Determining the potential difference of an electrochemical cell

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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

In conclusion …

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

  1. 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

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  1. 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

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  1. Give the conventional representation (cell diagram) that describes an electrochemical cell set up between the Zn2+(aq) / Zn(s) redox system and the Fe3+(aq) / Fe2+(aq) redox system.  Identify the salt bridge and the phase boundary in your answer.
Fe3+(aq) + e-  ⇌  Fe2+(aq)E⦵ = +0.77 V
Zn2+(aq) + 2e-  ⇌  Zn(s)E⦵ = -0.76 V

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  1. Write an equation to show the predicted reaction between the MnO4–(aq) / Mn2+(aq) redox system and the SO42-(aq) / SO32-(aq) redox system.  Identify the oxidising agent and the reducing agent.
SO42-(aq) + 2H+(aq) + 2e-  ⇌  SO32-(aq)  +  H2O(l)E⦵ = +0.17 V
MnO4–(aq) + 8H+(aq) + 5e-  ⇌  Mn2+(aq) + 4H2O(l)E⦵ = +1.51 V


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  1. A student set up electrochemical cell shown below, measuring an initial voltage of +0.18V at 298K.

(a) What is the purpose of the salt bridge?

(b) Calculate the electrode potential for the left hand half cell, given that E⦵for the Cu2+(aq) / Cu(s) half cell is +0.34V.  Explain why it is not also 0.34V.

(c) Give the conventional representation for the electrochemical cell.

(d) If the voltmeter is replaced by a motor, the EMF of the electrochemical cell slowly decreases to 0V in time.  Explain why this happens with reference to the changes that occur in the half cells.

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  1. We can measure the electrode potential for a silver wire immersed in 1.00 mol dm-3  silver nitrate at 298K using a standard hydrogen electrode.  Under these conditions, E⦵for Ag+(aq) / Ag(s) is +0.80V.

(a) Complete the cell diagram for this electrochemical cell:

(b)  Silver chloride precipitates out of solution when  1.00 mol dm-3  sodium chloride is added to the Ag+(aq) / Ag(s)  half cell.  Explain how the cell EMF will change, with reference to le Chatelier’s principle.

Answers

  1. Statements B and D are true.

EXAM TIP: Cancel out electrons, H+ ions and water on either side of your final answer.

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5. (a) The ions in the salt bridge are free to move and complete the circuit.

(b) E is not + 0.34V because the left half cell does not have a concentration of 1.00 mol dm-3 of Cu2+ ions.

EXAM TIP: We know that the concentration of Cu2+ in the lefthand cell is less than 1 mol dm-3, so the equilibrium position for the Cu2+ / Cu half cell must have shifted to the left which means there are more electrons and E must be less positive than 0.34V.

Cu2+ + 2e– ⇌ Cu E⦵ = 0.34 V (when [Cu2+] = 1 mol dm-3)

(d) The lefthand half cell is the oxidation half cell so Cu is oxidised to Cu2+ and electrons flow around the circuit to the righthand half cell. The [Cu2+] in the oxidation half cell increases. In the reduction (righthand) half cell Cu2+ is reduced to Cu – the [Cu2+] decreases. Eventually the [Cu2+] in each half cell will be the same and all redox reactions will stop. Electrons will no longer flow and so the EMF becomes zero.

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6.  (a)   Pt(s) | H2(g) | 2H+(aq)  ||  Ag+(aq) | Ag(s)

(b)  Ag+(aq)  +  e-  ⇌  Ag(s)   E⦵ = + 0.80V

Ag+ is removed from solution as it precipitates out, the position of equilibrium shifts to the left to oppose the change, producing more electrons and so E⦵ becomes less positive.                                

EXAM TIP: (E⦵ = Ecell / EMF in this example)