Question

The minimum voltage (in V) required to bring about the electrolysis of 1M copper (II) sulphate solution at 298 K is (Given E$^\circ_{Cu^{2+}/Cu}$ = 0.34V and E$^\circ_{H_2O/H^+}$ = -1.23V)

• A. 1.57
• B. 0.89
• C. -0.89
• D. -1.57

Hint:

In electrolysis, the cathode is where reduction occurs (more positive E$^\circ$), and the anode is where oxidation occurs (more negative E$^\circ$). The minimum voltage required is the absolute value of the cell potential for the non-spontaneous electrolysis reaction.

Answer 1

The Correct Option is B

Step 1: Identify the possible reactions at the cathode and anode.
In the electrolysis of aqueous CuSO$_4$ solution, the possible reduction reactions at the cathode are: % Option (A) Cu$^{2+}$(aq) + 2e$^-$ $\rightarrow$ Cu(s) \quad E$^\circ$ = +0.34 V % Option (B) 2H$^+$(aq) + 2e$^-$ $\rightarrow$ H$_2$(g) \quad E$^\circ$ = 0.00 V (in neutral or acidic solution, water reduction is often considered as 2H$_2$O + 2e$^-$ $\rightarrow$ H$_2$(g) + 2OH$^-$) Considering the given standard reduction potential for water in acidic conditions (although the solution is CuSO$_4$, water electrolysis is always a possibility): 2H$_2$O + 2e$^-$ $\rightarrow$ H$_2$(g) + 2OH$^-$ For which we are given E$^\circ_{H_2O/H^+} = -1.23$V, which is related to the oxidation of water. The reduction of water to hydrogen in neutral solution is: 2H$_2$O(l) + 2e$^-$ $\rightarrow$ H$_2$(g) + 2OH$^-$ (E$^\circ$ = -0.83 V at pH 7) The possible oxidation reactions at the anode are: % Option (A) 2SO$_4^{2-}$(aq) $\rightarrow$ S$_2$O$_8^{2-}$(aq) + 2e$^-$ \quad E$^\circ$ = +2.01 V % Option (B) 2H$_2$O(l) $\rightarrow$ O$_2$(g) + 4H$^+$(aq) + 4e$^-$ \quad E$^\circ$ = +1.23 V
Step 2: Determine the reactions that will occur based on standard electrode potentials.
At the cathode, the reduction with the higher standard reduction potential is favored. Comparing E$^\circ_{Cu^{2+}/Cu}$ = +0.34 V and E$^\circ_{H_2O/H^-}$ (related to hydrogen evolution), copper will be deposited. At the anode, the oxidation with the lower standard oxidation potential (higher standard reduction potential for the reverse reaction) is favored. Comparing the oxidation of sulfate and water, the oxidation of water is kinetically and thermodynamically more feasible under normal conditions. So, the overall electrolysis reaction is approximately: Cu$^{2+}$(aq) + H$_2$O(l) $\rightarrow$ Cu(s) + $\frac{1}{2}$O$_2$(g) + 2H$^+$(aq)
Step 3: Calculate the minimum voltage required for electrolysis.
The minimum voltage required for electrolysis is related to the standard cell potential (E$^\circ_{cell}$) of the non-spontaneous reaction: E$^\circ_{cell}$ = E$^\circ_{cathode}$ - E$^\circ_{anode}$ E$^\circ_{cathode}$ (reduction of Cu$^{2+}$) = +0.34 V E$^\circ_{anode}$ (oxidation of H$_2$O) = +1.23 V E$^\circ_{cell}$ = 0.34 V - 1.23 V = -0.89 V The minimum voltage required to drive this non-spontaneous reaction is the magnitude of E$^\circ_{cell}$ with a positive sign: Minimum voltage = +0.89 V Final Answer: \[ \boxed{+0.89} \]