# 20.3 Using E° Values

The usage of E° values

• The value of standard electrode potential, E° can be used in some ways:
1. To predict standard cell potentials, E°cell.
2. To determine the direction of electron flow
3. To predict the feasibility of reactions
4. To determine the strength of reducing and oxidising agents

Predicting standard cell potentials, E°cell

• The standard cell potential, E°cell of a cell consisting of two half-cells can be predicted by connecting them together using wires with a voltmeter

• Find from the Data Booklet the required equilibrium equation.
• Copy down the equilibrium equations and their corresponding E° values
• Use the formula E°cell = E°(bigger value) – E°(smaller value).

Determining the direction of electron flow • When the high-resistance voltmeter of a cell is removed, electrons will flow from one cell to The electron will flow from the half-cell with more electrons(anode) to the half-cell with less electrons(cathode).

• For example, if the voltmeter in the Zn/Zn²⁺ half-cell and Cu/Cu²⁺ half-cell is removed, electrons will flow from zinc electrode to copper electrode

• Due to the fact that the system is equilibrium, the decrease in electron concentration in the Zn/Zn²⁺ equilibrium will be accompanied by some changes. The same goes to the increase in electron concentration in Cu/Cu²⁺ half-cells. This comes from Le Chatelier’s principle • If the electrons continue to flow, the position of equilibrium keeps shifting, eventually producing a one-way reaction.
Zn(s) → Zn²⁺(aq) + 2e⁻      , releases electrons to the Cu/Cu²⁺ half-cell Cu²⁺(aq) + 2e⁻ → Cu(s)                      , accepts electrons from the Zn/Zn²⁺ half-cell
…combining these two equations:
Zn(s) + Cu²⁺(aq) → Zn²⁺(aq) + Cu(s)
• Electrons will always flow from the half-cell with lower(more negative or less positive) E° value to the half-cell with higher(less negative or more positive) E° value.
• Two more examples: Predicting the feasibility of  reactions(two  methods)

• Reactions with positive E° value is said to be thermodynamically feasible

• Sometimes this prediction fails because:
1. the actual conditions of the reaction are not standard  conditions.
2. the reaction has high activation energy.

• Remember:
1. E° value gives no information about the reaction rates
2. E° value applies to standard conditions only

• If you are given an equation, split the equation into two half-equations, the find the corresponding E° values from the Data Booklet.
• Add the two E° values together and determine whether the sum is positive or negative.
• Reverse the equation may be necessary, remember to reverse the sign too
• Examples • Alternatively, it can also be predicted by looking at the reactants and products of the equation

• Example 1: Will oxygen oxidise iron(II) hydroxide to iron(III) hydroxide under alkaline conditions?
1. Find from the Data Booklet the equations involving the reactants and products stated and write down their corresponding E° values 2. Determine the direction of electron Remember that electrons flow from the half-cell with more negative E° value to the half-cell with more positive E° value.
3. Determine the shift of position of equilibrium, and see whether it tallies with the reactants and products 4. From the direction of electron flow and shift of equilibrium, we can deduce that:
• iron(II) hydroxide will be oxidised to iron(III) hydroxide because the position of equilibrium of first reaction shifts to the left
• oxygen will accept electrons from the first reaction and will be reduced to hydroxide ions because the position of equilibrium of second reaction shifts to right
• Example 2: Will chlorine oxidise manganese(II) ions to manganate(VII) ions?
1. Find from the Data Booklet the equations involving the reactants and products stated and write down their corresponding E° values 2. Determine the direction of electron Remember that electrons flow from the half-cell with more negative E° value to the half-cell with more positive E° value.
3. Determine the shift of position of equilibrium, and see whether it tallies with the reactants and products 4. From the direction of electron flow and shift of equilibrium, we can deduce that:
• manganese(II) ions will not be oxidised to manganate(VI) ions because if it were to occur the position of equilibrium of first reaction will shift to the left. However, it shifts to the right because the first reaction has a more positive E° value.
• chlorine will not be reduced to chloride ions because if it were to occur the position of equilibrium of the second reaction will shift to the right
• However, it shifts to the left because the second reaction has a more negative E° value.

Determining the strength of oxidising and reducing agents

• From the last example, the scenario can also be interpreted as:
1. Chlorine is a weaker oxidising agent than manganate(VI) ion, hence it will not oxidise manganese(II) ion to manganate(VI) ion
2. Manganese(II) ion is a weaker reducing agent than chloride ion, hence it will not reduce chlorine to chloride ion
• From the second last example, the scenario can also be interpreted as:
1. Iron(II) hydroxide is a stronger reducing agent than hydroxide ion, hence it will reduce oxygen to hydroxide ion
2. Oxygen is a stronger oxidising agent than iron(III) hydroxide, hence it will oxidise iron(III) hydroxide to iron(III) hydroxide

Effect of temperature, pressure and concentration on E° value

• Since redox equilibria are equilibria, a change in temperature, concentration and pressure will have an effect on its position of equilibrium. This will also affect the value of E° because E° indicates the position of equilibrium of a redox equilibrium
• If a change causes the position of equilibrium to shift to the right, the value of E° increases.
• If a change causes the position of equilibrium to shift to the left, the value of E° decreases.