 25°C

$Pt | H_2(g) | H_2CO_3(aq.) + NaHCO_3(aq.) || CH_3COOH(aq.) | H_2(g) | Pt$ (1 bar) (100 ml, 0.2M) (100 ml 0.2M) (100 ml 0.1M) (1 bar) For $H_2CO_3 : Ka_1 = 10^{-7}, Ka_2 = 10^{-11}$ for $CH_3COOH : Ka = 10^{-5}$ Now some electrolyte is added in one of the compartment of the given cell in List-l and corresponding cell potential in List-ll is measured. Select the correct option matched correctly: $Take \frac{2.303RT}{F} = 0.06 Volts$

	List-l		List-II
(P)	If no electrolyte is added in either of the compartment of the cell, then	(1)	0.36 Volts
(Q)	If 100 ml 0.2 M NaOH is added in the anode compartment of the original cell	(2)	0.24 Volts
(R)	If 50 ml 0.1 M NaOH is added in the cathode compartment of the original cell then	(3)	0.3 Volts
(S)	If 100 mL 0.02 M HCl is added in the cathode compartment of the	(4)	0.12 Volts

• A. (P)-(2), (Q)-(1), (R)-(4), (S)-(3)

Answer 1

Answer: (A)

(P)-(2), (Q)-(1), (R)-(4), (S)-(3)

Answer 2

The cell is a concentration cell with hydrogen electrodes:

$Pt | H_2(g, 1 \text{ bar}) | \text{Anode solution} || \text{Cathode solution} | H_2(g, 1 \text{ bar}) | Pt$

The cell potential is given by $E_{cell} = E_{cathode} - E_{anode}$. For a hydrogen electrode, $E = - \frac{2.303RT}{F} pH$. Given $\frac{2.303RT}{F} = 0.06$ Volts. $E_{cell} = -0.06 \, pH_{cathode} - (-0.06 \, pH_{anode}) = 0.06 (pH_{anode} - pH_{cathode})$.

Initial conditions: Anode compartment: 100 ml 0.2 M $H_2CO_3$ + 100 ml 0.2 M $NaHCO_3$. This is a buffer solution of $H_2CO_3$ and $HCO_3^-$. Moles of $H_2CO_3 = 0.2 \, M \times 0.1 \, L = 0.02$ mol. Moles of $NaHCO_3 = 0.2 \, M \times 0.1 \, L = 0.02$ mol (providing $HCO_3^-$). Total volume = 200 ml = 0.2 L. Concentrations: $[H_2CO_3] = \frac{0.02}{0.2} = 0.1 \, M$, $[HCO_3^-] = \frac{0.02}{0.2} = 0.1 \, M$. $H_2CO_3$ is a weak acid with $K_{a1} = 10^{-7}$. $pK_{a1} = 7$. Using the Henderson-Hasselbalch equation: $pH_{anode} = pK_{a1} + \log \frac{[HCO_3^-]}{[H_2CO_3]} = 7 + \log \frac{0.1}{0.1} = 7$.

Cathode compartment: 100 ml 0.1 M $CH_3COOH$. Moles of $CH_3COOH = 0.1 \, M \times 0.1 \, L = 0.01$ mol. Volume = 100 ml = 0.1 L. Concentration: $[CH_3COOH] = 0.1 \, M$. $CH_3COOH$ is a weak acid with $K_a = 10^{-5}$. $[H^+]_{cathode} = \sqrt{K_a \times [CH_3COOH]} = \sqrt{10^{-5} \times 0.1} = \sqrt{10^{-6}} = 10^{-3} \, M$. $pH_{cathode} = -\log(10^{-3}) = 3$.

(P) No electrolyte is added. $pH_{anode} = 7$, $pH_{cathode} = 3$. $E_{cell} = 0.06 (7 - 3) = 0.06 \times 4 = 0.24$ V. (P) matches with (2).

(Q) 100 ml 0.2 M NaOH is added in the anode compartment. Initial moles: $H_2CO_3 = 0.02$, $HCO_3^- = 0.02$. Added NaOH moles = $0.2 \, M \times 0.1 \, L = 0.02$ mol. Reaction: $H_2CO_3 + OH^- \rightarrow HCO_3^- + H_2O$. Moles after reaction: $H_2CO_3 = 0.02 - 0.02 = 0$, $OH^- = 0.02 - 0.02 = 0$, $HCO_3^- = 0.02 + 0.02 = 0.04$. Total volume = 200 ml + 100 ml = 300 ml = 0.3 L. The solution contains $0.04$ mol of $HCO_3^-$ in 0.3 L. This is a solution of $NaHCO_3$. $HCO_3^-$ is an amphoteric species. The pH of a solution of $NaHCO_3$ can be approximated by $pH \approx \frac{pK_{a1} + pK_{a2}}{2}$ for $H_2CO_3$. $pK_{a1} = 7$, $pK_{a2} = 11$. $pH_{anode} \approx \frac{7 + 11}{2} = 9$. The cathode compartment is unchanged, $pH_{cathode} = 3$. $E_{cell} = 0.06 (9 - 3) = 0.06 \times 6 = 0.36$ V. (Q) matches with (1).

