The cell potential for the given cell at 298 | vedclass

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Question

(B) The cell reaction is: $H_{2(g)} + Cu^{2+}_{(aq)} \rightarrow 2 H^{+}_{(aq)} + Cu_{(s)}$Using the Nernst equation: $E_{cell} = E^{\ominus}_{cell} -

Vedclass · Accepted Answer

$7$

Vedclass · Answer

(B) The cell reaction is: $H_{2(g)} + Cu^{2+}_{(aq)} \rightarrow 2 H^{+}_{(aq)} + Cu_{(s)}$Using the Nernst equation: $E_{cell} = E^{\ominus}_{cell} - \frac{0.06}{n} \log \frac{[H^{+}]^2}{[Cu^{2+}]}$Here,$n = 2$,$E_{cell} = 0.31 \, V$,$E^{\ominus}_{cell} = E^{\ominus}_{Cu^{2+}/Cu} - E^{\ominus}_{H^{+}/H_2} = 0.34 - 0 = 0.34 \, V$.Given $pH = 3$,so $[H^{+}] = 10^{-3} \, M$.Substituting the values: $0.31 = 0.34 - \frac{0.06}{2} \log \frac{(10^{-3})^2}{[Cu^{2+}]}$$-0.03 = -0.03 \log \frac{10^{-6}}{[Cu^{2+}]}$$1 = \log \frac{10^{-6}}{[Cu^{2+}]}$$10^1 = \frac{10^{-6}}{[Cu^{2+}]}$$[Cu^{2+}] = 10^{-7} \, M$.Comparing with $10^{-x} \, M$,we get $x = 7$.

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