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(D) The half-cell reaction is: $O_{2(g)} + 4H^+_{(aq)} + 4e^- ightarrow 2H_2O_{(l)}$For $HA ightleftharpoons H^+ + A^-$,$[H^+] = \sqrt{K_a 	imes

Vedclass · Accepted Answer

$1.112$

Vedclass · Answer

(D) The half-cell reaction is: $O_{2(g)} + 4H^+_{(aq)} + 4e^- ightarrow 2H_2O_{(l)}$For $HA ightleftharpoons H^+ + A^-$,$[H^+] = \sqrt{K_a 	imes C} = \sqrt{10^{-4} 	imes 1} = 10^{-2} \ M$.Using the Nernst equation: $E = E^{\circ} - \frac{0.0592}{n} \log \frac{1}{P_{O_2} [H^+]^4}$.Here,$n = 4$,$P_{O_2} = 1 \ atm$,and $[H^+] = 10^{-2} \ M$.$E = 1.23 - \frac{0.0592}{4} \log \frac{1}{1 	imes (10^{-2})^4}$.$E = 1.23 - \frac{0.0592}{4} \log (10^8)$.$E = 1.23 - \frac{0.0592}{4} 	imes 8$.$E = 1.23 - 0.0592 	imes 2 = 1.23 - 0.1184 = 1.1116 \ V \approx 1.112 \ V$.

For a half-cell containing a $Pt$ rod immersed in a solution of $1 \ M$ $HA$,$O_{2(g)}$ is bubbled at $1 \ atm$. The standard reduction potential for water formation is $1.23 \ V$. Given a dissociation constant,$K_a = 1 \times 10^{-4}$ for $HA$,what is $E_{\text{Half-cell}}$ at $298 \ K$ in $V$?

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