If the concentration of a Zn^{2+} solution is | vedclass

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(A) The Nernst equation for the $Zn/Zn^{2+}$ electrode is given by: $E_{Zn/Zn^{2+}} = E^{\circ}_{Zn/Zn^{2+}} - \frac{0.059}{2} \log [Zn^{2+}]$.When th

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$0.03 \, V$ increase

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(A) The Nernst equation for the $Zn/Zn^{2+}$ electrode is given by: $E_{Zn/Zn^{2+}} = E^{\circ}_{Zn/Zn^{2+}} - \frac{0.059}{2} \log [Zn^{2+}]$.When the concentration is diluted $10$ times,the new concentration becomes $[Zn^{2+}]' = \frac{[Zn^{2+}]}{10}$.The new potential is $E' = E^{\circ}_{Zn/Zn^{2+}} - \frac{0.059}{2} \log \left( \frac{[Zn^{2+}]}{10} \right)$.$E' = E^{\circ}_{Zn/Zn^{2+}} - \frac{0.059}{2} (\log [Zn^{2+}] - \log 10)$.$E' = E^{\circ}_{Zn/Zn^{2+}} - \frac{0.059}{2} \log [Zn^{2+}] + \frac{0.059}{2} \log 10$.Since $\log 10 = 1$,we get $E' = E_{Zn/Zn^{2+}} + \frac{0.059}{2} = E_{Zn/Zn^{2+}} + 0.0295 \, V \approx E_{Zn/Zn^{2+}} + 0.03 \, V$.Thus,the potential increases by $0.03 \, V$.

If the concentration of a $Zn^{2+}$ solution is diluted $10$ times,what is the change in the potential of the $Zn/Zn^{2+}$ electrode?

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