?If \(\alpha=\frac{1}{2}\) , an electrode interface is unable to rectify alternating

Chapter 19, Problem P19D.6

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If \(\alpha=\frac{1}{2}\) , an electrode interface is unable to rectify alternating current because the current density curve is symmetrical about \(\eta=0\). When \(\alpha \neq \frac{1}{2}\) , the magnitude of the current density depends on the sign of the overpotential, and so some degree of ‘faradaic rectification’ may be obtained. (a) Suppose that the overpotential varies as \(\eta=\eta_{0} \cos \omega t\). Derive an expression for the mean flow of current (averaged over a cycle) for general \(\alpha\), and confirm that the mean current is zero when \(\alpha=\frac{1}{2}\) . In your calculations work in the limit of small \(\eta_{0}\) but to second order in \(\eta_{0} F / R T\) (that is, when expanding the exponentials in the Butler–Volmer equation, retain up to the second-order terms). (b) Calculate the mean direct current at \(25^{\circ} \mathrm{C}\) for a \(1.0 \mathrm{\ cm}^{2}\) hydrogen– platinum electrode with \(\alpha=0.38\) when the overpotential varies between \(\pm 10 \mathrm{\ mV}\) at 50 Hz.

Text Transcription:

alpha=1/2

eta=0

alpha neq 1/2

eta=eta_0 cos omega t

alpha

eta_0

eta_0 F/RT

25^circ C

1.0 cm^2

alpha=0.38

pm 10 mV