At 298 K,if the emf of the cell correspondin | vedclass

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(C) The cell reaction is $Zn_{(s)} + 2H^+_{(aq)} \rightarrow Zn^{2+}(0.01 \ M) + H_{2(g)}(1 \ atm)$.$E^o_{cell} = E^o_{cathode} - E^o_{anode} = E^o_{H

Vedclass · Accepted Answer

$9$

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(C) The cell reaction is $Zn_{(s)} + 2H^+_{(aq)} ightarrow Zn^{2+}(0.01 \ M) + H_{2(g)}(1 \ atm)$.$E^o_{cell} = E^o_{cathode} - E^o_{anode} = E^o_{H^+|H_2} - E^o_{Zn^{2+}|Zn} = 0 - (-0.76) = 0.76 \ V$.Using the Nernst equation: $E_{cell} = E^o_{cell} - \frac{0.06}{n} \log \frac{[Zn^{2+}] \cdot P_{H_2}}{[H^+]^2}$.Here $n = 2$,$E_{cell} = 0.28 \ V$,$[Zn^{2+}] = 0.01 \ M$,$P_{H_2} = 1 \ atm$.$0.28 = 0.76 - \frac{0.06}{2} \log \frac{0.01 	imes 1}{[H^+]^2}$.$0.28 - 0.76 = -0.03 \log \frac{0.01}{[H^+]^2}$.$-0.48 = -0.03 \log \frac{0.01}{[H^+]^2}$.$16 = \log \frac{0.01}{[H^+]^2} = \log 0.01 - \log [H^+]^2 = -2 - 2 \log [H^+]$.Since $pH = -\log [H^+]$,we have $16 = -2 + 2(pH)$.$18 = 2(pH) \implies pH = 9$.

At $298 \ K$,if the $emf$ of the cell corresponding to the reaction,$Zn_{(s)} + 2H^+_{(aq)} \rightarrow Zn^{2+}(0.01 \ M) + H_{2(g)}(1 \ atm)$ is $0.28 \ V$,then the $pH$ of the solution at the hydrogen electrode is (Given: $\frac{2.303 \ RT}{F} = 0.06 \ V$,$E^o_{Zn^{2+}|Zn} = -0.76 \ V$)

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