What is the potential of a cell containing tw | vedclass

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(C) The cell reaction is a concentration cell: $H^{+} (0.025 \ M) 	o H^{+} (10^{-8} \ M)$.Using the Nernst equation for a concentration cell at $298

Vedclass · Accepted Answer

$0.38$

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(C) The cell reaction is a concentration cell: $H^{+} (0.025 \ M) 	o H^{+} (10^{-8} \ M)$.Using the Nernst equation for a concentration cell at $298 \ K$:$E_{cell} = E^{\circ}_{cell} - \frac{0.0591}{n} \log \frac{[H^{+}]_{cathode}}{[H^{+}]_{anode}}$Since it is a hydrogen concentration cell,$E^{\circ}_{cell} = 0 \ V$ and $n = 1$.$E_{cell} = 0 - 0.0591 \log \frac{10^{-8}}{0.025}$$E_{cell} = -0.0591 \log (4 	imes 10^{-7})$$E_{cell} = -0.0591 (\log 4 + \log 10^{-7})$$E_{cell} = -0.0591 (0.602 - 7) = -0.0591 (-6.398) \approx 0.378 \ V \approx 0.38 \ V$.

What is the potential of a cell containing two hydrogen electrodes,the negative one in contact with $10^{-8} \ M \ H^{+}$ and the positive one in contact with $0.025 \ M \ H^{+}$? (in $V$)

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