Electrochemistry

Introduction:Few common objects based on electrochemistry that are helping the mankind:

Batteries

Pacemaker

pH meter

•Redox processes

•Thermodynamics of redox reactions

•Electrode potential

•Nernst equation

•Electrochemical cells

•Hydrogen fuel cells

We will study the following in this section:

Electrochemistry is all about the chemistry between two chemical species, where one loses electron and the other accepts that!

Such exchange of electrons is known as REDOX processes.

Balancing a redox reaction:

LEO GERmany

goes to

Losing Electron is Oxidation Gaining Electron is Reduction

OIL RIG

Oxidation Is Losing Reduction Is Gaining

Common acronyms that help us to solve problems:

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7

8

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Example:

Identify the species that are getting oxidized or reduced:

ClO3- + I- I2 + Cl-

NO3- + Sb Sb4O6 + NO

10

(in acidic medium)

11

12

13

electrons

14

15

DO it yourself!

MnO-4 (aq) + Fe2+ (aq) Mn2+ (aq) + Fe3+ (aq)

Balance the following reaction occurring in acidic media:

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Balancing the following reaction in basic medium:

BH4- + ClO3

- H2BO3- + Cl-

Oxidation half reaction:

BH4- + 3H2O H2BO3

- + 8e- + 8H+-1 +1

Reduction half reaction:

ClO3- + 6e- + 6H+ Cl- + 3H2O

-1+5

Let’s assume the reaction taking place in acidic medium!

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(BH4- + 3H2O H2BO3

- + 8e- + 8H+)

ClO3- + 6e- + 6H+ Cl- + 3H2O

× 3

× 4( )

Cancelling the electrons and adding

3BH4- + 4ClO3

- 3H2BO3- + 4Cl- + 3H2O

Lets assume the reaction taking place in basic medium!

BH4- + 8OH- H2BO3

- + 8e- + 5H2O

ClO3- + 6e- + 3H2O Cl- + 6OH-

(

(

)

)

× 3

× 4

3BH4- + 4ClO3

- 3H2BO3- + 4Cl- + 3H2O

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Assume that someone has predicted oxidation number of boron like this:

BH4- + ClO3

- H2BO3- + Cl-

-5 +3

Try it out to see whether there will be any change in the final balanced equation!

MnO-4 (aq) + Fe2+ (aq) Mn2+ (aq) + Fe3+ (aq)

Our next reaction was the following to be balanced in acidic medium

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Reduction half cell reaction:

MnO4-(aq) + 5e- + 8H+ Mn2+(aq) + 4H2O

5Fe2+(aq) 5F3+(aq) + 5e-

Oxidation half cell reaction:

MnO4-(aq) + 5Fe2+(aq) 8H+ Mn2+(aq) + 5Fe3+(aq) + 4H2O

Balancing redox reaction

Electrochemical cell

Galvanic cell Electrolytic Cells

Spontaneous reactions occur in galvanic (voltaic) cells

Non-spontaneous reactions occur in electrolytic cells

Galvanic cell

Chemical Cells Concentration Cells

Chemical cell with transference

Chemical cell without transference

Electrolyte Concentration Cell

Electrode Concentration Cell

Cell with transference

Cell without transference

Galvanic cell

Left electrode:   Zn(s) → Zn2+ + 2e– Oxidation ( Anodic )Right electrode:   Cu2+ + 2e–→ Cu(s) Reduction ( Cathodic)

The net reaction is the oxidation of zinc by copper(II) ions:

Zn(s) + Cu2+ → Zn2+ + Cu(s)

Half cell reactions:

Cell description conventions Zn(s) | Zn2+(aq) || Cu2+(aq) | Cu(s)

The two electrolytes (the solutions) must be in contact.

Violation of electroneutrality

The reaction stops after only a chemically insignificant amount has taken place.

In order to sustain the cell reaction, The charge carried by the electrons through the external circuit must be accompanied by a compensating transport of ions between the two cells.

This ionic transport involves not only the electroactive species Cu2+ and Zn2+, but also the counter-ions, which in this example are nitrate, NO3

-.

Transport of ions

Barrier between the two solutions can be a porous membrane, or a salt bridge, is used.

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Spontaneity of electrochemical cellsThe question is how can we know which electrochemical reactionswould produce electricity spontaneously when connected with a circuit?

• It depends whether the reaction is thermodynamically favorable!

