Redox and Electrode Potentials

Half equations in acidic conditions

• Water is in the products

• Balance hydrogen with H⁺ (reactants)

• Balance charges with e^- (reactants)

Half equations in the presence of water

• Water is in the reactants

• Balance hydrogen with H⁺ (products)

• Balance charges with e^- (products)

Redox titrations: Iron (II) ions and Potassium Manganate (VII)

• Overall equation: MnO₄^- + 8H⁺ +5Fe²⁺+ → Mn²⁺ +5Fe³⁺ + 4H₂O

• Purple → Pale pink

Percentage by mass

(Actual mass / Theoretical mass) x 100

In a similar experiment a student carried out the titration leaving the iron solution overnight, what change will we expect to see

• Fe²⁺ oxidised to Fe³⁺

• % of Fe²⁺ calculated would be lower

Redox titration: Iodine-thiosulfate

• Overall equation: I₂ + 2S₂O₃^2- → 2I^- + S₄O₆^2-

• Observation: Colourless → Pale Yellow

Describe and explain how the student should determine the end point of the titration correctly for iodine-thiosulfate

Add starch , colour change from blue-black to colourless

Standard electrode potential of a half cell

The emf when a half cell is connected to a standard hydrogen half-cell at 298K and 100KPa and concentration of 1 moldm^-3

Voltmeter

Measures the potential pushing power of electrons through the circuit but keeps the current at zero

Wire

Allows the movement of electrons

Electrodes

Where the half equations are taking place and are referred to as the half-cells

Salt bridge

• Filter paper soaked in a solution of KNO₃

• Allows the movement of ions to complete the circuit and compensate for changes in concentration in each half-cell

Why is KNO₃ a suitable solution

It does not interfere with the redox reaction

Example of an electrochemical cell

Standard Hydrogen Half-cell

Electrochemical series

• Right-hand side: Reducing agents

• Left-hand side: Oxidising agents

Strongest and weakest reducing agent

• Strongest: Li as it has the most negative E°

• Weakest: Ce3+ as it has the least negative E°

Strongest and weakest oxidising agent

• Strongest: Ce4+ as it has the most positive E°

• Weakest: Li+ as it has the least positive E°

Template to explain why a reaction is feasible

• E°_/_>E°_/_

• More positive electrode system shifts right and _is reduced to _

• More negative electrode system shifts left and _ is oxidised to _

E° cell calculation

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

E° cell = E°(reduction) - E°(oxidation) - use this to explain why a reaction is not feasible