Chemistry, 02.10.2020 14:01 ROYALDRAGONX

Explain the electrolysis of cuso4
using carbon electrode

Answers: 2

Show answers

Answers

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 $10^5\; \rm Pa$ , andA 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:

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

Answer from: Quest

They carry out the exchange of gases as we breathe ( oxygen and carbon dioxide).

Answer from: Quest

the ideal gas law equation is as follows

pv = nrt

where p - pressure

v - volume

n - number of moles

r - universal gas constant

t - temperature

so if the pressure, volume and temperature are already known

we are left with n and r

since r is the universal gas constant that has a known fixed value then r too is known

so we are left with 'n'

once we know temperature volume and pressure

we can find the number of moles of gas present using the ideal gas law equation

this is not my own answer, this is an answer for the same question by the user blahblahmali. i do not claim ownership of this account.

Another question on Chemistry

Chemistry, 21.06.2019 20:00

When a comet collides with earth, it adds material to our planet and causes great damage. therefore, a collision like this is a a. destructive force b. constructive force c. geologic process and event d. constructive and destructive force

Answers: 1

Answer

Chemistry, 22.06.2019 02:10

What approach is required to balance the objectives of sustainable development? balancing the objectives of sustainable development requires a(n) .

Answers: 3

Answer

Chemistry, 22.06.2019 20:00

State one important difference between a physical change and a chemical change?

Answers: 1

Answer

Chemistry, 23.06.2019 00:30

What is calcium oxide+diphosphorus pentoxide--> calcium phosphate balanced