That essentially means, we have to predict in which direction, an electrochemical reaction would proceed according to the thermodynamic considerations!

Therefore, in the next few slides we will go through few fundamental concepts and parameters of thermodynamics as briefly and as simply as possible…..

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Gibbs free energy (ΔG):

J. W. GibbsUSA, 19th century

iron iron oxide (rust)The change is spontaneous in open atmosphere.

So we say that under the above condition, iron oxide isthermodynamically more stable than iron.

In order to express the above phenomenon in the language of thermodynamics, according to the work of J. W. Gibbs, we state the Gibbs energy (ΔG) of the final state must be less than the initial state.

Or, ΔG (iron oxide) < ΔG (iron) and therefore, ΔG = ΔG (final) – ΔG (initial) = negative

Please note that ΔG = -ve doesn’t mean that any reaction will always happen in that direction & ΔG = +ve doesn’t mean that reaction will not happen in that direction.

For a spontaneous chemical transformation, ΔG = -ve

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Chemical reaction, equilibrium and ΔG

aA (aq) + bB (aq) cC (aq) + dD (aq)Consider the following chemical reaction occurring reversibly at const. T and P,

At equilibrium,

K = {C}c

eq {D}deq

{A}aeq {B}b

eq Please note: {C} denotesthe activity of species C

At any time,

Q = {C}c {D}d {A}a

{B}b Q: reaction quotient

If at any time Q > K, the reaction shifts towards left side!

If at any time Q < K, the reaction shifts towards right side!If Q = K, equilibrium is reached.

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Chemical equilibrium

{C}ceq {D}d

eq {A}a

eq {B}beq

{C}c {D}d {A}a

{B}b =

or, in other words chemical equilibrium is reached when

At this condition, ΔG = 0 is true for the reaction

aA (aq) + bB (aq) cC (aq) + dD (aq) [at equilibrium]

Q = = K

and ΔG ≠ 0 is true for the same reaction at any time other than equilibrium!

aA (aq) + bB (aq) cC (aq) + dD (aq) [not at equilbrium]

Reaction is running reversibly but not under equilibrium so ΔG ≠ 0

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Chemical equilibriumPrevious scenario can be explained mathematically like this:

eq

ξreactant end product end

equilibrium point

δGδξ

< 0δGδξ

> 0

δGδξ

= 0

Free energy does not change with extent of reaction at equilibrium point

under constant T & P

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Gibbs energy (ΔG) and reaction quotient (Q):

ΔG = ΔG0 + RTlnQ

There exists a fundamental equation for a reversible chemical reaction:

ΔG0 is defined as the standard free energy of the reaction at 25o C, at 1 atm pressure and all solutes (non-ionic) have concentration = 1 M (if aqueous) under pH = 7.0

At equilibrium, ΔG = 0 and Q = K, so the above equation becomes

ΔG0 = - RTlnK

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ΔG, Q & electrochemical cell potential (E):

In mechanics, ‘potential’ implies the ability/capacity of performing a work.

Similarly, free energy (ΔG) is defined is the maximum amount of non-expansionwork (Wa) that can be extracted from a closed system, where Wa is maximum when the closed system undergoes a completely reversible change.

In an electrochemical reaction, this non-expansion work is directly related to theelectrochemical potential (E) or the capacity of electrochemical work done by thereaction.

Wa, max = ΔG

-nFE = ΔGHere, n = number of moles of electron transferred in the cell reaction F = Faraday’s constant of charge of 1 mol electrons = e-NA

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ΔG = ΔG0 + RTlnQWe know,

or, -nFE = -nFE0 + RTlnQ

or, E = E0 – (RT/nF)lnQ The Nernst equation!

Gibbs energy and Nernst equation

Gibbs energy accounts for a cell’s voltage!

E0 is the cell potential at standard condition, known as standard cell potential

E is known as cell potential or electromotive force (EMF) or cell voltage

If the above equation is applied for a half cell reaction , thenE becomes the half cell potential or simply the electrode potential.

E0 becomes standard half cell potential or simply the standard electrode potential

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Working form of Nernst equation:

E = E0 – (RT/nF)lnQ

Ecell = E0cell – (RT/nF)lnQ

So, we can write

Ecell = E0cell – 2.303(RT/nF)log10Q

R = 8.314 Jmol-1K-1

T = 298 K

F = 96485 Cmol-1

2.303(RT/nF)

0.059 Vn

=Ecell = E0cell – (0.059 V/n)log10Q

It is interesting to note that when a reversible electrochemical reaction is at equilibrium,

ΔG = 0, so Ecell = o That means cell cannot perform any external work!

