The solution of CuSO4, in which copper rod is immersed, is diluted to 10 times. The reduction electr — Electrochemistry Chemistry Question

💡 Solution & Explanation

The reduction reaction is $Cu^{2+} + 2e^- \rightarrow Cu$. $E = E^\circ - (0.0591/2) \log(1/[Cu^{2+}])$. Diluting 10 times makes the new concentration $[Cu^{2+}]/10$. The change in the log term is $\log(10) = 1$. The change in potential is $-(0.0591/2) \times 1 = -0.02955$ V. It decreases by approximately 0.0295 V.