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Question

Chemistry Question on Electrochemistry

If the standard electrode potential of Cu2+/CuCu^{2+}/ Cu electrode is 0.34 V, what is the electrode potential of 0.01 M concentration of Cu2+Cu^{2+}?(T = 298 K)

A

0.399 V

B

0.281 V

C

0.222 V

D

0176 V

Answer

0.281 V

Explanation

Solution

Cu2++2e>CuCu^{2+} + 2e^- {->} Cu At 298K298\,K E=E00.0591nlog1[Mn+]E = E^0 - \frac{0.0591}{n} log \frac{1}{[M^{n+}]} ECu2+/Cu0.05912log1102E_{{Cu}^{2+}/Cu} - \frac{0.0591}{2} log \frac{1}{10^2} =(0.340.0591)V= (0.34 - 0.0591) V =0.2809V = 0.2809\,V