So, E0cell = (0.059 V/n)log10 K

We will be able to measure the equilibrium constant (K) of that reversible electrochemical reaction at equilibrium provided we know the E0

cell !

Types of electrodes:

1. Metal –Metal ion electrode:

Metals in equilibrium with its ions in solution

Zn(s) | Zn2+(aq) Cu2+(aq) | Cu(s) Cd(s) | Cd2+

M(s) → Mn+ + ne–

Amalgam electrode : Sodium amalgam Electrode (Na(Hg) | Na+) Na(Hg) → Na+ + e–

1) More readily reversible2) Do not react with water3) Avoid impurities formation4) Pt wire dipping into the amalgam pool

Gas electrodesInvolve a gaseous species such as H2, O2, or Cl2.

A typical reaction of considerable commercial importance isCl–(aq) → ½ Cl2(g) + e–

Similar reactions involving the oxidation of Br2 or I2 also take place at platinum surfaces

2. Non-metal electrodes:Non metal and its corresponding ion in solution in the presence of an unreactive metal ( Pt)

Hydrogen electrode:Pt(s), H2(g) | H+(c) ½ H2(g) → H+ + e–

Oxygen Electrode:Pt(s), O2(g, po2) | OH-(c) ½ O2(g) + H2O + 2e– → 2OH-

Consists of a silver wire covered with a thin coating of silver chloride, which is insoluble in water, dipped into a solution containing the same anion as that of the salt.

The electrode reaction consists in the oxidation and reduction of the silver: Ag(s) + Cl–(aq) → AgCl(s) + e–

The half cell would be represented as Ag (s)| AgCl(s)|Cl–(aq)

This kind of electrode finds very wide application in electrochemical measurements, as we shall see later.

3. Metal –Insoluble salt electrode:

Calomel electrode: Hg/Hg2Cl2(s)/KCl Hg(s) + Cl–(aq) → ½ Hg2Cl2(s) + e–

Reference Electrodes

Pb-PbSO4 electrode: Used in lead storage batteryPb/PbSO4, H2SO4

Sb/Sb2O3/OH- Sb (s) + 2OH- → ½ Sb2O3 + 3/2 H2O + e–

Metal-Metal Oxide electrode:

Mercury-mercury oxide electrodes: Hg /HgO(s)/OH-

Hg (l) + 2OH- → HgO + H2O + 2e–

4. Oxidation-reduction electrode: (Ion-ion electrodes)

Two ions containing the same element, but in different oxidation states are present in solution and an inert metal like Pt is immersed in it.

Pt(s) | Fe3+(aq), Fe2+(aq) || Fe2+(aq) → Fe3+ (aq) + e–

Quinhydrone electrode: Quinone-Hydroquinone Electrode

Electrodes with organic species:

O

O

+ 2H+ + 2e

OH

OH

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Relation between cell potentials and half cell potentials

Ecell = E0cell – (0.059 V/n)log10Q (Nernst equation)

At equilibrium, ΔG = 0 and Ecell = 0 so no work will done by the cell reaction.

Ox (1) + Red (2) Red (1) + Ox (2) (total cell reaction)

Oxidation half reaction(anode)

Red (2) Ox (2) + ne- Ox (1) + ne- Red (1)

Reduction half reaction (cathode)

Let’s predict the spontaneity of the electrochemical reaction in general form:

Eox = E0ox – (0.059 V/n)log10[a ox(2)/a red(2)] Ered = E0

red – (0.059 V/n)log10[a red(1)/a 0x(1)]

How can we express the overall cell potential Ecell?

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Cell potential and half cell potentials

The reaction quotient Q for the overall reaction

Ox (1) + Red (2) Red (1) + Ox (2) is [aRed(1) aOx (2)]

[aOx(1) aRed (2)]Q =

Ecell = E0cell – 0.059 V

n log 10

[aRed(1) aOx (2)]

[aOx(1) aRed (2)]

So the Nernst equation for the overall reaction becomes

How the two half cell reactions finally combine to yield the above Nernst equation?

Ecell = Ered – Eox ?

