Question

In the electrolysis of copper sulfate solution using copper electrodes, which one of the following reactions takes place at the cathode?

• A. $2H_{2}O+2e^{-}\to H_{2}O+2O{H}^{-}$
• B. $Cu\to C{u}^{2+}+2e^{-}$
• C. $2H^{+}+2e^{-}\to H_2$
• D. $C{u}^{2+}+2e^{-}\to Cu$

Answer 1

Answer: (D)

$C{u}^{2+}+2e^{-}\to Cu$

${Cu}^{2+}+2e^{-}⟶Cu{E}_{red}=0.34H^{+}+e^{-}⟶\frac{1}{2}{H}_2{E}_{red}=0E_{{Cu}^{2+}/Cu}>E_{H^{+}/H_2}$

$\therefore$ reduction of ${Cu}^{2+}/Cu$ occurs at cathode.