If the Zn^{2+}/Zn electrode is diluted to 100 | vedclass

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(B) The electrode potential for $Zn^{2+}/Zn$ is given by the Nernst equation: $E = E^0 - \frac{0.0591}{n} \log \frac{1}{[Zn^{2+}]}$.When the concentra

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Decrease of $59 \ mV$

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(B) The electrode potential for $Zn^{2+}/Zn$ is given by the Nernst equation: $E = E^0 - \frac{0.0591}{n} \log \frac{1}{[Zn^{2+}]}$.When the concentration is diluted $100$ times,the new concentration becomes $[Zn^{2+}]' = \frac{[Zn^{2+}]}{100}$.The change in potential is $\Delta E = E' - E = - \frac{0.0591}{2} \log \frac{1}{[Zn^{2+}]/100} - (- \frac{0.0591}{2} \log \frac{1}{[Zn^{2+}]})$.$\Delta E = - \frac{0.0591}{2} (\log \frac{100}{[Zn^{2+}]} - \log \frac{1}{[Zn^{2+}]}) = - \frac{0.0591}{2} \log(100) = - \frac{0.0591}{2} 	imes 2 = -0.0591 \ V$.Thus,the electrode potential decreases by $59 \ mV$.

If the $Zn^{2+}/Zn$ electrode is diluted to $100$ times,what is the change in the electrode potential?

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