(R) 50 ml 0.1 M NaOH is added in the cathode compartment. Initial moles: $CH_3COOH = 0.01$. Added NaOH moles = $0.1 \, M \times 0.05 \, L = 0.005$ mol. Reaction: $CH_3COOH + OH^- \rightarrow CH_3COO^- + H_2O$. Moles after reaction: $CH_3COOH = 0.01 - 0.005 = 0.005$, $OH^- = 0$, $CH_3COO^- = 0.005$. Total volume = 100 ml + 50 ml = 150 ml = 0.15 L. The solution is a buffer containing $CH_3COOH$ and $CH_3COO^-$. $[CH_3COOH] = \frac{0.005}{0.15} = \frac{1}{30} \, M$. $[CH_3COO^-] = \frac{0.005}{0.15} = \frac{1}{30} \, M$. $K_a$ for $CH_3COOH = 10^{-5}$, $pK_a = 5$. $pH_{cathode} = pK_a + \log \frac{[CH_3COO^-]}{[CH_3COOH]} = 5 + \log \frac{1/30}{1/30} = 5 + \log(1) = 5$. The anode compartment is unchanged, $pH_{anode} = 7$. $E_{cell} = 0.06 (7 - 5) = 0.06 \times 2 = 0.12$ V. (R) matches with (4).

(S) 100 mL 0.02 M HCl is added in the cathode compartment. Initial moles: $CH_3COOH = 0.01$. Added HCl moles = $0.02 \, M \times 0.1 \, L = 0.002$ mol. HCl is a strong acid. It will react with the conjugate base if present, but the original cathode solution is just a weak acid. Adding a strong acid to a weak acid solution will suppress the dissociation of the weak acid and the pH will be primarily determined by the strong acid, unless the concentration of the strong acid is much lower than the weak acid's $K_a$ or the concentration of the weak acid. Total volume = 100 ml + 100 ml = 200 ml = 0.2 L. Concentration of added HCl in the mixture = $\frac{0.002 \, mol}{0.2 \, L} = 0.01 \, M$. The cathode solution contains $0.1 \, M$ $CH_3COOH$ and $0.01 \, M$ HCl. HCl is a strong acid, so it dissociates completely: $HCl \rightarrow H^+ + Cl^-$. This contributes $0.01 \, M$ of $H^+$. $CH_3COOH \rightleftharpoons H^+ + CH_3COO^-$, $K_a = 10^{-5}$. Let x be the concentration of $CH_3COOH$ that dissociates. $[CH_3COOH] = 0.1 - x$. $[H^+] = 0.01 + x$. $[CH_3COO^-] = x$. $K_a = \frac{[H^+][CH_3COO^-]}{[CH_3COOH]} = \frac{(0.01 + x)x}{0.1 - x} = 10^{-5}$. Since $K_a$ is small and we are adding a strong acid, $x$ is expected to be small. Assume $0.1-x \approx 0.1$ and $0.01+x \approx 0.01$. $\frac{0.01 \times x}{0.1} \approx 10^{-5} \Rightarrow 0.1 x \approx 10^{-5} \Rightarrow x \approx 10^{-4}$. This value of $x$ ($10^{-4}$) is small compared to 0.01 and 0.1, so the approximation is valid. $[H^+]_{cathode} = 0.01 + x = 0.01 + 10^{-4} = 0.0101 \, M$. $pH_{cathode} = -\log(0.0101) \approx -\log(10^{-2}) = 2$. Using a calculator: $pH_{cathode} = -\log(0.0101) \approx 1.9956 \approx 2$. The anode compartment is unchanged, $pH_{anode} = 7$. $E_{cell} = 0.06 (7 - 2) = 0.06 \times 5 = 0.30$ V. (S) matches with (3).

Summary of matches: (P) - (2) (Q) - (1) (R) - (4) (S) - (3)

The correct option matching (P)-(Q)-(R)-(S) with (1)-(2)-(3)-(4) is (2)-(1)-(4)-(3).