Redox Equilibria. Redox equilibria When a metal electrode is placed into a solution of one of its...

Redox Equilibria

Redox equilibria

• When a metal electrode is placed into a solution of one of its salts two things can happen;

• 1) Metal ions go into solution;

• M(s) → M+(aq) + e-

• 2) metal ions come out of solution.

• M+(aq) + e- → M(s)

• The result is a redox equilibrium;

• M+(aq) + e- ⇌ M(s)

Eg; Cu in a solution of CuSO4.

Copper ions might come out of solution;

Cu2+(aq) +2e- → Cu(s)

Or copper atoms might go into solution;Cu(s) → Cu2+

(aq) +2e-

Giving an eqm; Cu2+ (aq) +2e- ⇌ Cu(s)

Similar equilbria exist between;1) Ions of the same metal in different oxidation states with a platinum or graphite electrode.

Eg; A mixed solution of iron (ii) and iron (iii)

Fe3+(aq) +e- ⇌ Fe2+

(aq)

With an inert graphite electrode.

NB Platinum is inert and under normal circumstances will not form ions.

H2 at 101kPa and 298K

Platinum wire

Porous platinum

2) Ions of a gas in contact with particles of that gas absorbed onto the surface of a platinised platinum electrode.

IE; A platinum electrode on which a black, spongy layer of more platinum has been deposited by electrolysis.

Acidic solution

The charge on the metal electrode depends on the position of the equilibrium;

M+(aq) + e- ⇌ M(s)

If the forward reaction is favoured electrons are “removed” so the electrode will have a positive charge.

If the backward reaction is favoured; Electrons are “released”, giving a negative charge.

Electrode chargesThe potential difference between the electrode and solution depends upon;

Silver

Iron1) The metal involved

2) The concentration of the solution

0.05M

0.1M

Electrochemical cells• Electrochemical cells convert chemical

energy into electrical energy.• They consist of two half cells, each of which

typically has a metal electrode dipping into a salt solution.

• The half cells are connected by a salt bridge, which allows charge to be transferred between the half cells without allowing the solutions to mix.

• As current flows electrons will be transferred from the more reactive to the less reactive metal.

• This is the basic principle behind batteries!

Eg; The Daniel Cell

Consists of two half cells;

A zinc electrode in a solution of zinc sulphate.

A copper electrode in copper sulphate.

The two half cells are then connected via a piece of filter paper soaked in KNO3 – which

avoids the complication of using a metal wire.

Salt bridge

Cu half cell

Zn half cell

EMF

Redox reactions in electrochemical cells.

Oxidation takes place in the zinc half cell;

Reduction takes place in the copper half cell;

Zn(s) → Zn 2+(aq) + 2e- Cu 2+ (aq) +2e- →Cu (s)

When the circuit is completed electrons flow from the more negative to the more positive electrode.

Zinc will enter into the solution.

Copper will come out of the solution and be deposited on its electrode.

Eg in the Daniel Cell from zinc to copper.

Electrode potentials

• Zn(s) → Zn2+(aq) + 2e-

• The equilibrium will be far to the RHS.

• Zinc will acquire a –ve charge.

• It is said to have a –ve electrode potential.

• Its value is a measure of the ability of the metal to act as a reducing agent (ie to release electrons).

• Zinc itself will be oxidised.

Potential difference

• It is not possible to measure the electrode potential of a half cell directly.

• Only the potential difference (aka voltage) between two half cells.

• One cell, the hydrogen cell, is used as a reference and allocated a potential of 0.

• As potential difference depends on temperature and concentration standard conditions are specified.

• Namely 298 K,101 kPa and a concentration of 1 mol dm-3.

Standard hydrogen electrode

H2 at 101kPa and 298K

Platinum wire

[H2SO4] = 0.5 moldm-3

Porous platinum

Standard CellsOne half cell is the standard hydrogen cell…

The solution is 1M with respect to the metal ion.

… the other is a test cell.

• At the standard hydrogen electrode;

• ½H2 (g) → H+ (aq) + e-

• The pd is referred to as the standard electrochemical potential, EΘ.

• For the standard hydrogen electrode EΘ = 0.• In the test cell;

• M+(aq) + e- → M(s)

• EΘ is the pd measured by the voltmetre in the standard cell (aka electromotive force – emf)

• Electrodes with –ve EΘ will be better reducing agents than hydrogen.

• NB as [M+] increases EΘ becomes more positive as reduction is more likely.

