spontaneity of redox reactions

Spontaneity of Redox Reactions How is related to other thermodynamic quantities such as G and K. In an electrochemical cell, chemical energy is converted to electrical energy. Electrical energy in this case is the product of the emf of the cell and the total electrical charge (in coulombs) that passes through the cell:

The total energy charge is determined by the number of moles of electrons(n) that pass through the circuit. By definition

Where F is Faraday constant, is the electrical charge contained in 1 mole of electrons. Expts. Show 1 faraday is equivalent to 96,485.3 coulombs or 96,500 coulombs ( roughly).

Chang, R.(2002). Chemistry 7th ed.Mc graw Hill. bkk


1F = 96,500 C/molBecause 1J = 1 C X 1VWe can also express the units of faraday as 1 F = 96,500 J/V. molThe measured emf is the max. voltage that a cell can achieve. This value is used to calculate the max. amount of electrical energy obtained from the chemical reaction. This energy is used to do the electrical work (Wele) henceWmax =W ele= - n F Ecell

The –ve sign on the right hand side indicates that the electrical work done by the system on the surroundings.G is the free energy available to do work , hence G = Wmax

We can now write G = -n F Ecell both n and F are positive quantities and G is negative quantity for a spontaneous process, so Ecell must be positive quantity. Under standard state conditions the above eqn. becomesG0 = -n F E0

cell ………………..(1)


From our study on second law of thermodynamics we know the relation between free energy and equilibrium constant (K)G = - RT ln K …………(2)Combining eqn. (1) and (2) -n F E0

cell= - RT ln K

Solving for E0cell = …………(3)

when T = 298K, eqn (3) can be simplified by substituting for R and F:

E0cell = =

the eqn can be rewritten using base 10 logarithm

E0cell =


Thus, if any of the three quantities G , K or E0cell is known, the

other quantities can be calculated.


Eg. Calculate the equilibrium constant for the following reaction at 25 C:Sn(s) +2Cu2+ (aq) Sn2+(aq) +2 Cu+ (aq)Solution : to calculate the K first calculate E0

cell of the Sn2+/Sn and Cu2+ /Cu couples. The half cell reaction are Oxidation: Sn(s) Sn2+(aq) + 2e-

Reduction: 2Cu2+ (aq) +2e- 2 Cu+ (aq)From table we findE0 Sn

2+/Sn = -0.14 V and E0 Cu

2+/Cu

+ = 0.15 V. Thus

E0 cell = E0 Cu2+

/Cu+

- E0 Sn2+

/Sn=0.15 V – (-0.14)= 0.29V

E0 cell =

In the overall reaction we find n=2. ThereforelnK =22.6 K=e22.6 K = 6.5 X 109

Practice: Calculate the equilibrium constant for the following reaction at 25 C : Fe2+(aq) +2Ag(s) Fe(s) +2Ag+(aq)


Calculate the standard free energy change for the following reaction at 25˚ C.2Au(s) + 3 Ca2+ (1M) 2Au3+(1M) + 3Ca(s)To calculate G0 for the reaction we can use G0 = -n F E0

, first we need to calculate E0 .

Calculate G0 for the following reaction at 25˚ C.2 Al 3+ (aq) + 3Mg(s) 2Al(s) + 3Mg 2+ (aq)


The Effect of Concentration on Cell EmfThe discussion so far were on redox reactions in their standard state conditions, but standard states are difficult to maintain and sometimes impossible. Nevertheless, there is a mathematical relationship between the emf of a cell and the concentration of reactants and products in a redox reaction under non standard state conditions.

The Nernst EquationConsider a redox reaction of the type aA + bB cC + dDfrom equation G = G˚ + RT ln Q ( from thermodynamics)Because G = nFE and G˚ = nFE˚, the equation can be expressed as nFE = nFE˚ + RT ln Q dividing this eqn by –nF, we get

E=E˚ lnQ (The Nernst Equation) where Q is the reaction quotient


E = E˚ ln Q At 298 K, the Nernst equation becomes E = E˚ ln Q (nat log) or E = E˚ log Q (base 10 log)

At equilibrium, there is no net transfer of electrons, so E=0, Q=K, where K is the equilibrium.The Nernst equation enables us to calculate E as a function of reactant and product concentrations in a redox reaction. For example, for the Daniell cell Zn(s) + Cu 2+ (aq) Zn2+ (aq) +Cu(s)

E = E˚ ln


If the ratio [Zn2+/Cu2+] is less than 1, ln [Zn2+/Cu2+] is a negative number, so the second term on the right- hand side of the above equation is positive. Under this condition E is greater than the standard emf E˚. If the ratio is greater than 1, E is smaller than E˚.