Ecell = E0x – Ered ?

Ecell = Ered + Eox ?} Any one of such

combinations has to be true!

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= E0red + E0

ox –(0.059 V/n)log10[a red(1)/a ox(1)] - (0.059 V/n)log10[a ox(2)/a red(2)]

Ecell = Ered + Eox

= (E0ox – (0.059 V/n)log10[a ox(2)/a red(2)]) + (E0

red – (0.059 V/n)log10[a red(1)/a ox(1)])

Cell potential and half cell potentials

= E0red + E0

ox – 0.059 V n log 10

[aRed(1) aOx (2)]

[aOx(1) aRed (2)]

= E0cell – (0.059 V/n)log10Q

So we get

E0cell = E0

red + E0ox

Now lets come back to the question of spontaneity of the reaction

Ox (1) + Red (2) Red (1) + Ox (2)

and also, Ecell = Ered + Eox

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Spontaneity and cell potential-nFE = ΔG (relationship between Gibbs energy and electrochemical potential)

From the above diagram, we can see that an electrochemical reaction will proceedspontaneously from left to right when ΔG < 0 and Ecell > 0

Ox (1) + Red (2) Red (1) + Ox (2)

The reaction will proceed as above till the equilibrium is reached!

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ElectrodesFollowing are the examples of few common types of electrodes used in Galvanic cells

Note that each reaction taking place at electrode represents a half-cell reactionand the half cell potential (Ered or Eox) is termed as Galvani potential difference

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Electrode potential

Standard cell potential (E0cell) can be obtained by two ways:

E0cell = RTlnK

nF

The cell potential (Ecell) of a Galvanic cell can be measured experimentally

Ecell = E0cell – (0.059 V/n)log10Q

The cell potential (Ecell) of a Galvanic cell can also be predicted theoretically

(provided we know E0cell and activities)

One way is (provided we know K)

The other way is E0cell = E0

red + E0ox (provided we know E0

red & E0ox)

How to know E0red & E0

ox ?There is no way we can measure E0

red/ox for redox half cell reactions!

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Standard electrode potentialA convention is adopted to assign the standard half cell or electrode potentials

H+ (aq) + e- (Pt) ½ H2 (g)

The standard half cell potential or Eo red of the above the reaction

is set as 0.0 volt (by definition)!

EH+/H = E0H+/H –

0.059 V n log 10

1

[aH+] (activity of pure H2 gas is unity)

Under standard condition of [aH+] = 1.0 and at 298 K and 1 atm pressure,the above equation becomes

EH+/H = E0H+/H = 0.0 Volt

So the standard reduction potential of hydrogen electrode is zero volt

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Standard reference potential and standard half cell potentials

Using the standard hydrogen electrode (SHE), we can measure the standard half cellpotentials of other redox half cell reactions!

X n+

X m+

SHE

salt bridge

Under standard condition, Ecell of the Galvanic cell will bethe E0

half cell of the unknown half cell!

reference electrode

Ecell reading

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Standard reduction potentials and IUPAC conventions

With respect to SHE, any redox couple will have its standard half cell potentialeither positive or negative.

That means some half cells will under go oxidation and some will undergo reduction.

To avoid confusion, according to IUPAC, the standard half cell or electrode potentialsare expressed as reduction potential only!

The convention is

Oxidized species + ne- Reduced species (under standard condition)

Eo half cell = E0 standard reduction potential = Eo

standard electrode potential

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

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

(E0 red = + 0.40 V)

(E0 red = - o.76 V)

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Standard electrode potentials (basically the std. reduction potential at 298 K and under std. condition)

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Standard electrode potentials (make yourself familiar with such table)

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Significance of standard electrode potentials

salt bridge

Anod

e

O2 (g) + 4H+ + 4e- 2H2O (l)Zn2+ (aq) + 2e- Zn (s)

(E0 = + 0.40 V)

(E0 = - o.76 V)

Which chamber should contain the zinc electrode to make a spontaneous cell?

Cath

ode

an ox! red cat!

at anode oxidation takes place reduction takes place at cathode

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Standard cell potential and standard electrode potential

Previously we saw:

E0cell = E0

red + E0ox

But here we are talking about std. reduction potential only! So the std. reductionpotential at anode needs to be sign reversed to represent oxidation.

