Electrode Potential Calculations (OCR A Level Chemistry)

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Calculating Standard Cell Potential

  • Once the Eof 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 (Ecell) can be calculated by subtracting the less positive Efrom the more positive Evalue
    • The half-cell with the more positive Evalue will be the positive pole
      • By convention this is shown on the right hand side in a conventional cell diagram, so is termed  Eright

    • 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  Eleft

Ecell = Erightꝋ Eleftꝋ   

    • Since oxidation is always on the left and reduction on the right, you can also use this version

Ecell = Ereductionꝋ Eoxidation

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)      E= +0.34 V

Zn2+(aq) + 2e- ⇌ Zn(s)      E= −0.76 V

Electrochemistry Calculations - Electrochemical Cell, downloadable AS & A Level Chemistry revision notes

   Answer

   Step 1: Calculate the standard cell potential. The copper is more positive so must be the right hand side.

EcellErightꝋ - Eleftꝋ   

Ecell = (+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 Eis more positive than the Evalue of the Zn2+ / Zn half-cell

Examiner Tip

A helpful mnemonic for remembering redox in cells

Lio the Lion, downloadable AS & A Level Chemistry revision notes

 

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 Evalues 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 Ecell is positive

  • For example, two half-cells in the following electrochemical cell are:

Cl2 (g) + 2e- ⇌ 2Cl- (aq)        E = +1.36 V

Cu2+ (aq) + 2e- ⇌ Cu (s)        E = +0.34 V

  • Cl2 molecules are reduced as they have a more positive E 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 E 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 E value of +1.02 V ((+1.36) - (+0.34))
  • The backward reaction is not feasible (not spontaneous) as it has a negative Evalue of -1.02 ((+0.34) - (+1.36))

Reaction Feasibility (1), downloadable AS & A Level Chemistry revision notesPrinciples of Electrochemistry - Reaction Feasibility (2), downloadable AS & A Level Chemistry revision notes

A reaction is feasible when the standard cell potential E is positive

Examiner Tip

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

Author: Richard

Expertise: Chemistry

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.