CAIE A-Level Chemistry 24.2 Electrode and Cell Potentials
Practise standard electrodes, cell potentials, redox feasibility, concentration effects and electrochemical ΔG.
- Syllabus
- 2028–2030
- Course
- Chemistry 9701
- Level
- A2
Practise standard electrodes, cell potentials, redox feasibility, concentration effects and electrochemical ΔG.
Define the term standard cell potential.
the potential difference between two half-cells/two electrodes (in a cell)
under standard conditions of 1 atm.,298 K, (all) solutions being 1 moldm−3
Draw a fully labelled diagram of the experimental set-up you could use to measure the standard electrode potential of the Pb2+(aq)/Pb(s) electrode. Include the necessary chemicals.

8 marking points, any 2 points for each mark
H2 / hydrogen
correct delivery system for H2Pb2+ (aq)
Pb electrode
Pt electrode
H+(aq) solution
salt bridge
voltmeter/V labelled
The E⊖ for a Pb2+(aq)/Pb(s) electrode is -0.13 V .
Suggest how the E for this electrode would differ from its E⊖ value if the concentration of Pb2+(aq) ions is reduced. Indicate this by placing a tick (✓) in the appropriate box in the table.

Explain your answer.
more negative
shifts Pb2+(+2e−)⇋Pb equilibrium / reaction to the left
Car batteries are made up of rechargeable lead-acid cells. Each cell consists of a negative electrode made of Pb metal and a positive electrode made of PbO2. The electrolyte is H2SO4(aq).
When a lead-acid cell is in use, Pb2+ ions are precipitated out as PbSO4( s) at the negative electrode.
Complete the half-equation for the reaction taking place at the positive electrode.
PbO2( s)+SO42−(aq)+4H++2e−→PbSO4( s)+2H2O
The diagrams show how the voltage across two different cells changes with time when each cell is used to provide an electric current.


Suggest a reason why
- the voltage of the lead-acid cell changes after several hours,
- the voltage of the fuel cell remains constant.
reagents/ PbO2/H2SO4 and used up/concentration decreases
as fuel/hydrogen is being continuously supplied/fuel has not run out
Some electrode potentials are shown in Table 3.1.

Table 3.1
Complete the diagram to show a standard hydrogen electrode.
Label your diagram. Identify all substances. You do not need to state standard conditions.

hydrogen, delivery system, H+, platinum, [1]
An electrochemical cell is set up using an Fe3+/Fe2+ electrode and a standard hydrogen electrode.
Identify the positive electrode in the electrochemical cell and the direction of electron flow in the external circuit.
positive electrode
Electrons flow from the electrode to the electrode.
iron . . . . . hydrogen . . . . . iron [1]
The vanadium-containing species in the electrode reactions given in Table 3.1 are V,V2+, V3+,VO2+ and VO2+.
Identify one vanadium-containing species that does not react with Fe2+ ions under standard conditions.
Use data from Table 3.1 to explain your answer.
(for specified V2+,V3+ or VO2+ ) E⊖ is more positive than / above -0.44 AND more negative than / below 0.77 V [1]
Identify all the vanadium-containing species that will react with Fe2+ ions under standard conditions.
V and VO2+[1]
Write an equation for one of the possible reactions identified in (ii).
V+Fe2+→V2++FeVO2++2H++Fe2+→VO2++H2O+Fe3+[1]
Another electrochemical cell is set up using an Fe3+/Fe2+ electrode and an alkaline ClO−/Cl− electrode.
The concentration of Fe3+ is 1000 times greater than the concentration of Fe2+ in the Fe3+/Fe2+ electrode. All other conditions are standard.
Use the Nernst equation to calculate the E value of the Fe3+/Fe2+ electrode.
Show your working.
Nernst: E=0.77+(0.059/z)log[ox]/[red][1]
0.947 [1]
Write an equation for the reaction that occurs in the cell, under these conditions.
2Fe3++Cl−+2OH−→2Fe2++ClO−+H2O[1]
Another electrochemical cell is set up using an Fe2+/Fe electrode and an alkaline ClO−/Cl− electrode under standard conditions.
Calculate the value of ΔG⊖ for the cell.
ΔG⊖=kJmol−1
E⊖cell =1.33 V[1]ΔG⊖=−nE⊖cell F[1]−257[1]
Iron(II) chloride, FeCl2, is oxidised by chlorine to form iron(III) chloride, FeCl3, under standard conditions.

Table 3.2
Use Table 3.2 and other data to calculate the Gibbs free energy change, ΔG⊖, for this reaction.
Show your working.
ΔS=−179[1]ΔG=ΔH−TΔS[1]
-74.7
Hypophosphorous acid is an inorganic acid.
The conjugate base of hypophosphorous acid is H2PO2−.
H2PO2−is a strong reducing agent. It can be used to reduce metal cations without the need for electrolysis.
equation 1HPO32−+2H2O+2e−⇌H2PO2−+3OH−E⊖=−1.57 V
In an experiment, an alkaline HPO32−/H2PO2−half-cell is constructed with [H2PO2−]=0.050 moldm−3.
All other ions are at their standard concentration.
Predict how the value of E of this half-cell differs from its E⊖ value.
Explain your answer.
electrode potential E would become more positive / less negative (than E⊖ )
lower [H2PO2−]AND shifts equilibrium to the right-hand side
The Cr3+/Cr half-cell has a standard electrode potential of -0.74 V .
An electrochemical cell consists of an alkaline HPO32−/H2PO2−half-cell and a Cr3+/Cr half-cell.
Calculate the standard cell potential, Ecell ⊖.
+1.57−0.74=(+)0.83( V)
Complete the diagram in Fig. 2.1 to show how the standard electrode potential of the Cr3+/Cr half-cell can be measured relative to that of the standard hydrogen electrode.
Identify the chemicals, conditions and relevant pieces of apparatus.

Fig. 2.1

any three [1] any six [2] all nine [3]
Label Fig. 2.1 to show:
- which is the positive electrode
- the direction of electron flow in the external circuit.
Pt electrode positive AND flow of electrons anticlockwise (to the SHE)
H2PO2−reduces Ni2+ to Ni in alkaline conditions.
Use equation 1 to construct the ionic equation for this reaction.
equation 1HPO32−+2H2O+2e−⇌H2PO2−+3OH−
H2PO2−+3OH−+Ni2+→HPO32−+2H2O+Ni