Worked example 1
Using the Data Booklet, predict the spontaneity of the following pair of solutions when they are mixed.
Solutions: Acidified aqueous sodium dichromate(VI) and aqueous iron(II) sulfate
Step 1: To make it easier to pinpoint the ions or molecules involved, rewrite the substances in their chemical formulas: Na₂Cr₂O₇ and FeSO₄
Ions present : Na⁺, H⁺, Cr₂O₇²⁻, Fe²⁺, SO₄²⁻
H⁺ is present as an ion as Na₂Cr₂O₇ is acidified. We ignore water as a potential oxidising or reducing agent.
Step 2: Select from the data booklet, half equations with the ions highlighted above.
Do not flip the equations, copy them directly from the data booklet to prevent confusion in step 3, later on. Underlining the reagent would help you better identify the oxidising or reducing agent for the next step.
|1. Cr₂O₇²⁻ + 14H⁺ + 6e⁻ ⇌ Cr³⁺ + 7H₂O||+1.33|
|2. 2H⁺ + 2e⁻ ⇌ H₂||0.00|
|3. Na⁺ + e⁻ ⇌ Na||-2.71|
|1. Cr₂O₇²⁻ + 14H⁺ + 6e⁻ ⇌ Cr³⁺ + 7H₂O|
|2. 2H⁺ + 2e⁻ ⇌ H₂|
|3. Na⁺ + e⁻ ⇌ Na|
|4. Fe³⁺ + e⁻ ⇌ Fe²⁺||+0.77|
|5. Fe²⁺ + 2e⁻ ⇌ Fe||-0.44|
|6. SO₄²⁻ + 4H⁺ + 2e⁻ ⇌ SO₂ + 2H₂O||+0.17|
|7. S₂O₈²⁻ + 2e⁻ ⇌ 2SO₄²⁻||+2.01|
|4. Fe³⁺ + e⁻ ⇌ Fe²⁺|
|5. Fe²⁺ + 2e⁻ ⇌ Fe|
|6. SO₄²⁻ + 4H⁺ + 2e⁻ ⇌ SO₂ + 2H₂O|
|7. S₂O₈²⁻ + 2e⁻ ⇌ 2SO₄²⁻|
Step 3: Reject and identify the two appropriate half-equations for the reaction. Oxidising agents (undergoing reduction) should be found on the left side of the half-equation and have the most positive Eθ. Reducing agents (undergoing oxidation) should be found on the right side of the half-equation and have the least positive Eθ.
Cr₂O₇²⁻ is the oxidising agent as it appears on the left side of the equation and has a more positive Eθ than H⁺ and Na⁺. Hence, reject equations 2 and 3.
Since Cr₂O₇²⁻ is the oxidising agent, the ions in FeSO₄ would act as the reducing agent and would be on the right side of the half-equation. Thus equations 5 and 6 should be rejected, since Fe²⁺ and SO₄²⁻ appear on the left side. Since the Eθ of equation 4 is more negative than that of 7, choose Fe²⁺ as the oxidising agent instead of SO₄²⁻. Hence, reject equation 7.
Step 4: Calculate the Eθcell
Eθcell = +1.33 - (+0.77) = +0.56 V
Since +0.56 V > 0, the reaction is spontaneous.
Only one ion is preferentially discharged, even if there is more than one ion at the electrode.
Three factors will determine which ion is preferentially discharged:
- 1. Position of ion in the electrochemical series
Cations with the most positive Eθ are preferentially reduced. Anions with the most negative Eθ are preferentially oxidised.
- 2. Concentration of ion
Some ions present in high concentration may be preferentially discharged despite its position in the electrochemical series.
E.g. In saturated NaCl(brine), Cl⁻ is preferentially oxidised even though the Eθ (O₂/H₂O) is more negative than Eθ(Cl₂/Cl⁻).
- 3. Nature of electrodes
While inert electrodes (graphite and platinum) do not affect the discharge of ions, reactive electrodes can participate in the oxidation process. Metals such as copper would be oxidised in place of the anions in the solution.
Electrolysis of aqueous solutions
It is easy to determine which ions are involved in the redox equation when it is the electrolysis of a single substance in the molten state. But when an aqueous substance is involved, the electrolysis of water must be considered as well.
In the ‘O’ level syllabus, the electrolysis of water involves the discharge of H⁺ and OH⁻ ions at the cathode and anode respectively. Water is assumed to self-dissociate, producing H⁺ and OH⁻ ions. While this is true, the concentration of both ions in pure water, as we have learnt in Acid-Base Equilibria, is very low at 10⁻⁷ mol dm⁻³ (25℃).
Hence more accurately, at the cathode, water can be reduced to form hydrogen gas.