Predict whether the following reaction would proceed spontaneously as written at 298 K:Co(s) + Fe 2+(aq) Co2+ (aq) + Fe(s) given that [Co2+ ]= 0.15 M and [Fe2+ ]= 0.68 M.

Reasoning and solution : We can use the value of E for the redox process to determine the spontaneity of the reaction. To do so we need to know E˚ and ln Q in the Nernst equation. The half reaction reactions are

Oxidation: Co(s) Co2+ (aq) + 2e- Reduction: Fe 2+(aq) + 2e- Fe(s) The values of E˚Co

2+/Co= - 0.28V and E˚ Fe

2+/Fe= -0.44V. There fore the standard

emf is E˚ = E˚ Fe

2+/Fe - E˚Co

2+/Co= -0.44V-(-0.28V) = -0.16V


Lets find the value of E

E = E˚ - = - 0.16 V -

= - 0.16+ 0.019V = - 0.14VBecause E is negative, the reaction is not spontaneous in the direction written.

Practice ques: Will the following reaction occur spontaneously at 25˚C, given that the [Fe 2+]=0.06M and [Cd2+]= 0.010M?

Cd(s) + Fe 2+(aq) → Cd2+(aq) + Fe(s)


Now suppose you want to know what ratio of [Co2+]/[Fe2+] the reaction will would be spontaneous we can use the equation

E = E˚ - We set E=0, which corresponds to the equilibrium situation 0 = - 0.16 V -

= -12.5

= e-12.5= K =4 X 10 -6

Thus for the reaction to be spontaneous, the ratio [Co2+]/[Fe2+] must be smaller 4 X 10 -6.


Consider the electrochemical cell shown below. In a certain experiment, the emf(E) of the cell is found to be 0.54V. Suppose that the [Zn2+]= 1.0M and pH2

=1 atm. Calculate [H+].

Write the redox reaction to find n, we know the standard reduction potential of E˚ Zn2+ /Zn= 0.76V, substitute the values. ( ans= 2X10-4M)

It may be useful to realize here that an electrochemical cell which involves the measurement of H+ ions can be used to measure pH. This principle is used in pH meter.

Home work: What is the emf of a cell consisting of a Cd2+/ Cd half cell and a Pt/H+/H2 half cell if [Cd2+ ] = 0.20M, [H+]= 0.16M and pH2

=0.80atm?

This can be solved using the Nernst equation. Note that the conc. of gas expressed in atm.


Concentration cells Electrode potentials depend on ion concentration, it is possible to

construct a cell from two half cell of the SAME material but differing in ion conc. Such a cell is called concentration cell.

Two zinc electrodes in two different aq. Solution of zinc sulphate. Eg 1.0M and 0.1M

There fore reduction occur in 1.0M compartment and oxidation occur in 0.1M compartment. The cell diagram


And the half-reactions are Oxidation: Zn(s) → Zn 2+ (0.10M) +2e-

Reduction: Zn 2+ (1.0M) +2e-→ Zn(s) Overall: : Zn 2+ (1.0M) → Zn 2+ (0.10M) The emf of the cell is

The of this cell is zero ( same electrode, same ions) so the emf E of the cell is

The potential is small which eventually becomes zero when the conc. Becomes same.

A biological cell can be compared to conc. Cell. Membrane potential develop due to unequal conc of same ions on different sides of membrane. Nerve impulses, heart beat, communication in living cells.


Batteries An electrochemical cell or a series of combined electrochemical cells that

can be used as a source of direct electric current at a constant voltage. The operation of battery is similar in principle to the elctrochemical cells (galvanic, daniell, voltaic), a battery has the advantage of being completely self contained . ( doesn’t require salt bridge or other components).

Types of batteries The dry cell battery or Leclanche’ cell The mercury battery The lead storage battery The solid –state lithium battery Fuel cells


Corrosion: Deterioration of metals by an electrochemical process.Corrosion causes enormous damage-buildings, bridges, ships, automobiles etc.

Examples-Rusting of iron, tarnish on silver, green patina on copper or brass etc.What happens in rusting?In the formation of rust on iron, oxygen and water must be present. The reaction

involved are quite complex and not completely understood, the main steps are believed to be as follows.

The metal surface acts as the anode where oxidation occursFe (s) → Fe2+ (aq) + 2e- E=- 0.45 VThe electrons given up by iron reduce atmospheric oxygen to water at cathode

which is another region of the same metal surface:O2 (g) + 4H+(aq) + 4e- → 2H2O(l) E= 1.23V (note that the E of the reduction half reaction will be less than 1.23V because the

H+ is not 1 M but even at pH 7 the reduction half reaction is 0.81 V, which means the cell potential is positive and the reaction is spontaneous.)

