Predicting the feasibility of redox reactions using standard electrode potentials

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 I_2(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 I_2(aq)

Cu_(s) + I_2(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

1. 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) I_2(aq) / 2I^–_(aq) and Cr₂O₇^2- _(aq) / Cr³⁺_(aq)

(ii) Fe³⁺_(aq) / Fe²⁺_(aq) and H₂O_2(aq) / 2H₂O_(l)

(iii) IO₃^–_(aq) / ½I_2(aq) and Br_2(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

–

2. 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.

–

3. A sample of a nickel-copper alloy was dipped in sulphuric acid. Use the standard electrode potential data below to predict and explain what happened.

2H+(aq) + 2e-  ⇌   H2(g)	E⦵ =  0.00 V
Ni2+(aq) + 2e-  ⇌  Ni(s)	E⦵ = -0.25 V
Cu2+(aq) + 2e-  ⇌  Cu(s)	E⦵ = +0.34 V

–

4. Use the electrochemical data below to identify an acid that is able to oxidise copper.  Explain your reasoning and write an equation for the redox reaction that occurs.

2H+(aq) + 2e-  ⇌   H2(g)	E⦵ =  0.00 V
SO42-(aq) + 4H+(aq) + 2e-  ⇌  SO2(g)  + 2H2O(l)	E⦵ = +0.17 V
Cu2+(aq) + 2e-  ⇌  Cu(s)	E⦵ = +0.34 V
NO3-(aq) + 4H+(aq) + 3e- ⇌ NO(aq) + 2H2O(l)	E⦵ = +0.96 V

–

5. With reference to the standard electrode data below, explain why zinc reacts with acidified chromate(VI) ions to form chromium(II) ions in two steps. Write a fully balanced equation for each reaction that occurs. 

Zn2+(aq)  + 2e-  ⇌  Zn(s)	E⦵ = -0.76 V
Cr2O72-(aq) + 14H+(aq) + 6e-  ⇌  2Cr3+(aq) + 7H2O(l)	E⦵ = +1.33 V
Cr3+(aq) + e-  ⇌  Cr2+(aq)	E⦵ = -0.42 V

Answers

1. (i) Cr₂O₇^2- _(aq) + 6I^–_(aq) + 14H⁺_(aq) ⇌ 2Cr³⁺_(aq) + I_2(aq) + 7H₂O_(l)

(ii) H₂O_2(aq) + 2H⁺_(aq) + 2Fe²⁺_(aq) ⇌ 2H₂O_(l) + 2Fe³⁺_(aq)

(iii) 2IO₃^–_(aq) + 10Br^–_(aq) + 12H+_(aq) ⇌ I_2(aq) + 5Br_2(aq) + 6H₂O_(l)

–

2. The planned oxidation of chloride ions is not feasible because the IO₃^–_(aq) / ½I_2(aq) half cell has a less positive standard electrode potential than the Cl_2(g) / 2Cl^–_(aq) half cell. This means that the IO₃^–_(aq) / ½I_2(aq) half cell will be the oxidation half cell and the Cl_2(g) / 2Cl^–_(aq) half cell will be the reduction half cell.

EXAM TIP: be careful with your wording! E^⦵ belongs to a specified half cell, not an oxidising agent or H+ ions.

–

3. The Ni²⁺ / Ni half cell has a less positive E^⦵ than the H⁺ / H₂ half cell, so Ni will be oxidised to Ni²⁺ by the H⁺ from the acid. Ni²⁺ moves into solution.  H⁺ ions cannot oxidise Cu to Cu²⁺ as the Cu²⁺ / Cu half cell has a more positive E^⦵ than the H⁺ / H₂ half cell.

–

4. For copper to be oxidised the Cu²⁺ / Cu half cell must have a less positive E^⦵ than the E^⦵ for the acid half cell. The NO₃^–, H⁺ / NO half cell has an E^⦵ of + 0.96V compared with + 0.34V for the Cu²⁺ / Cu half cell, so nitric acid would oxidise copper.

–

5. In the first step, Zn reduces Cr₂O₇^2- to Cr³⁺ as the Zn²⁺ / Zn half cell has a less positive standard electrode potential compared with the Cr₂O₇^2-, H⁺ / Cr³⁺ half cell. 

3Zn_(s) + Cr₂O₇^2-_(aq) + 14H⁺_(aq)  ⇾  3Zn²⁺ + 2Cr³⁺_(aq) + 7H₂O_(l) 

In the second step, Zn reduces Cr³⁺ to Cr²⁺ as E^⦵for the Cr³⁺ / Cr²⁺ half cell is more positive than the E^⦵for the Zn²⁺ / Zn half cell.

Zn_(s) + 2Cr³⁺_(aq) ⇾  Zn²⁺_(aq) + 2Cr²⁺_(aq)