We can link any two half cells together to form an electrochemical cell.
A high resistance voltmeter prevents current (electrons) flowing from the oxidation half cell to the reduction half cell and so measures the maximum potential difference – this is the electromotive force (e.m.f) of the cell, represented by the symbol Ecell. Ecell only depends on the concentration of the solutions, the temperature, the pressure of a gas etc. If the measurements are taken under standard conditions of 298K, 100kPa and 1.00 mol dm-3, Ecell becomes E⦵cell.
We can use a cell diagram as shorthand to represent the diagram above:
The anode half cell (where oxidation takes place) is written on the left hand side and the cathode half cell (where reduction takes place) is written on the right hand side, the cell components are essentially listed in the same order that the overall redox reaction happens.
If the half cell represents a reaction between two ions in solution (with a platinum electrode) then we would write it as:
And the hydrogen half cell / electrode can be shown as:
Using standard electrode potential data to calculate Ecell
Consider an electrochemical cell constructed from Ag+(aq) / Ag(s) and Mg2+(aq) / Mg(s) half cells. For each half cell we can look up the value of the standard electrode potential, E⦵, in a data book.
Ag+(aq) + e– ⇾ Ag(s) E⦵ = +0.80V
Mg2+(aq) + 2e– ⇾ Mg(s) E⦵ = -2.37V
The Mg2+(aq) / Mg(s) half cell is the more negative, which means that it will form the oxidation (negative) half cell. Mg(s) has a tendency to release electrons (it is oxidised) making it a strong reducing agent.
The Ag+(aq) / Ag(s) half cell is the more positive, which means that it will form the reduction (positive) half cell. Ag+ ions have a tendency to accept electrons and are reduced to form Ag(s). This makes Ag+ ions an oxidising agent.
Note how careful we have to be in using language and terminology here – it is meaningless to say that ‘silver’ is a reducing agent, we need to be clear we are describing the behaviour of silver ions 🤔.
Ecell⦵ = E⦵(more positive) – E⦵(less positive)
So, silver ions will be reduced to silver metal and magnesium metal will be oxidised to magnesium ions in this example:
Mg(s) + 2Ag+(aq) ⇾ Mg2+(aq) + 2Ag(s)
Ecell⦵ = E⦵Ag+/Ag – E⦵Mg2+/Mg = +0.80 – (-2.37) = +3.17 V
So, just to recap …
When we construct an electrochemical cell from two half cells, the half cell with the more positive E⦵ will always be the reduction half cell and the half cell with the less positive E⦵ will always be the oxidation half cell. The reaction can only proceed in one direction.
We can use this to determine the feasibility of redox reactions without having to carry out the actual experiment, which is kind of useful.
Practice questions
- Draw a fully labelled diagram to show how an electrochemical cell could be set up to measure the maximum potential difference between a Ni2+(aq)/ Ni(s) half cell and an I2(aq) / 2I–(aq) half cell. Calculate the standard cell potential for this cell.
| Ni2+(aq) + 2e– ⇌ Ni(s) | E⦵ = – 0.25V |
| I2(aq) + 2e- ⇌ 2I–(aq) | E⦵ = + 0.54V |
- A cell is constructed from two redox potentials:
Cu2+(aq) / Cu(s) E⦵ = +0.34V
Ag+(aq) / Ag(s) E⦵ = +0.80V
Which of the following statements are true?
A. The cell potential is +1.14V
B. The silver electrode increases in mass
C. Cu2+ is reduced by Ag
D. Cu is oxidised by Ag+
E. Cu2+ is oxidised by Ag
Answers
- Statements B and D are true.
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