ELECTROCHEMISTRY

Electrode potential and half cellHalf cell

2 possible electron transfer reactions occur:M(s) Mn+(aq) + ne- (Oxidation)Mn+(aq) + ne- M(s) (Reduction)Reversible reaction occurs.When the reversible reaction reached equilibrium,

there is a potential difference between the metal and aqueous solution – electrode potential.

Mn+(aq) + ne- M(s), E = electrode potential

Electrode potential may be positive or negative, depending on the position of equilibrium.

Factors affecting electrode potential, E:1)Temperature2)Pressure3)Nature of metal (position in electrochemical series)4)Concentration of ions.If [Mn+(aq)] increased, equilibrium shift to the right.

More electrons removed from electrode electrode becomes more positive.

Because electrode potential is affected by the 4 factors above, therefore electrode potential must be measured under standard condition (298K, 1 atm, 1.00 mol dm-3).

The electrode potential measured under standard condition is called standard electrode potential, Eθ.

Standard Hydrogen Electrode (S.H.E)

*Consist of a platinised Pt electrode *in a solution of 1.00 mol dm-3 H+(aq)

ions. *H2 gas at the pressure of 1 atm is

bubbled over the platinum electrode.

Electrode is alternately bathed in first H+(aq) and then H2(g).

Hydrogen is adsorbed on the Pt and equilibrium is established.

½ H2(g) H+(aq) + e- E = 0.00 VHas a electrode potential of zero.Is used to compare the electrode potentials of other

half-cells (as a reference electrode).Function of Pt electrode:1)As an inert metal.2)Allow’s H2 gas to be adsorbed onto its surface.

3)Covered by loosely deposited layer of finely divided Pt – increase surface area equilibrium achieved faster.

Cell Diagram

(-) electrode (+) electrodeOxidationMore electrons at

surface of electrode.Mn+(aq) + ne-

M(s)Equilibrium shift to left.

ReductionElectrons removed from

electrode.Mn+(aq) + ne- M(s)Equilibrium shift to

right.

Writing a cell diagram:

Metal electrode(s) metal ion(aq) metal ion(aq) Metal electrode(s)

E.g :Zn(s) Zn2+(aq) Cu2+(aq) Cu(s)

Measuring E of metals or non-metals in contact of their ions in aqueous solution.Standard electrode potential, E Is

the electromotive force (e.m.f) of a cell made up of a test electrode connected to a standard hydrogen electrode under standard condition (1 mol dm-3 concentration of H+ at 25C and 1 atm.)

Unit : Volts

Measuring E of metals or non-metals in contact of their ions in aqueous solution.

Standard electrode potential (E) of Zn

Reaction at electrodes: Oxidation : Reduction : Overall reaction :

Ecell = - 0.76 V (Zn electrode is negative – oxidation occurs and electron produced)

Cell diagram :

Standard electrode potential (E) of Mg

Reaction at electrodes: Oxidation :Reduction :Overall reaction :

Ecell = - 2.37 V (Mg electrode is negative – oxidation occurs and electron produced)

Cell diagram :

Standard electrode potential (E) of Cu

Reaction at electrodes: Oxidation :Reduction :Overall reaction :

Ecell = +0.34 V (Cu electrode is positive – reduction

occurs and electron removed)Cell diagram :

Standard electrode potential (E) of Cl2

Reaction at electrodes: Oxidation :Reduction :Overall reaction :

Ecell = +1.36 V

Cell diagram :**the least oxidised species is placed next to the inert

electrode.

Measuring E of ions of the same element in different oxidation states.

Standard electrode potential (E) of Fe3+/Fe2+

Reaction at electrodes: Oxidation :Reduction :Overall reaction :

Ecell = +0.77 V

Cell diagram :**the least oxidised species is placed next to the inert electrode.

Standard electrode potential (E) of Cr3+/Cr2+

Reaction at electrodes: Oxidation :Reduction :Overall reaction :

Ecell = - 2.37 V (Zn electrode is negative – oxidation

occurs and electron produced)Cell diagram :

Uses of standard electrode potential (E)

1) To determine strength of both oxidising and reducing agents.

Electrochemical series:

2) To predict the direction of electron flow from a simple cell.

Electron flow from the negative electrode (more negative E value) to the positive electrode (more positive E value)

Electrochemical cell

Two half-cells joined by a salt bridge containing KCl or KNO3.

Function of salt bridge : 1) Complete the circuit.2) Maintains electrical neutrality in the cell.

Salt bridge (KCl)

Electrochemical cellNegative electrode Positive electrode

Metal with larger negative E - Zn.

Oxidation occurs.Metal dissolves in the

solution :Zn(s) Zn2+(aq) + 2e-

Electron flows through the wire and voltmeter to Cu electrode.

Metal with larger positive E - Cu.

