Question

Explain the electrolysis of cuso4
using carbon electrode
​

Answer 1

Answer:

If this solution is under standard conditions, then it will be possible to explain the products of this electrolysis ($\rm Cu \, (s)$, $\rm O_2\, (g)$, $\rm H^{+}\; (aq)$) with reference to the standard reduction potentials.

Explanation:

Assume that this solution is under standard conditions, including:

• A concentration of $1.00\; \rm mol \cdot L^{-1}$,
• A pressure of $10^5\; \rm Pa$, and
• A temperature of $25\; \rm ^\circ C$.

This $\rm CuSO_4$ solution (in water) would include the following species:

• $\rm Cu^{2+}$ ions,
• $\rm {SO_4}^{2-}$ ions, and
• $\rm H_2O$ molecules.

(The carbon electrode is relatively inert and typically won't take part in the chemical reaction.)

Consider the electrolysis of this solution as the sum of the reduction half-reaction and the oxidation half-reaction. Look up the standard reduction potential for half-reactions involving these species:

$\rm {\bf Cu^{2+} \; (aq)} + e^{-} \rightleftharpoons Cu^{-}\; \rm (aq)$. Standard reduction potential: $E^\circ = 0.159\; \rm V$.

$\rm {\bf Cu^{2+} \; (aq)} + 2\; e^{-} \rightleftharpoons Cu\; \rm (s)$. Standard reduction potential: $E^\circ = 0.3419\; \rm V$.

$\rm {\bf {SO_4}^{2-}} + {\bf H_2O\; (l)} + 2\; e^{-} \rightleftharpoons {SO_3}^{2-} + 2\; OH^{-}$. Standard reduction potential: $E^\circ = -0.936\; \rm V$.

$\rm 2\; {\bf H_2O\, (l)} + 2\; e^{-} \rightleftharpoons H_2\, (g) + 2\; OH^{-}\, (aq)$. Standard reduction potential: $E^{\circ} = -0.828\; \rm V$.

$\rm O_2 \, (g) + 4\; H^{+}\, (aq) + 4\; e^{-} \rightleftharpoons 2\; {\bf H_2O\, (l)}$. Standard reduction potential: $E^{\circ} = 1.229\; \rm V$.

(Species found in this solution are shown in bold typeface.)

Reduction Half-reaction

Note, that in the first four reactions, species from this solution are all on the same side as $\rm e^{-}$. In other words, in those four reactions, these species would gain electrons and are reduced. These reactions are thus plausible reduction half-reactions. The standard electrode potential of each half reaction will be the same as the standard reduction potential. Because reduction takes place at the cathode, these potentials are denoted as $E^\circ(\text{cathode})$.

$\rm {\bf Cu^{2+} \; (aq)} + e^{-} \rightleftharpoons Cu^{-}\; \rm (aq)$. $E^\circ(\text{cathode}) = 0.159\; \rm V$.

$\rm {\bf Cu^{2+} \; (aq)} + 2\; e^{-} \rightleftharpoons Cu\; \rm (s)$. $E^\circ(\text{cathode}) = 0.3419\; \rm V$.

$\rm {\bf {SO_4}^{2-}} + {\bf H_2O\; (l)} + 2\; e^{-} \rightleftharpoons {SO_3}^{2-} + 2\; OH^{-}$. $E^\circ(\text{cathode}) = -0.936\; \rm V$.

$\rm 2\; {\bf H_2O\, (l)} + 2\; e^{-} \rightleftharpoons H_2\, (g) + 2\; OH^{-}\, (aq)$. $E^{\circ}(\text{cathode}) = -0.828\; \rm V$.

The only reduction half-reaction that will take place will be the one with the most positive (least negative) electrode potential. Among these four plausible half-reaction, that reaction would be the reduction of $\rm Cu^{2+}\; (aq)$ to $\rm Cu\, (s)$ (elemental copper,) $\rm {\bf Cu^{2+} \; (aq)} + 2\; e^{-} \rightleftharpoons Cu\; \rm (s)$, with an electrode potential of $0.3419\; \rm V$. That explains why metallic copper will deposit on the cathode.

Oxidation Half-reaction

On the other hand, the reaction $\rm 2\; {\bf H_2O\, (l)} \rightleftharpoons O_2 \, (g) + 4\; H^{+}\, (aq) + 4\; e^{-}$ seems to be the only one that involves species from this solution on the opposite side of $\rm e^{-}$. The actual reaction would take place in the other direction:

$\rm 2\; {\bf H_2O\, (l)} \rightleftharpoons O_2 \, (g) + 4\; H^{+}\, (aq) + 4\; e^{-}$.

The corresponding electrode potential will be the opposite of the standard reduction potential. Because this reaction corresponds to oxidation, it will happen at the anode.

$E^\circ(\text{anode}) = -E^{\circ} = -1.229\; \rm V$.

That explains why $\rm O_2\, (g)$ will be produced at the anode.

Overall Reaction

Multiply coefficients in the reduction half-reaction by two and combine the two half-reactions. Make sure that electrons are eliminated from both sides of this equation:

$\rm 2\; Cu^{2+}\, (aq) + 2\; H_2O\, (l) \to 2\; Cu\, (s) + O_2\, (g) + 4\; H^{+}\, (aq)$.

$\rm {SO_4}^{2-}\, (aq)$ would be the spectator ion. Add that to the reaction to obtain the chemical equation of this electrolysis:

$\rm 2\; CuSO_4\, (aq) + 2\; H_2O\, (l) \rightleftharpoons 2\; Cu\, (s) + O_2\, (g) + 2\; H_2SO_4$.