Calculate the emf of the half-cell given belo | vedclass

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(B) The half-cell reaction is: $H_2(g) 	o 2H^+(aq) + 2e^-$.Using the Nernst equation for the oxidation potential: $E = E^\circ - \frac{0.059}{n} \log

Vedclass · Accepted Answer

$0.109$ $V$

Vedclass · Answer

(B) The half-cell reaction is: $H_2(g) 	o 2H^+(aq) + 2e^-$.Using the Nernst equation for the oxidation potential: $E = E^\circ - \frac{0.059}{n} \log Q$.Here, $n = 2$, $E^\circ = 0 	ext{ V}$, $[H^+] = 0.02 	ext{ M}$, and $P_{H_2} = 2 	ext{ atm}$.$Q = \frac{[H^+]^2}{P_{H_2}} = \frac{(0.02)^2}{2} = \frac{0.0004}{2} = 0.0002 = 2 	imes 10^{-4}$.$E = 0 - \frac{0.059}{2} \log(2 	imes 10^{-4})$.$E = -0.0295 	imes (\log 2 + \log 10^{-4}) = -0.0295 	imes (0.3010 - 4) = -0.0295 	imes (-3.699) \approx 0.109 	ext{ V}$.

Calculate the emf of the half-cell given below: $Pt(s) | H_2(g, 2 \text{ atm}) | HCl(aq, 0.02 \text{ M})$, $E^\circ_{H^+/H_2} = 0 \text{ V}$. (Given: $\frac{2.303RT}{F} = 0.059$, $\log 2 = 0.3010$)

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