redox II 14

Define oxidation and reduction in terms of electron transfer and changes in oxidation number

Oxidation= loss of electrons, increase in oxidation number.

Reduction= gain of electrons, reduction in oxidation number

Define standard electrode potential of a half cell (E cell)

The emf (voltage) of a cell measured under standard conditions of 298K, 100KPa, 1mol dm-3 concentration of ions in solution, when a half cell is connected to the standard hydrogen electrode.

What is the absolute potential difference

The potential difference between the metal and the solution.

E.g. magnesium and copper, the potential difference is greater with magnesium. This is because the equilibrium position is further left for magnesium because its tendency to release electrons is greater. Causes a greater negative charge on the metal, and more positive ions in the solution.

Why is it impossible to measure the absolute potential difference between a metal electrode and its solution

One terminal of the voltmeter has to be connected to the metal electrode, which is done with wires so is easy.

The other terminal of the voltmeter has to be connected to the solution, which would mean having to dip another piece of metal into the solution. The 2nd piece would create its own potential difference. So you’d be measuring the potential difference between tie pieces of metal, and not original metal and the solution.

How are the problems of measuring the absolute potential difference between a metal electrode and its solution resolved

Create a reference electrode, and measure the difference in potential difference between that and the metal electrode.

The Standard hydrogen electrode is the reference electrode.

How does the standard hydrogen electrode work

Consists of hydrogen gas at 100KPa, bubbled over porous platinum foil, in a solution of HCl with a [H+] of 1 mol dm-3, at 298K.

Platinum foil porous to increase surface area, so equilibrium between H2(g) and H+(aq) is established quickly.

½ H2(g) ⇌ H+(aq) + e-

Equation to calculate standard emf (E cell)

E cell = Ecell reduced - ecell oxidised.

Most negativitie is oxidised

How changes from standard conditions change electrode potentials

Equilibrium position changes depending on reaction conditions. So standard conditions used to ensure we get the same values which allow us to make comparisons

Why is potassium nitrate used in salt bridges

it doesn’t react to form precipitates with ions in the half-cells

way to remember cell notation

ROOR.

Reduced | oxidised || oxidised | reduced.

Single line shows a physical change, double line shows a salt bridge

When do you need a platinum electrode. E.g. using Fe 3+ / Fe 2+ and Mg 2+ / Mg cell

When using 2 non metals as a half cell.

Mg (s) | Mg 2+ (aq) || Fe 3+ (aq), Fe 2+ (aq) | Pt (s)

What’s a key requirement for a reaction to be thermodynamicly feasible

E cell must be positive

Why, even though E cell is positive, so feasible, could a reaction not go (2 reasons)

1) Non- standard conditions.

E.g. 2Fe + O2 + 2H2O ⇌ 2Fe 2+ + 4OH- (Fe 2+ reaction more negative).

Increasing [O2] shifts equilibrium to right, making it easier for O2 to gain electrons. The electrode potential for this half cell will become more positive, so the full cell potential will be higher.

Increasing [Fe] shifts equilibrium to the left. Less electrons will be used up. The electrode potential of Fe2+/ Fe becomes less negative, so the full cell potential will be lower.

2) kinetically unfavourable.

Slow enough rate of reaction that it appears there’s no reaction (e.g. rusting).

If reaction has a high activation energy, it may stop the reaction happening

Disproportionation reaction meaning

Oxidation and reduction simultaneously occurring on the same element

What’s the relationship between E cell and entropy

E cell is directly proportional to the total entropy change. The larger the cell potential, the larger the entropy change during the cell reaction.

E cell directly proportional to delta S total

What is the relationship between E cell and equilibrium constant (K)

E cell is directly proportional to Ln K

How rechargeable batteries work (energy storage cells e.g. lithium ion battery)

Plugging them in to supply a current. This current forces electrons to flow in the opposite way. All we do is reverse the overall discharge equation to show a battery recharging.

Discharge= Li + CoO2 ⇌ Li+[CoO2]-

Recharge= Li+[CoO2]- ⇌ Li + CoO2

What is a fuel cell

An electrochemical cell in which the chemical energy of a fuel cell is converted directly into electrical energy. It differs from other electrochemical cells as it has a continuous supply of reactants to produce a steady electric current

How do hydrogen fuel cells work

Hydrogen gas flows into the negative terminal while oxygen gas flows onto the positive terminal.

Reactions in acidic conditions=

H2(g) → 2H+ + 2e-

½ O2(g) + 2H+ + 2e- → H2O(l) (catalysed by nickel/ nickel (II) oxide)

Reactions in alkaline conditions=

2H2O(l) + 2e- → H2(g) + 2OH-(aq)

How biofuel cells work

Alcohol like methanol can be used in place of hydrogen in cells.

CH3OH(l) + H2O(l) → CO2(g) + 6H+(aq) + 6e-

1 ½ O2(g) + 6H+(aq) + 6e- → 3H2O(l)