The correct representation of Nernst's equati | vedclass

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Question

(A) The Nernst equation for the electrode reaction $M^{n+} + ne^- \rightarrow M(s)$ is given by: $E_{M^{n+}/M} = E^o_{M^{n+}/M} - \frac{RT}{nF} \ln \f

Vedclass · Accepted Answer

$E_{M^{n+}/M} = E^o_{M^{n+}/M} + \frac{0.0591}{n}\log ([M^{n+}])$

Vedclass · Answer

(A) The Nernst equation for the electrode reaction $M^{n+} + ne^- \rightarrow M(s)$ is given by: $E_{M^{n+}/M} = E^o_{M^{n+}/M} - \frac{RT}{nF} \ln \frac{1}{[M^{n+}]}$ Using $\ln x = 2.303 \log x$,we get: $E_{M^{n+}/M} = E^o_{M^{n+}/M} - \frac{2.303RT}{nF} \log \frac{1}{[M^{n+}]}$ $E_{M^{n+}/M} = E^o_{M^{n+}/M} + \frac{2.303RT}{nF} \log [M^{n+}]$ At $298 \ K$,the value of $\frac{2.303RT}{F} \approx 0.0591 \ V$. Therefore,$E_{M^{n+}/M} = E^o_{M^{n+}/M} + \frac{0.0591}{n} \log [M^{n+}]$. Thus,option $A$ is correct.

The correct representation of Nernst's equation for the reduction of a metal ion $M^{n+}$ to metal $M$ is:

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