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Question: For a hald cell containing a Pt rod immersed in a solution of 1MHA, $O_2(g)$ is bubbled at 1 atm. Th...

For a hald cell containing a Pt rod immersed in a solution of 1MHA, O2(g)O_2(g) is bubbled at 1 atm. The stnadard reduction potential for water formation is 1.23 V. Given a dissociation constant, Ka=1×10−4K_a = 1 \times 10^{-4} for HA, what is Ehalfcellat298KE_{half cell at 298K} in V?

Answer

1.11 V

Explanation

Solution

The half-cell reaction for water formation from oxygen is: O2(g)+4H+(aq)+4e−⟶2H2O(l)O_2(g) + 4H^+(aq) + 4e^- \longrightarrow 2H_2O(l)

The standard reduction potential is given as Eo=1.23 VE^o = 1.23 \text{ V}. The number of electrons transferred, n=4n = 4. The pressure of oxygen gas, PO2=1 atmP_{O_2} = 1 \text{ atm}.

First, we need to determine the concentration of H+H^+ ions in the 1M HA solution. HA is a weak acid with a dissociation constant Ka=1×10−4K_a = 1 \times 10^{-4}. The dissociation equilibrium for HA is: HA(aq)⇌H+(aq)+A−(aq)HA(aq) \rightleftharpoons H^+(aq) + A^-(aq)

Let the initial concentration of HA be C=1 MC = 1 \text{ M}. Let xx be the concentration of H+H^+ ions at equilibrium. At equilibrium: [HA]=C−x=1−x[HA] = C - x = 1 - x [H+]=x[H^+] = x [A−]=x[A^-] = x

The acid dissociation constant KaK_a is given by: Ka=[H+][A−][HA]K_a = \frac{[H^+][A^-]}{[HA]} 1×10−4=(x)(x)1−x1 \times 10^{-4} = \frac{(x)(x)}{1 - x}

Since KaK_a is small (1×10−41 \times 10^{-4}) and the initial concentration is relatively large (1 M1 \text{ M}), we can assume x≪1x \ll 1. Thus, 1−x≈11 - x \approx 1. 1×10−4=x211 \times 10^{-4} = \frac{x^2}{1} x2=1×10−4x^2 = 1 \times 10^{-4} x=1×10−4=1×10−2 Mx = \sqrt{1 \times 10^{-4}} = 1 \times 10^{-2} \text{ M} So, the concentration of H+H^+ ions, [H+]=1×10−2 M[H^+] = 1 \times 10^{-2} \text{ M}.

Next, we calculate the reaction quotient QQ for the half-cell reaction: Q=1PO2[H+]4Q = \frac{1}{P_{O_2} [H^+]^4} (since the activity of pure liquid water is 1) Q=1(1 atm)(1×10−2 M)4Q = \frac{1}{(1 \text{ atm}) (1 \times 10^{-2} \text{ M})^4} Q=11×10−8=108Q = \frac{1}{1 \times 10^{-8}} = 10^8

Finally, we use the Nernst equation to find the half-cell potential (EhalfcellE_{half cell}) at 298K: Ehalfcell=Eo−0.0591nlog⁡QE_{half cell} = E^o - \frac{0.0591}{n} \log Q Ehalfcell=1.23 V−0.05914log⁡(108)E_{half cell} = 1.23 \text{ V} - \frac{0.0591}{4} \log (10^8) Ehalfcell=1.23−0.05914×8E_{half cell} = 1.23 - \frac{0.0591}{4} \times 8 Ehalfcell=1.23−0.0591×2E_{half cell} = 1.23 - 0.0591 \times 2 Ehalfcell=1.23−0.1182E_{half cell} = 1.23 - 0.1182 Ehalfcell=1.1118 VE_{half cell} = 1.1118 \text{ V}

Rounding to two decimal places, Ehalfcell≈1.11 VE_{half cell} \approx 1.11 \text{ V}.