Electrochemical Series

Li+(aq) + e- → Li(s) -3.03 V

Ca+2(aq) + 2e- → Ca(s) -2.87 V

Al+3(aq) + 3e- → Al(s) -1.66 V

Zn+2(aq) + 2e- → Zn(s) -0.76 V

Pb+2(aq) + 2e- → Pb(s) -0.13 V

2H+(aq) + 2e- → H2(g) 0.0 V

Cu+2(aq) + 2e- → Cu(s) +0.34 V

Ag+(aq) + e- → Ag(s) +0.8 V

The most negative reduction potentials are written at the top of the series.

The best oxidising agents at the bottom left.

The best reducing agents are at the top right.

Redox equilibria

are written as

reduction.

Cell diagramsCell diagrams are a shorthand used to represent electrochemical cells. To write one first put down the reactions that are taking place;

H2 → 2H+ + 2e- Cu2+ + 2e- → Cu

Separate the oxidised and reduced forms by I and leave out the electrons;

Pt H2 I 2H+ Cu2+ I Cu

Use a double line, II, for the salt bridge

Pt H2 I 2H+ Cu2+ I CuII

Caculating the emfs of cellsThe emf of a cell can be calculated from the EΘ of the half cells.

Zn (s) I Zn2+ (aq) Cu 2+ (aq) I Cu (s)

Zn2+(aq) + 2e- ⇌ Zn (s) Cu2+

(aq) + 2e- ⇌ Cu (s)

EΘ = - 0.76V EΘ = +0.34V

Look up the EΘ values;

EMF = EΘrhc - EΘlhc

= 0.34 – (-0.76) = 1.1V

The cell emf has the sign of the RH electrode – which is the more positive one.

• Alternatively instead of using both half equations in the reduced forms, reverse the half cell that involves oxidation;

• Zn2+(aq) + 2e- →Zn (s) EΘ = - 0.76V

• Zn (s) → Zn2+(aq) + 2e- EΘ = +0.76V

• Then add the of EΘ the other half cell.

• Cu2+ (aq) + 2e- → Cu (s) EΘ = +0.34V

• EMF = +0.76 + +0.34 = +1.1V

What would be the emf of a cell formed by connecting up the following half cells?

Cu+2(aq) + 2e- → Cu(s) Ag+

(aq) + e- → Ag(s)

EΘ = +0.34 V

EΘ = +0.8 V

Cu+2 (aq) + 2e- → Cu (s)

Ag+ (aq) + e- → Ag (s)

Overall EMF = 0.8 – 0.34 = 0.46V

EΘ = +0.8 V

EΘ = +0.34 V

What would be the emf of a cell formed by connecting up the following half cells?

Cu+2(aq) + 2e- → Cu(s) Fe3+

(aq)+ e- → Fe2+ (aq)

EΘ = +0.34 V EΘ = +0.77 V

Overall EMF = 0.77 – 0.34 = 0.43V

Cu+2 (aq) + 2e- → Cu(s)

Fe3+(aq) + e- → Fe2+

(aq)EΘ = +0.34 V

EΘ = +0.77 V

Predicting whether reactions will take place.

• EΘs can be used to predict which redox reactions are possible.

• The more positive EΘ the more likely the reaction.

• Generally if EΘ ≥ +0.4V a reaction is feasible.

• But in practice the rate might be so slow that a reaction will not happen.

• EΘ is defined for standard conditions, if concentrations are increased the reaction may become possible.

Can iodine oxidise bromide ions?

• 1) Write out the equation;

• I2 + 2Br - → 2I- + Br2• 2) Separate this into half equations;

• I2 + 2e- → 2I- EΘ = +0.54 V

• 2Br - → Br2 + 2e- EΘ = -1.09 V

• 3) Add 0.54 +(-1.09) = -0.55V.• 4) -0.55 < 0.4 so the reaction is not

possible.

Competition for electrons

• EΘ of a half cell is a measure of its oxidising or reducing power. IE its ability to compete for electrons.

• Generally stronger oxidising agents have the more positive EΘs.

• Eg; I2 + 2e- → 2I- EΘ = +0.54 V

• Cl2 + 2e- → 2Cl- EΘ = +1.36 V

• Therefore chlorine is a stronger oxidising agent than iodine.

• The more negative EΘs the stronger is the reducing agent. Eg;

• Pb2+ + 2e- ⇌ Pb EΘ = -0.13V

Ca2+ + 2e- ⇌ Ca EΘ = -2.87V

• Therefore calcium is a stronger reducing agent than lead.