Reduction occurs.Electron from Zn half

cell taken by Cu2+(aq).Cu metal formed and

depositied at Cu electrode.

Cu2+(aq) + 2e- Cu(s)Cell reaction :

Oxidation : Zn(s) Zn2+(aq) + 2e-

Reduction : Cu2+(aq) + 2e- Cu(s)Overall : Zn(s) + Cu2+(aq) Zn2+(aq) + Cu(s)

Electrons flow from negative electrode through the wire and voltmeter to positive electrode.

This cause a potential difference – electromotive force (e.m.f)

e.m.f is the maximum potential difference of two metals.

It gives a quantitative measure of the likelihood of the redox reaction taking place in the cell.

Standard Cell Potential (e.m.f), E

cell.

* Is the measure of tendency of electrons to flow through the external circuit

* under standard conditions of 25C, 1 atm and 1.00 mol dm-3 concentration.

* It is the maximum potential difference between the electrodes.

Example :

Cell diagram :

Standard Cell Potential (e.m.f), E

cell.

Calculating Ecell : Example 1

Cell reactions : E/VOxidation :Reduction :Overall :

Note : a) when reaction is reversed, the sign is changed.

b) Eθ is not dependent on the num. of electrons transferred.

Formula method :Ecell = Ered

- Eox

Example 2A cell consisting of half-cells : Cu in CuSO4(aq)

and Ag in AgNO3(aq). Write the cell diagram, reaction equation in each electrode, predict the electron flow and calculate the standard cell potential.

Exercises

Uses of Eθcell

To predict the feasibility of a reaction.Reactions with positive Eθ

cell are thermodynamically feasible.

E.g:

Mg2+(aq) + 2e- Mg(s) E θ = -2.37 VCu2+(aq) + 2e- Cu(s) E θ = +0.34 V

Half equations:Oxidation :Reduction :Overall :*E θ

cell > 0 , therefore reaction is feasible.

E.g: Is this reaction feasible?Pb(s) + 2Cr3+(aq) Pb2+(aq) + 2Cr2+(aq)

Limitations of prediction made using Eθ

cell value. Using Eθ

cell to predict feasibility of reaction might not be suitable when:

a) The reaction has very high activation energy (kinetically slow).

Eθ value give no information about reaction rates.

b) The reactions are not in standard conditions.

Eθ value apply to standard conditions only (298K, 1 atm, all concentrations at 1.00 mol dm-3).

b) Reactions are not in standard conditions

E.g : Oxidation of concentrated HCl by MnO2

MnO2 + 4H+ + 2e- Mn2+ + 2H2O Eθ = +1.23 V

2Cl- Cl2 + 2e- Eθ = -1.36 V

MnO2 + 4H+ + 2Cl- Mn2+ + Cl2 + 2H2O Eθcell = -0.13

V*Reagents are heated.Eθ

cell is negative, reaction predicted not feasible.But the reactions is not in standard condition : [H+]

≠ 1.00 mol dm-3, temperature ≠ 25°C.Cl2 can be prepared by heating MnO2 with

concentrated HCl.

a) The reaction has very high activation energy

E.g 1 :

Cu2+(aq) + 2e- Cu(s) Eθ = +0.34 V

H2(g) 2H+(aq) + 2e- Eθ = 0.00 V

Cu2+(aq) + Zn(s) Zn2+(aq) + Cu(s) E θ cell = +0.34 V

Eθcell value is positive, reaction predicted feasible.

However, the rate of reaction is too slow (due to high Ea). Hence in practice, this reaction does not occur at standard condition.

The prediction fails because Eθcell value gives no information

about the kinetics of the reaction.E.g. 2 : Oxidation of methanal to methanoic acid by K2Cr2O7.

3HCHO + Cr2O72- + 8H+ 2Cr3+ + 3HCO2H + 4H2O E θ

cell = +1.27V

E θ cell is positive, reaction predicted feasible, but in practice no

reaction occur (high Ea).Heating is required for reaction to occur.

Exercise1. Water can be oxidised to oxygen according to the

equation:

2H2O(l) O2(g) + 4H+(aq) + 4e-

Given the following E values, what can be use to oxidise water in this way?

Exercise

Effect of concentration of ions on E θ cell

value.

E.g 1 : Fe3+(aq) + e- Fe2+(aq) Eθ = +0.77 V

If [Fe3+] > 1.00 mol dm-3, equilibrium will shift to the right causing Eθ to be more positive.

If [Fe3+] < 1.00 mol dm-3, equilibrium will shift to the left causing Eθ to be more negative.

If [Fe2+] > 1.00 mol dm-3, equilibrium will shift to the left, Eθ will be more negative.

If [Fe2+] < 1.00 mol dm-3, equilibrium will shift to the right, Eθ will be more positive.

