Electrochemical Cells and Cell Potential

Learning Objectives (OCR Spec 5.2.3)

By the end of this lesson you should be able to:

• Calculate E°_cell from two standard electrode potentials using E°_cell = E°_cathode - E°_anode
• Predict the feasibility of a redox reaction from electrode potentials
• Construct overall equations by combining half-equations
• Recognise that ΔG and E°_cell are related (ΔG = -nFE_cell)
• Explain the operation of a hydrogen fuel cell as a commercial application

Calculating E°_cell

When two half-cells are joined, the overall cell emf is:

E°_cell = E°_cathode - E°_anode

or equivalently:

E°_cell = E°(more positive) - E°(more negative)

where the cathode is where reduction occurs (the more positive electrode) and the anode is where oxidation occurs.

The resulting E°_cell will always be positive if the cell is constructed "correctly" (i.e. with the more positive electrode labelled as the cathode).

Worked Example 1: Daniell Cell

Zinc half-cell: E°(Zn^2+/Zn) = -0.76 V

Copper half-cell: E°(Cu^2+/Cu) = +0.34 V

The more positive electrode is copper, so it is the cathode:

E°_cell = E°_cathode - E°_anode = (+0.34) - (-0.76) = +1.10 V

Half-equations:

• Anode (oxidation, reverse the more negative): Zn(s) -> Zn^2+(aq) + 2e-
• Cathode (reduction, keep as written): Cu^2+(aq) + 2e- -> Cu(s)

Overall equation (add the half-equations, electrons cancel):

Zn(s) + Cu^2+(aq) -> Zn^2+(aq) + Cu(s) E°_cell = +1.10 V

Worked Example 2: Feasibility of Chlorine Oxidising Iron(II)

Cl2(g) + 2e- <=> 2Cl-(aq), E° = +1.36 V

Fe^3+(aq) + e- <=> Fe^2+(aq), E° = +0.77 V

Which is more positive? Cl2/Cl-. So Cl2 is the oxidising agent (reduced at the cathode), and Fe^2+ is the reducing agent (oxidised at the anode).

E°_cell = 1.36 - 0.77 = +0.59 V

Positive, so the reaction is thermodynamically feasible. Overall:

Half-equations:

• Cathode: Cl2(g) + 2e- -> 2Cl-(aq)
• Anode (reversed, doubled to balance electrons): 2Fe^2+(aq) -> 2Fe^3+(aq) + 2e-

Overall:

Cl2(g) + 2Fe^2+(aq) -> 2Cl-(aq) + 2Fe^3+(aq) E°_cell = +0.59 V

This is why chlorine water can oxidise Fe^2+ salts to Fe^3+ in the laboratory.

Worked Example 3: Will Cu Displace Zn?

Can Cu(s) reduce Zn^2+ to Zn?

The proposed reaction is: Cu(s) + Zn^2+(aq) -> Cu^2+(aq) + Zn(s). For this, Cu would have to be the anode (oxidised) and Zn^2+/Zn the cathode (reduced).

E°_cell = E°_cathode - E°_anode = (-0.76) - (+0.34) = -1.10 V

Negative, so the reaction is NOT feasible. The reverse (the Daniell cell reaction) is feasible.

Predicting Feasibility: The Rule

A reaction is feasible if its E°_cell is positive.

• Large positive E°_cell (> +0.3 V): strongly feasible.
• Small positive or negative E°_cell: reaction may not proceed noticeably (thermodynamics may indicate feasibility, but kinetics may not cooperate).
• Negative E°_cell: not feasible in that direction.

E°_cell must be > 0 V for a reaction to be thermodynamically feasible. In practice, OCR uses a rough cutoff of about +0.3 V to say "significantly feasible" because smaller positive values may not reliably correspond to observable reactions under real (non-standard) conditions.

Connecting to Gibbs Free Energy

There is a direct link between the cell emf and the Gibbs free energy change of the cell reaction:

ΔG° = -nFE°_cell

where:

• n = number of moles of electrons transferred per mole of reaction
• F = Faraday constant = 96485 C mol^-1
• E°_cell is in volts
• ΔG° comes out in joules (divide by 1000 for kJ)

For the Daniell cell: n = 2, E°_cell = 1.10 V

ΔG° = -2 x 96485 x 1.10 = -212267 J mol^-1 = -212 kJ mol^-1

Negative, so the reaction is thermodynamically feasible - consistent with the positive E°_cell.

Key takeaway: E°_cell > 0 <=> ΔG° < 0 <=> reaction feasible.

This is a beautiful unification of thermodynamics and electrochemistry.

Effect of Non-Standard Conditions

E° values assume 1 mol dm^-3, 298 K, 100 kPa. Real conditions may differ, and cell emf shifts accordingly. Qualitatively:

• Increasing the concentration of a reactant (left-hand side of the half-equation) shifts the equilibrium right (reduction), increasing E.
• Decreasing the concentration of a reactant shifts the equilibrium left, decreasing E.

You can reason about shifts using Le Chatelier's principle on the half-cell equilibrium. For example, for Cu^2+ + 2e- <=> Cu, increasing [Cu^2+] pushes the equilibrium right, making the electrode more positive.

Temperature also matters; higher T changes equilibrium constants and so changes E.

Fuel Cells - Hydrogen Fuel Cell