Electrode Potential Calculations (OCR A Level Chemistry A): Revision Note

Exam code: H432

Richard Boole

Written by: Richard Boole

Reviewed by: Caroline Carroll

Updated on

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

Standard cell: a zinc rod in 1.00 mol dm-3 zinc sulfate solution and a copper rod in 1.00 mol dm-3 copper(II) sulfate solution, joined by a KNO3 salt bridge and a high-resistance voltmeter.

   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

Cartoon of a lion roaring 'ROOR!', a mnemonic: Lio the Lion, Left Is Oxidation, and the cell order Reduced/Oxidised (salt bridge) Oxidised/Reduced.

 

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)          

  • Cu metal is oxidised as it is on the right of the half-equation with the 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))

Cell of a copper rod in 1.00 mol dm-3 Cu2+ solution and a platinum electrode in 1.00 mol dm-3 Cl- solution with chlorine gas at 100 kPa, joined by a KNO3 salt bridge; the voltmeter reads +1.02 V and electrons flow from the copper to the chlorine electrode.
Reaction feasibility: Cu2+/Cu (E = +0.34 V, less positive, oxidation occurs, Cu to Cu2+) and Cl2/Cl- (E = +1.36 V, more positive, reduction occurs), giving Cl2 + Cu → 2Cl- + Cu2+ (or CuCl2) with E = +1.02 V.

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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Richard Boole

Author: Richard Boole

Expertise: Curriculum Expert

Richard has taught Chemistry for over 15 years as well as working as a science tutor, examiner, content creator and author. He wasn’t the greatest at exams and only discovered how to revise in his final year at university. That knowledge made him want to help students learn how to revise, challenge them to think about what they actually know and hopefully succeed; so here he is, happily, at SME.

Caroline Carroll

Reviewer: Caroline Carroll

Expertise: Head of Content Delivery

Caroline graduated from the University of Nottingham with a degree in Chemistry and Molecular Physics. She spent several years working as an Industrial Chemist in the automotive industry before retraining to teach. Caroline has over 12 years of experience teaching GCSE and A-level chemistry and physics. She is passionate about delivering high-quality resources to help students achieve their full potential.