2H₂O + 2e⁻ ⇌ H₂ + 2OH⁻
Whereas at the anode, water can be oxidised to form O₂(g).
2H₂O ⇌ O₂ + 4H⁺ + 4e⁻
Unless the electrolyte is acidic or alkaline, which will result in high concentrations of H⁺ and OH⁻ ions, assume that H⁺ and OH⁻ are not present for the purpose of predicting the reactions occurring at the electrodes.
Worked example 2
Give the ion-electron equations for the electrode reactions and the balanced overall equation in the electrolysis of molten NaCl using graphite electrodes and aqueous NaCl using platinum electrodes.
Set-up for molten NaCl:
Species present: Cl⁻ (l), Na⁺ (l)
At the anode,
2Cl⁻ (l) → Cl₂(g) + 2e⁻
At the cathode,
Na⁺ (l) + e⁻ → Na (l)
Overall equation: 2NaCl(l) → Cl₂(g) + 2Na(l)
Set up for aqueous NaCl:
Species present: Cl⁻ (aq), Na⁺ (aq), H₂O(l)
At the anode, all possible reactions:
|Cl₂ + 2e⁻ ⇌ 2Cl⁻||+1.36|
|O₂ + 4H⁺ + 4e⁻ ⇌ 2H₂O||+1.23|
|Cl₂ + 2e⁻ ⇌ 2Cl⁻|
|O₂ + 4H⁺ + 4e⁻ ⇌ 2H₂O|
Since the Eθ (O₂/H₂O) is more negative than the Eθ (Cl₂/Cl⁻), H₂O is preferentially oxidised.
Thus the reaction at the anode,
2H₂O (l) → O₂ (g) + 4H⁺ (aq) + 4e⁻
At the cathode, all possible reactions:
Since the Eθ (H₂O/H₂) is more positive than the Eθ (Na⁺/Na), H₂O is preferentially reduced,
Thus the reaction at the cathode would be as such:
2H₂O (l) + 2e⁻ → H₂(g) + 2OH⁻ (aq)
Hence, overall equation would be: 2H₂O (l) → 2H₂(g) + O₂(g)
|2H₂O + 2e⁻ ⇌ H₂ + 2OH⁻||-0.83|
|Na⁺ + e⁻ ⇌ Na||-2.71|
|2H₂O + 2e⁻ ⇌ H₂ + 2OH⁻|
|Na⁺ + e⁻ ⇌ Na|
Electrolytic purification of copper
This real life application of the electrolytic cell is commonly tested. When crude copper is extracted, it contains impurities such as silver and zinc. As we know, copper is used in making electrical wires and purifying copper improves its electrical conductivity.
Take note of the following things,
- 1. The impure copper is made the anode,
- 2. While pure copper is made the cathode.
- 3. Aqueous copper(II) sulfate is the electrolyte
- 4. Electron flows from the impure copper (anode) to the pure copper (cathode)
The set up is as such:
Paying attention to the half-equations and Eθ of the possible oxidation reactions:
|S₂O₈²⁻ + 2e⁻ ⇌ 2SO₄²⁻||+2.01|
|O₂ + 4H⁺ + 4e⁻ ⇌ 2H₂O||+1.23|
|Ag⁺ + e⁻ ⇌ Ag||+0.80|
|Cu²⁺ + 2e⁻ ⇌ Cu||+0.34|
|Zn²⁺ + 2e⁻ ⇌ Zn||-0.76|
|S₂O₈²⁻ + 2e⁻ ⇌ 2SO₄²⁻|
|O₂ + 4H⁺ + 4e⁻ ⇌ 2H₂O|
|Ag⁺ + e⁻ ⇌ Ag|
|Cu²⁺ + 2e⁻ ⇌ Cu|
|Zn²⁺ + 2e⁻ ⇌ Zn|
By regulating the operating voltage such that it is just sufficient for Cu to be oxidised, pure copper can be obtained.
At the anode, as Eθ (Zn²⁺/Zn) is more negative then Eθ (Cu²⁺/Cu), Zn will be oxidised. Hence both Cu and Zn dissolve into the solution as cations and migrate to the cathode.
Meanwhile, since the Eθ (Ag⁺/Ag) is more positive than Eθ (Cu²⁺/Cu), it will not be oxidised and remains as silver metal. As copper around the silver dissolves, it will drop off the electrode and form anode sludge.
Since the Eθ (S₂O₈²⁻/SO₄2-) and Eθ (O₂H₂O) are too positive, SO₄²⁻ and H₂O will remain unoxidised.
At the cathode, since Eθ (Cu²⁺ /Cu) is more positive than Eθ (Zn²⁺/Zn), Cu²⁺ will be preferentially reduced and deposited on the pure copper cathode. Zn²⁺ will remain in the solution.