The overall reaction is 2Fe (s) + O2(g)+ 4H+(aq)→ 2Fe2+(aq) +2H2O(l)E= Ecat - Eand

= 1.23V-(-0.44V)= 1.67 V


Notice the reaction occurs in acidic medium : the H+ ions are supplied in part by the reaction of atmospheric CO2 with H2O to form H2CO3. the Fe2+ ions is further oxidized by oxygen.

Fe2+ (aq) + O2(g) + (4 + 2x) H2O(l)→ 2Fe2O3. xH2O + 8H+ (aq) The hydrated form of iron(III) oxide is called rust. As the amount of water

associated with iron oxide varies, we represent the formula as 2Fe2O3. xH2O


The electrical circuit is completed by the migration of electrons and ions causing rusting rapidly. The fast rusting of automobiles in cold placed is due to salts (NaCl or CaCl2) spread on the roads to melt ice.

Aluminum has greater tendency to oxidize than Iron (compare reduction potential). We might expect to see lot of corroded aluminum like soda cans, our cooking pots etc. How ever this doesn’t happen because Al forms insoluble Al2O3 layer when it is exposed in air. This layer is not porous ( like iron) and protects the underlying metal.

Copper forms a layer of green substance called patina (CuCO3) which protects the underlying metal

Silver forms silver sulfide when it comes in contact with food stuff.


Methods adopted to protect metals from corrosion.Most methods are targeted to prevent rust formation.Painting metal surface with paint. But sooner the paint is scratched the iron

is exposed and rusting starts. Metals such as chromium, tin or zinc afford a more durable surface coating.

The steel used in making automobiles, for example, is coated by dipping it in a bath of molten zinc, a process known as galvanizing.

As the potentials indicate Fe 2+(aq) + 2e- Fe(s) E˚= -0.45V Zn 2+ (1.0M) +2e-→ Zn(s) E˚= -0.76 VZinc is oxidized more easily than iron, and therefore when the metal

corrodes, zinc is oxidizes instead of iron. An incipient oxidation of iron would reverse the process immediately since Zn can reduce Fe2+.

As long as zinc and iron are in contact , Zn protects iron from oxidation even if zinc layer is scratched.


Cathodic protection is The technique of protecting a metal from corrosion by connecting it to

second metal that is more easily oxidized. It is unnecessary to cover the entire surface of the metal with second

metal, as in galvanizing. All it requires is an electrical contact with the second metal. An under steel pipeline can be protected by connecting it through an insulated wire to a stake of magnesium, which acts as a sacrificial anode and corrodes instead of the iron. In effect the arrangement is a galvanic cell in which the easily oxidized magnesium acts as the anode and the pipeline behaves as the cathode and the moist soil acts as the electrolyte.

Anode: Mg(s) Mg2+ (aq) + 2e- E˚= 2.37V* Cathode: O2(g) + 4 H+(aq) +4e- 2H2O(l) E˚ =1.23V* *written as oxidation potential PROBLEM Mg is often attached to the hull of ships to protect to steel from rusting .

Write a balanced equation for the corrosion reactions that occur (a) in the presence of Mg and (b) in the absence of Mg.


Electrolysis Process in which electrical energy is used to cause a nonspontaneous

chemical reaction to occur. An electrolytic cell is an apparatus used for carrying out electrolysis. The same principles underlie electrolysis and the processes that take place in a electrochemical cell.

Electrolysis of molten sodium chloride. Sodium chloride is an ionic compound which can be electrolyzed to form

sodium metal and chlorine. The cell used for large scale electrolysis of NaCl is called a Down’s cell.


The reactions at the electrodes are Anode(oxidation) : 2Cl-(l)→ Cl2(g)+ 2 e-

Cathode(reduction: 2Na+(l) +2e-→ 2Na(l) Overall; 2Na+(l) + 2Cl-(l)→ 2Na(l)+ Cl2(g) This process is the major source of pure sodium metal and

chlorine gas. Theoretical aspects show that the E˚ value for the overall process is about -4V, which means the reaction is non spontaneous process. There fore a

minium voltage of 4V must be supplied by the battery to carry out the reaction. Inpractice however a higher voltage is required because of inefficiencies in the electrolytic process because of over voltage.


