Question

Explain the electrolysis of cuso4 using carbon electrode

asked Jan 16, 2021 149k views

1 Answer

Pratik Bhoir · Answer 1 · 2021-01-22T21:14:36+0000

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.

Step-by-step 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:

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