Learning check
Once you have watched the video, check your learning with this quiz.
The reaction between copper and chlorine
Copper reacts with chlorine gas and forms cupric chloride, CuCl2.
Cu(s) + Cl2(g) → CuCl2(s)
- The reaction is spontaneous and exothermic.
- The electrons perform work.
- What happens if we run the reaction ”backwards”?
- Electrolysis = splitting (lysis) with the help of electricity.
Electrolysis of cupric chloride, CuCl2
Electrolysis of cupric chloride.
Cathode reaction (reduction)
- Electrons are pushed into the cathode
- Aqueous copper ions are reduced to solid copper:
- Cu2+(aq) + 2e– → Cu(s)
Anode reaction (oxidation)
- Electron shortage
- Chloride ions pulled towards the anode, oxidized to chlorine gas:
- Cl–(aq) → Cl2(g) + 2e–
Production of sodium hydroxide
Sodium hydroxide is produced by electrolysis of sodium chloride.
Anode reaction
Similar to the electrolysis of cupric chloride:
- Electron shortage in the anode
- Chloride ions pulled towards the anode, oxidized to chlorine gas:
- Cl–(aq) → Cl2(g) + 2e–
Cathode reaction
Two possibilities:
| 1. Na+(aq) + e– → Na(s) | \(e^0_{\text{Na}} = -2.71 \text{V}\) |
| 2. 2H2O + 2e– → H2(g) + 2OH–(aq) | \(e^0_{\text{H}_2\text{O}} = -0.83 \text{V}\) |
It can only be the second one, because:
- Any sodium atoms that form would immediately react with water:
- 2Na(s) + 2H2O → 2Na+(aq) + 2OH–(aq) + H2(g)
- The standard electrode potential for water is greater than the electrod potential for sodium, making the water a stronger oxidizing agent.
- \(e^0_{\text{H}_2\text{O}} > e^0_{\text{Na}}\) ⇒ Reaction 2 takes place rather than reaction 1.
Total reaction
2NaCl(aq) + 2H2O → Cl2(g) + H2(g) + 2NaOH(aq)
The concentrated sodium hydroxide solution is let out through a pipe.
