Using electrode potentials

The problem and background knowledge

I think the main problem with questions involving oxidation and reduction and standard electrode potentials is that teachers often teach cell convention – and many teach it wrongly. This can be seen when examiners mark scripts as students often use the phrases ‘left hand side’ or ‘right hand side’ and have no clue about what they are referring to. Equations like Etotal = ELHS – ERHL are often used in completely the wrong way. A cell will produce the same amount of electricity whichever way round the two half-cells are physically connected so it really does not matter which is on ‘the right’ or ‘the left’. The use of ‘anode’ and cathode’ can also be confusing as the anode is the positive electrode in an electrolysis cell and yet is the negative electrode in a voltaic (or electrochemical) cell.

Cell convention is not specifically on the IB syllabus1 and to my mind adds a totally unnecessary complication to what is a very simple problem. I also prefer to use positive and negative electrode and do not refer to anode or cathode although the syllabus does use anode and cathode so students will need to know these words. It is probably best if they learn that oxidation always occurs at the anode. Students really only need to understand and learn three basic facts about half-equations and electrochemical cells.

1. The half-equation represents an equilibrium reaction and can go either way. The E⦵ value is not an enthalpy term and does not change value if the equation is written the other way round.

2. The standard electrode potential is a measure of the potential difference obtained when a standard half-cell (1.00 mol dm-3 concentration, 298 K and 1 atmosphere pressure) is connected to a standard hydrogen half-cell (using a standard hydrogen electrode). By convention (and stress that it is only a convention) if the value for E⦵ is negative electrons flow from the half-cell to the standard hydrogen electrode and if the value is positive the electrons flow from the hydrogen half-cell to the other half-cell.

3. When half-cells are connected (either using a salt bridge or when they react when brought together) the electrons flow from the more negative half-cell to the more positive half-cell.

Armed with this knowledge, students should be able to answer correctly any IB questions on redox chemistry that involves standard electrode potentials.

Footnote

1 In fact in sub-topic 9.2 under 'guidance it states "For voltaic cells, a cell diagram convention should be covered." However this would appear to be for giving a shorthand version of a diagram of a cell such as Zn(s)/Zn2+(aq) || Cu2+(aq)/Cu(s) rather than as a way of calculating the total EMF produced by the redox reaction between zinc metal and copper(II) ions.

Examples

Let’s look at a simple example first. What will be oxidised and what will be reduced when a Mn(s)/Mn2+(aq) half-cell with E⦵ = – 1.18 V is connected to a Ni(s)/Ni2+(aq) half-cell with E⦵ = – 0.26 V and what will be the total EMF of the cell?

To answer this simply get the students to write down the two half-equations. It really does not matter which they write first and which way round they write them, although probably they should use the cell convention. For example:

Mn(s) ⇌ Mn2+(aq) + 2e− E⦵ = – 1.18 V

Ni(s) ⇌ Ni2+(aq) + 2e− E⦵ = – 0.26 V

Since the manganese half-cell is the more negative the electrons will flow from this half-cell to the nickel half-cell. That is the manganese half-cell is giving up electrons and the nickel half-cell is receiving electrons so the actual reactions occurring are:

Mn(s) → Mn2+(aq) + 2e­− E⦵ = – 1.18 V

Ni2+(aq) + 2e− → Ni(s) E⦵ = – 0.26 V

So the overall reaction will be

Mn(s) + Ni2+(aq) → Mn2+(aq) + Ni(s)

That is, the manganese metal is oxidised and the nickel(II) ions are reduced. The total EMF is the quantitative difference between the two values and will always be positive for the spontaneous reaction. In this case it will equal the difference between – 1.18 V and – 0.26 V which is 0.92 V.

Exactly the same principle can be applied when any two half-cells are joined together, either, as in cell, using a salt bridge or in a test-tube type reaction.

For example, can potassium dichromate(VI) oxidise hydrochloric acid to chlorine?

To answer this get the students to write the two half-equations. Dichromate(VI) is in equilibrium with chromium(III) ions in its half-cell and chloride ions are in equilibrium with chlorine in the second half-cell.

Cr2O72− + 14H+(aq) + 6e− ⇌ 2Cr3+(aq) + 7H2O E⦵ = + 1.36 V

2Cl−(aq) ⇌ Cl2 + 2e− E⦵ = + 1.36 V

Not only do both have positive E⦵ values but they also have exactly the same value. This means that no electrons will flow from one cell to the other so no reaction will occur.

In other data books, including the previous IB data booklet, the value for the dichromate/chromium(III) half-cell is + 1.33 V. If this value is used then electrons will flow from the chromium half-cell (less positive) to the chlorine half-cell (more positive) so the spontaneous half-equations will be:

2Cr3+(aq) + 7H2O → Cr2O72−(aq) + 14H+(aq) + 6e− E⦵ = + 1.33 V

Cl2(aq) + 2e− → 2Cl−(aq) E⦵ = + 1.36 V

By multiplying the second equation by three to balance the number of electrons we can get the overall spontaneous equation. Stress to students when you do this that, unlike enthalpy change values, the standard electrode value is not multiplied and remains unchanged. The overall equation is thus:

2Cr3+(aq) + 3Cl2(aq) + 7H2O → Cr2O72−(aq) + 14H+(aq) + 6Cl−(aq) E⦵ total = 0.03 V

This means that chlorine can oxidise chromium(III) ions and that dichromate cannot spontaneously oxidise chloride ions. The total EMF for the spontaneous reaction will be 0.03 V . This answer is interesting as it is only true that dichromate cannot spontaneously oxidise chloride ions under standard conditions. If the conditions (e.g. concentrations) are altered then the E⦵ values are so close that they may switch under different conditions with the chlorine half-cell becoming the more negative. For this reason hydrochloric acid is not used to acidify potassium dichromate(VI) when the dichromate is being used as an oxidizing agent in analytical chemistry.

Try a couple more and then click on the 'eye' to reveal the answer.1. What will be the total EMF and what will be oxidized and what will be reduced when a Cu(s)/Cu2+(aq) half-cell is connected to a Pb(s)/Pb2+(aq) half-cell using a salt bridge and in which direction will the electrons flow in the external circuit?

Lead, Pb, will be oxidized and copper(II) ions will be reduced. The EMF will be 0.47 V (using the values from Section 24 in the IB data booklet). Electrons will flow from the lead half-cell to the copper half-cell.

2. Will hydrogen peroxide act as an oxidising agent or a reducing agent when it reacts with an acidified solution of potassium permanganate?

The two relevant hydrogen peroxide half-equations are

O2(aq) + 2H+(aq) + 2e− ⇌ H2O2(aq) E⦵ = + 0.68 V

and

H2O2(aq) + 2H+(aq) + 2e− ⇌ 2H2O(aq) E⦵ = +1.78 V

The permanganate/manganese(II) half-cell has the more positive standard electrode potential (+ 1.51 V compared to + 0.68 V) so permanganate ions will be the oxidizing agent and hydrogen peroxide the reducing agent. If students do this reaction they will see bubbles of oxygen being evolved. (You could also deduce that hydrogen peroxide can spontaneously oxidise manganese(II) ions with a total EMF of 0.27 V although this was not asked for.)

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