If a solution of Cu^{+2}/Cu at 298 , K is dil | vedclass

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(B) The reduction reaction is: $Cu^{+2} + 2e^- 	o Cu$According to the Nernst equation: $E_{Cu^{+2}/Cu} = E^{\circ}_{Cu^{+2}/Cu} - \frac{0.059}{2} \lo

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$59 \, mV$ decrease

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(B) The reduction reaction is: $Cu^{+2} + 2e^- 	o Cu$According to the Nernst equation: $E_{Cu^{+2}/Cu} = E^{\circ}_{Cu^{+2}/Cu} - \frac{0.059}{2} \log \frac{1}{[Cu^{+2}]}$When the solution is diluted $100$ times,the concentration $[Cu^{+2}]$ becomes $[Cu^{+2}]/100$. The new potential $E'$ is:$E' = E^{\circ}_{Cu^{+2}/Cu} - \frac{0.059}{2} \log \frac{1}{[Cu^{+2}]/100}$$E' = E^{\circ}_{Cu^{+2}/Cu} - \frac{0.059}{2} (\log \frac{1}{[Cu^{+2}]} + \log 100)$$E' = E_{Cu^{+2}/Cu} - \frac{0.059}{2} 	imes 2$$E' = E_{Cu^{+2}/Cu} - 0.059 \, V$Thus,the electrode potential decreases by $0.059 \, V$ or $59 \, mV$.

If a solution of $Cu^{+2}/Cu$ at $298 \, K$ is diluted $100$ times,how will the electrode potential change?

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