Electrolysis of water. Study on your own ELECTROLYSIS of of Aqueous Sodium Chloride This is the most complicated electrolysis. In the electrolysis of

aqueous NaCl, the solution contains several species that could be oxidized or reduced. The possible oxidations reaction at anode are

2Cl-(aq)→Cl2(g) + 2e-

2H2O(l)→O2(g) + 4H+ (aq) +4e-

Comparing their standard electrode potential Cl2(g) + 2e-→2Cl-(aq) E= 1.36V O2(g) + 4H+ (aq) +4e-→2H2O(l) E= 1.23V The E are not very different but the values suggest that

H2O should be preferentially oxidized at the anode. But in the actual experiment the gas liberated is Cl2 and not O2. In studying electrolytic process we sometimes find that the voltage required for a reaction is higher than the electrode potential indicates. Overvoltage is the difference between the electrode potential and the actual voltage required to cause electrolysis. The over voltage for O2 is quite high. Therefore under normal operating conditions Cl2 is formed instead of O2.


The reductions that might occur are 2H+(aq) + 2e- →H2(g) E= 0.00V 2H2O(l)+2e- → H2(g) + 2OH- (aq) E= -0.83V Na+ (aq) + e-→ Na(s) E= -2.71V Reaction 3 is ruled out because it has a very negative

potential. Reaction 1 is preferred over reaction 2 but at pH 7 both are equally probable. Reaction 2 is used to describe the cathode reaction because the concentration of H+ is too low ( 1 X10-7 M) to make reaction 1 reasonable

Thus the half cell reactions are Anode (oxidation): 2Cl-(aq)→Cl2(g) + 2e-

Cathode (reduction) 2H2O(l)+2e- → H2(g) + 2OH- (aq) Over all: 2H2O(l)+ 2Cl-(aq)→ H2(g) + Cl2(g) +2OH- (aq) The over all reaction shows that as the Cl- ion concentration

decreases OH- ion increases. In addition to H2 and Cl2 gas produced, a useful by product NaOH is obtained by the aqueous solution at the end of electrolysis


Note that from the analysis of electrolysis : cat ions are likely to be reduced at the cathode and the anions likely to be oxidized at anode and in aqueous solutions water itself may be oxidized or reduced. The out come depends on the nature of other species.

Quantitative aspects of Electrolysis Developed primarily by Faraday- he observed that the mass of the product

formed ( or reactant consumed) at an electrode is proportional to both the amount of electricity transferred at the electrode and the molar mass of the substance in question.

For E.g. In the electrolysis of molten NaCl, The cathode reaction tells us that one Na+ ion accepts an electron from the electrode. To reduce one mole of Na+ ions, we must supply Avogadro’s number (6.02 X 1023) of electrons to the cathode. On the other hand, the stoichiometry of the anode reaction shows that the oxidation of two Cl- ion yields one chlorine molecule. Therefore, the formation of 1 mole of Cl2 results in the transfer of 2 mols of electrons from Cl- ions to the anode. Similarly, it takes 2 moles of electrons to reduce 1 mole of Mg2+ ions and three moles of electrons to reduce 1 mole of Al3+ ions.


In an electrolysis experiment, we measure the current ( in amperes, A) that passes through an electrolytic cell in a given period of time. The relationship between charge ( in coulombs, C) and current is

1C= 1A X 1s That is, a coulomb is a quantity of electrical charge passing any point in 1

second when the current is 1 ampere. Steps involved in calculating the quantities of subs produced in electrolysis.

For example: Consider electrolysis for molten CaCl2 in an electrolytic cell. Suppose a current of 0.452 A is passed through the cell for 1.50 hr. How much product will be formed at cathode and anode.

Solution write the half reactions.Anode ( oxidation): 2Cl-(l) → Cl2(g) + 2 e-

Cathode ( reduction): Ca2+(l) + 2 e-→ Ca (l)Over all : Ca2+(l) + 2Cl-(l) → Ca (l)+ Cl2(g)


The quantities of calcium metal and chlorine gas formed depends depends on current X time, or charge;

C= ? C= 0.425A X 1.5 hr X 3600s/hr X1 C /1A.s= 2.44 X 103 C

1 mol e- =96,500 C and 2 mol of e- are required to reduce 1mol of Ca2+ ions, the mass of Ca metal formed at the cathode is calculated as follows

No. of mols of electron = 2.44 X 103 C / 96,500 C mol-1 2 mols of e- reduce 1 mole of Ca2+

Therefore, 2.44 X 103 C / 96,500 C mol-1 will reduce

=1/2 X 2.44 X 103 C / 96,500 C mol-1 =0.01264 mols

Grams of Ca formed = 0.01264 mols X40.08g/mol =0.5067g

spontaneity of redox reactions

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