Effect of concentration of ions on Eθcell

value.E.g 2 :Cr2O7

2-(aq) + 14H+(aq) + 6e-(aq) 2Cr3+(aq) + 7H2O(l)

E θ = +1.33 V.What will happen to the Eθ value and the strength of the Cr2O7

2- solution as an oxidising agent if :

a) [Cr2O72-(aq) ] were to be increased.

b) [H+(aq)] were to be decreased.c) [Cr3+(aq) ] were to be increased.

Electrolysis

Is a process in which current is passed through the electrolyte (in molten or aqueous solution).

Anode(+) : oxidation occurs.Anions go the anode and are discharged.X- X + e-

Cathode (-) : reduction occurs.Cations go to the cathode and are discharged.M+ + e- MOnly 1 element at a time can be deposited at

each electrode.E.g : electrolysis of molten NaCl.At anode : 2Cl-(l) Cl2(g) + 2e- (oxidation)At cathode : Na+(l) + e- Na(s) (reduction)

Selective discharge of ions. Factors affecting selective discharge during

electrolysis :1. Position in the Redox series. Cations with the most positive (or less

negative) Eθ value is discharged first. Anions with the most negative (or less

positive) Eθ value is discharged first. E.g : What is discharged during electrolysis of a

mixture of CuSO4(aq) and NaI(aq)?

2. Relative concentration of ions. Ions of higher concentration will be discharged. E.g 1 : Electrolysis of brine (concentrated NaCl). At the anode:

Concentration of Cl- ion is much higher. Hence Cl- ion is discharged at the anode instead

of OH- ions. Hydrogen ions are discharged at the cathode

instead of Na.

The Eθ value of the forward reaction is negative, therefore the reaction is not feasible.

In electrolysis of dilute NaCl, the OH- ion is discharged instead (OH- ion is more negative)

3. Nature of the electrodes.

Quantitative Electrolysis Faraday’s first law : Q = I x tQ = amount of electric charge that flows through

an electric circuit, unit : C.I = current, unit : AT = time, unit : sThe mass of substance discharged at an electrode

in electrolysis is directly proportional to the amount of electric charge.

m α Q m = mass of substance liberated, unit : g

m α I x t

Example When a current of 0.400 A is passed through an

electrolyte for 30 minutes at 25°C, 2.00 g of metal is produced at the cathode. What mass of metal would be produced at the cathode if:

a) A current of 0.800 A is passes through the electrolyte for 30 minutes at 25°C.

b) A current of 0.200 A is passes through the electrolyte for 2.0 hours at 25°C.

c) A current of 0.400 A is passes through the electrolyte for 30 minutes at 60°C.

Faraday constant, F, is the amount of charge on 1 mole of electrons.

F = L x e where L = Avogadro constant

e = charge on the electron (1.60 x 10-19C)

1 F = 96 500 C mol-1

Faraday’s second law : Mn+(aq) + ne+ M(s)nF of charge is needed to form 1 mole of

M(s)

Questions:1. 30.0 g of metal samarium (Sm) was discharged by

electrolysis with a current of 2.4 A flowing for 24 125 s. What is the formula of samarium ions?

[Sm = 150] Hint : I mole of electron gives 1 F.

2. A current of 4.00 A was passed through copper(II) sulphate solution 965 s. what mass of copper was discharged at the cathode?[F = 96 500C mol-1; Cu = 63.5]

3. Calculate the volume of oxygen gas produced in cm3 (measured at s.t.p) when 2.00 A is passed through dilute H2SO4, for 1830 s. using platinum electrodes.

4. A current was passed through the circuit shown in figure above. In the experiment, 5.4 g of silver metal was discharged at the cathode in electrolysis cell B. What mass of copper was discharged at the cathode in cell A?[Cu = 63.5; Ag = 108]

The Avogadro’s constant (L) can be found from the results of an electrolysis experiment using the relationship, F = L x e

e.g : Nov 2008, paper 5.

Method :The cathode is cleaned and weighed before being placed in

the copper(II) sulphate solution.The circuit is completed and the current set at 0.3 A by adjusting the variable resistor.

The current is maintained at 0.3 A for exactly 40 minutes at which point the circuit is broken.

The cathode is removed from the solution and carefully washed with distilled water to remove any copper(II) sulphate solution.

Distilled water is removed from the cathode by rinsing it with propanone in which the water dissolves.The cathode is finally dried by allowing the propanone to evaporate from its surface.

The cathode is reweighed and placed back in the solution.A constant current of 0.3 A is passed for a further 40

minutes when the rinsing, drying and weighing are repeated.

Initial mass of cathode : 115.74g Mass of cathode after 40 min : 115.97g Mass of copper metal discharged :_______gCalculate a value for the Avogadro constant (L).[charge on the electron = 1.60 x 10-19 C; Cu = 63.5]