Cell Potential & Thermodynamic Favorability (College Board AP® Chemistry)
Study Guide
Written by: Philippa Platt
Reviewed by: Stewart Hird
Cell Potential & Thermodynamic Favorability
Standard electrode potential
The position of equilibrium and therefore the electrode potential depends on factors such as:
Temperature
Pressure of gases
Concentration of reagents
So, to be able to compare the electrode potentials of different species, they all have to be measured against a common reference or standard
Standard conditions also have to be used when comparing electrode potentials
These standard conditions are:
An ion concentration of 1.00 mol dm-3
A temperature of 298 K
A pressure of 100 kPa
Standard measurements are made using a high resistance voltmeter so that no current flows and the maximum potential difference is achieved
The electrode potentials are measured relative to a standard hydrogen electrode
The standard hydrogen electrode is given a value of 0.00 V, and all other electrode potentials are compared to this standard
This means that the electrode potentials are always referred to as standard electrode potential (Eo)
The standard electrode potential (Eo) is the potential difference ( sometimes called voltage) produced when a standard half-cell is connected to a standard hydrogen cell under standard conditions
For example, the standard electrode potential of bromine suggests that relative to the hydrogen half-cell it is more likely to get reduced, as it has a more positive Eo value
Br2 (l) + 2e– ⇌ 2Br– (aq) Eo = +1.09 V
2H+ (aq) + 2e– ⇌ H2 (g) Eo = 0.00 V
The standard electrode potential of sodium, on the other hand, suggests that relative to the hydrogen half-cell it is less likely to get reduced as it has a more negative Eo value
Na+ (aq) + e– ⇌ Na (s) Eo = -2.71 V
2H+ (aq) + 2e– ⇌ H2 (g) Eo = 0.00 V
Once the Eo 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 half-cell with the more positive c will be the positive pole
By convention, this is shown on the right-hand side in a conventional cell diagram, so is termed Erighto
By convention, this is shown on the left-hand side in a conventional cell diagram, so is termed Elefto
The half-cell with the less positive Eꝋ value will be the negative pole
How to calculate Ecello
Voltmeters measure potential on the right-hand side of the cell and subtract it from the potential on the left-hand side of the cell
If both E° values are known, the Ecell° of the cell formed can be calculated if the right-hand electrode and left-hand electrode are specified
For example, if in a cell the Erighto = silver electrode (+0.80V) and LHS is copper electrode (+0.34 V), then
Ecello = Erighto - Elefto
Ecello = +0.80 - 0.34 = +0.46 V
Since oxidation is always on the left and reduction on the right, you can also use this version
Ecello = Ereductiono - Eoxidationo
Worked Example
Calculating the standard cell potential
Calculate the standard cell potential for the electrochemical cell below. The half-equations are as follows:
Cu2+ (aq) + 2e- ⇌ Cu (s) Eo = +0.34 V
Zn2+ (aq) + 2e- ⇌ Zn (s) Eo = −0.76 V
Answer:
The copper is more positive so it must be the right-hand side.
Ecello = Erighto - Elefto
Ecello = (+0.34) - (-0.76) = +1.10 V
The voltmeter will therefore give a value of +1.10 V
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
The Eo values of a species indicate how easily they can get oxidised or reduced
The reaction will tend to proceed in the forward direction
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 backward direction
A reaction is feasible (likely to occur) when the Ecello is positive
The less positive the value, the easier it is to oxidise the species on the right of the half-equation
For example, two half-cells in the following electrochemical cell are:
Cl2 (g) + 2e- ⇌ 2Cl- (aq) Eo = +1.36 V
Cu2+ (aq) + 2e- ⇌ Cu (s) Eo = +0.34 V
Cl2 molecules are reduced as they have a more positive Eo 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 Eo 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 Eo value of +1.02 V
Calculated from (+1.36) - (+0.34)
The backward reaction is not feasible (not spontaneous) as it has a negative Eo value of -1.02
Calculated from (+0.34) - (+1.36)
A cell containing Cu2+/Cu and Cl2/2Cl–
A reaction is feasible when the standard cell potential Eo is positive
Cell Potential & Gibbs Free Energy
The standard free energy change can be calculated using the standard cell potential of an electrochemical cell
ΔGo = - n x Ecello x F
ΔGo = standard Gibbs free energy
n = number of electrons transferred in the reaction
Ecello = standard cell potential (V)
F = Faraday constant (96,485 C mol-1)
Worked Example
Calculating the standard Gibbs free energy change
Calculate the standard Gibbs free energy change for the following electrochemical cell:
2Fe3+ (aq) + Cu2+ (aq) 2Fe2+ (aq) + Cu (s)
Answer:
Step 1: Determine the two half-equations and their Eꝋ using the Data booklet:
Fe3+ (aq) + e- ⇌ Fe2+ (aq) Eo = +0.77 V
Cu2+ (aq) + 2e- ⇌ Cu (s) Eo = +0.34 V
Step 2: Calculate the Ecello
Ecello = Eredo - Eoxo
Ecello = (+0.77) - (+0.34)
Ecello = +0.43 V
Step 3: Determine the number of electrons transferred in the reaction
The Cu2+/Cu has a smaller Eo value which means that it gets oxidised
It transfers two electrons to two Fe3+ ions
Each Fe3+ ion accepts one electron so the total number of electrons transferred is two
Step 4: Substitute the values into the standard Gibbs free energy equation
ΔGo = - n x Ecello x F
ΔGo = -2 x (+0.43) x 96,485
ΔGo = -82,977.1 J mol-1 = -83 kJ mol-1
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