Electrode Potential Calculations (OCR A Level Chemistry A) : Revision Note
Calculating Standard Cell Potential
Once the Eꝋ of a half-cell is known, the potential difference or voltage or emf of an electrochemical cell made up of any two half-cells can be calculated
These could be any half-cells and neither have to be a standard hydrogen electrode
The standard cell potential (Ecellꝋ) can be calculated by subtracting the less positive Eꝋ from the more positive Eꝋ value
The half-cell with the more positive Eꝋ value will be the positive pole
By convention this is shown on the right hand side in a conventional cell diagram, so is termed Erightꝋ
The half-cell with the less positive Eꝋ value will be the negative pole
By convention this is shown on the left hand side in a conventional cell diagram, so is termed Eleftꝋ
Ecellꝋ = Erightꝋ - Eleftꝋ
Since oxidation is always on the left and reduction on the right, you can also use this version
Ecellꝋ = Ereductionꝋ - Eoxidationꝋ
Worked Example
Calculating the standard cell potential
Calculate the standard cell potential for the electrochemical cell below and explain why the Cu2+ / Cu half-cell is the positive pole. The half-equations are as follows:
Cu2+(aq) + 2e- ⇌ Cu(s) Eꝋ = +0.34 V
Zn2+(aq) + 2e- ⇌ Zn(s) Eꝋ = −0.76 V

Answer
Step 1: Calculate the standard cell potential. The copper is more positive so must be the right hand side.
Ecellꝋ = Erightꝋ - Eleftꝋ
Ecellꝋ = (+0.34) - (-0.76)
= +1.10 V
The voltmeter will therefore give a value of +1.10 V
Step 2: Determine the positive and negative poles
The Cu2+ / Cu half-cell is the positive pole as its Eꝋ is more positive than the Eꝋ value of the Zn2+ / Zn half-cell
Examiner Tips and Tricks
A helpful mnemonic for remembering redox in cells

Lio the lion goes Roor!
Lio stands for 'Left Is Oxidation' and he is saying ROOR because that is the order of species in the cell:
Reduced/Oxidised (salt bridge) Oxidised/Reduced
Feasibility & Standard Cell Potential
Feasibility
The Eꝋ values of a species indicate how easily they can get oxidised or reduced
The more positive the value, the easier it is to reduce the species on the left of the half-equation
The reaction will tend to proceed in the forward direction
The less positive the value, the easier it is to oxidise the species on the right of the half-equation
The reaction will tend to proceed in the backward direction
A reaction is feasible (likely to occur) when the Ecellꝋ is positive
For example, two half-cells in the following electrochemical cell are:
Cl2 (g) + 2e- ⇌ 2Cl- (aq) Eꝋ = +1.36 V
Cu2+ (aq) + 2e- ⇌ Cu (s) Eꝋ = +0.34 V
Cl2 molecules are reduced as they have a more positive Eꝋ value
The chemical reaction that occurs in this half cell is:
Cl2 (g) + 2e- → 2Cl- (aq)
Cu2+ ions are oxidised as they have a less positive Eꝋ value
The chemical reaction that occurs in this half cell is:
Cu (s) → Cu2+ (aq) + 2e-
The overall equation of the electrochemical cell is (after cancelling out the electrons):
Cu (s) + Cl2 (g) → 2Cl- (aq) + Cu2+ (aq)
OR
Cu (s) + Cl2 (g) → CuCl2 (s)
The forward reaction is feasible (spontaneous) as it has a positive Eꝋ value of +1.02 V ((+1.36) - (+0.34))
The backward reaction is not feasible (not spontaneous) as it has a negative Eꝋ value of -1.02 ((+0.34) - (+1.36))


A reaction is feasible when the standard cell potential Eꝋ is positive
Examiner Tips and Tricks
You may have to apply your understanding (from the above worked example) to questions with more than 2 equations
The process is still the same in terms of identifying the most positive / least negative value as the reduction reactions
This reaction will also contain the oxidising agent on the left hand side
Limitations of Eθ to predict reactions
The thermodynamic feasibility of a reaction can be deduced from the electrode potential, however, it gives no information about the rate of reaction
As standard electrode potentials are measured using solutions, we have to consider the le Châtelier's effect on concentration using non-standard conditions
For example, the redox equilibrium equation and standard electrode potential for the V3+ | V2+ system are:
V3+ (aq) + e- ⇌ V2+ (aq) Eθ = +0.26 V
If the concentration of V3+ (aq) is greater than 1.0 mol dm-3, then the equilibrium will shift to the right
This will remove electrons from the system, therefore, making the electrode potential less negative
If the concentration of V2+ (aq) is greater than 1.0 mol dm-3, then the equilibrium will shift to the left
This will add electrons to the system, therefore, making the electrode potential more negative
Any change to the concentration will cause a change to the electrode potential and, therefore, to the overall cell potential
This is true of any change to the conditions that results in non-standard conditions
Another, more basic limitation is the fact that many redox reactions are not aqueous
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