A redox reaction consists of two half reactions – an oxidation and a reduction reaction.
If we take a pair of half reactions we can use the standard electrode potential, E⦵, for each to determine which will form the oxidation half cell and which will form the reduction half cell. If we know this we can predict the direction / feasibility of the reaction.
The key is that the more positive of the two E⦵ will always form the reduction half cell and the more negative will always form the oxidation half cell.
E.g. Could Cu(s) reduce I2(aq)?
- Use a data table to look up the half reaction and E⦵ for copper and iodine
- The E⦵ data tells us that Cu will be oxidised to Cu2+ and I2 will be reduced to 2I–
- The answer is YES! Cu(s) will reduce I2(aq)
Cu(s) + I2(aq) ⇾ Cu2+(aq) + 2I–(aq)
E.g. Could Fe2+(aq) reduce Zn2+(aq)?
- Use a data table to look up the half reaction and E⦵ for iron(II) ions and zinc
- The E⦵ data tells us that Fe3+ will be reduced to Fe2+ and Zn will be oxidised to Zn2+
- The reaction that would occur is
Fe3+(aq) + Zn(s) ⇾ Fe2+(aq) + Zn2+(aq)
- The answer is NO! Fe2+(aq) cannot reduce Zn2+(aq). The reaction can only proceed in the direction shown in the equation above.
If we predict that a reaction is feasible, it may in reality have an exceptionally slow rate of reaction due to a high activation enthalpy (in which case a catalyst may be helpful). If we predict that a redox reaction is not feasible, there’s nothing we can do – electrons will never flow spontaneously from a more positive potential to a less positive one.
If you are not yet that confident writing redox equations from half equations, then start here 😊.
Practice questions
- Use the standard electrode data in the table below to write a balanced overall equation for the reactions that happen when the following half cells are connected:
(i) I2(aq) / 2I–(aq) and Cr2O72- (aq) / Cr3+(aq)
(ii) Fe3+(aq) / Fe2+(aq) and H2O2(aq) / 2H2O(l)
(iii) IO3–(aq) / ½I2(aq) and Br2(aq) / 2Br–(aq)
| Half reaction | E⦵ / V |
|---|---|
| Cr2O72- (aq) + 14H+(aq) + 6e- ⇌ 2Cr3+(aq) + 7H2O(l) | + 1.36 |
| Fe3+(aq) + e- ⇌ Fe2+(aq) | + 0.77 |
| Cl2(g) + 2e- ⇌ 2Cl–(aq) | + 1.36 |
| I2(aq) + 2e- ⇌ 2I–(aq) | + 0.54 |
| Br2(aq) + 2e- ⇌ 2Br–(aq) | + 1.07 |
| H2O2(aq) + 2H+(aq) + 2e- ⇌ 2H2O(l) | + 1.78 |
| IO3–(aq) + 6H+(aq) + 5e- ⇌ ½I2(aq) + 3H2O(l) | + 1.19 |
- A student plans to oxidise chloride ions to chlorine with iodate(V) ions. Explain whether this redox reaction is feasible with respect to the standard electrode data.
Answers
- (i) Cr2O72- (aq) + 6I–(aq) + 14H+(aq) ⇌ 2Cr3+(aq) + I2(aq) + 7H2O(l)
(ii) H2O2(aq) + 2H+(aq) + 2Fe2+(aq) ⇌ 2H2O(l) + 2Fe3+(aq)
(iii) 2IO3–(aq) + 10Br–(aq) + 12H+(aq) ⇌ I2(aq) + 5Br2(aq) + 6H2O(l)
- The planned oxidation of chloride ions is not feasible because the IO3–(aq) / ½I2(aq) half cell has a less positive standard electrode potential than the Cl2(g) / 2Cl–(aq) half cell. This means that the IO3–(aq) / ½I2(aq) half cell will be the oxidation half cell and the Cl2(g) / 2Cl–(aq) half cell will be the reduction half cell.
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