Calculate pH of HCl solution at 298,K tempera | vedclass

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(B) The cell reaction is: $H_2(g) + Cu^{2+}(aq) \rightarrow 2H^+(aq) + Cu(s)$.Using the Nernst equation: $E_{cell} = E^{\circ}_{cell} - \frac{0.0591}{

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(B) The cell reaction is: $H_2(g) + Cu^{2+}(aq) ightarrow 2H^+(aq) + Cu(s)$.Using the Nernst equation: $E_{cell} = E^{\circ}_{cell} - \frac{0.0591}{n} \log Q$.Here,$n = 2$,$E^{\circ}_{cell} = 0.34\,V$,$E_{cell} = 0.45\,V$,and $Q = \frac{[H^+]^2}{P_{H_2} \cdot [Cu^{2+}]}$.Substituting the values: $0.45 = 0.34 - \frac{0.0591}{2} \log \frac{[H^+]^2}{1 \cdot 0.02}$.$0.11 = -0.02955 \log \frac{[H^+]^2}{0.02}$.$-3.722 = \log \frac{[H^+]^2}{0.02}$.$10^{-3.722} = \frac{[H^+]^2}{0.02}$.$[H^+]^2 = 0.02 	imes 1.896 	imes 10^{-4} = 3.792 	imes 10^{-6}$.$[H^+] = \sqrt{3.792 	imes 10^{-6}} \approx 1.947 	imes 10^{-3}$.$pH = -\log[H^+] = -\log(1.947 	imes 10^{-3}) \approx 2.71$.

Calculate $pH$ of $HCl$ solution at $298\,K$ temperature for the following cell: $Pt_{(s)} \mid H_2 \,(1\,bar) \mid HCl\,(xM) \parallel Cu^{2+}\,(0.02\,M) \mid Cu_{(s)}$. Given that the standard cell potential $E^{\circ}_{cell} = 0.34\,V$ and the measured cell potential $E_{cell} = 0.45\,V$.

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