So according to the IUPAC convention,

E0cell = E0

cathode - E0anode

(old convention)

(New convention)

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

(E0 = + 0.40 V)

(E0 = - o.76 V)

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

at which electrode?

at which electrode?

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Significance of standard electrode potentials

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

Zn2+ (aq) + 2e- Zn (s)(E0

= + 0.40 V)

(E0 = - o.76 V)

salt bridge

Anod

e

Cath

ode

Rule of thumb: Half reaction with higher Eo will always oxidize the other

2Zn2+ (aq) + 4e-2Zn (s)

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

E0cell = E0

cathode - E0anode

= + 0.40 - (- 0.76) = +1.16 V Is that cell reaction spontaneous?

What is the expression of Ecell in this case?

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Calculating std. cell potential from st. electrode potential

Calculate E0cell of the following reactions pairs assuming standard conditions

and spontaneous cell reactions. All values are provided in std. reduction potential.

Formula: E0cell = E0

cathode - E0anode

Pb2+ (aq) + 2e- Pb (s) (E0 = - o.13 V)

Zn2+ (aq) + 2e- Zn (s) (E0 = - o.76 V)

Example I

K+ (aq) + e- K (s) (E0 = - o.83 V)

2H2O (l) + 2e- 2OH- (aq) + H2 (g) (E0 = - 2.93 V)

Example II

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How to represent a Galvanic cell (battery) on paper?

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

Zn2+ (aq) + 2e- Zn (s)(E0

= + 0.40 V)

(E0 = - o.76 V)

salt bridge

Anod

e

Cath

ode

2Zn2+ (aq) + 4e-2Zn (s)

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

It’s a pretty demanding job to draw a cell diagram like below, every time!

Scientifically we can represent the cell in the following manner (IUPAC):

Zn (s) Zn2+ (aq) O2 (g) H+ H2O (l) Pt (s)

cathodeelectrode

right hand side

anodeelectrode

left hand side(red cat right)(an ox left)

salt bridge phase boundary

phase boundary

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Systematic cell notations:

Previous cell can be represented in detailed manner like below:

Zn (s) Zn2+ (aZn2+ = 1, aq) O2 (g) H+ (aH+ = 1, aq) H2O (l) Pt (s)

Example: III

Write down the complete cell reaction from the notation below:

Example: IV

Write down the cell notation from the half cell reactions below:

K+ (aq) + e- K (s) (E0 = - o.83 V)

2H2O (l) + 2e- 2OH- (aq) + H2 (g) (E0 = - 2.93 V)

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Example: IV

Write down the half and complete cell reactions from the notation below:

Metal/insoluble salt electrode:

Ag (s) Cl- (aq) AgCl(s) Ag+ (aq) Ag (s)

AgCl (s) + e- Ag (s) + Cl- (aq)

Ag+ (aq) + e- Ag (s) (E0 = + o.8 V)

(E0 = +0.22 V)

Ag (s) + Cl- (aq) AgCl (s) + e-

Ag+ (aq) + e- Ag (s)

What is the complete cell reaction and standard cell potential?Ag+ (aq) + Cl- (aq) AgCl (s)

E0cell = E0

cathode - E0anode = (0.8 – 0.22) V= 0.58 V

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Determination of solubility product

Nernst equation:

Ecell = E0cell – 0.059 V log 10

[aAg+ (aq)] [aI- (aq)]

Ag+ (aq) + I- (aq)AgI (s)

Cell reaction:

[a AgI (s)]

Consider the cell: Ag+ (aq) Ag (s)AgI (s)I- (aq)Ag (s)

If the above cell reaction is reversible then the Keq is called the solubility product

Ag+ (aq) + I- (aq)AgI (s) Keq = Ksp =[aAg+ (aq)]eq [aI- (aq)]eq

[a AgI (s)]eq

We know, an electrochemical cell at equilibrium means Ecell = 0

So, E0cell = 0.059 V log10 Ksp

AgI (s) + e- Ag (s) + I- (aq)

Ag+ (aq) + e- Ag (s) (E0 = + o.8 V)

(E0 = -0.15 V)

Ksp = anti log E0

cell

0.059 V=or,

In this way we can determine, the Ksp for a sparingly soluble salt used in electrodes

-

Reduced species -> oxidized species + ne- Oxidation at anode

Oxidized species + ne- -> reduced species Reduction at cathode

Ecell = oxidation potential + reduction potential

Measured against standard hydrogen electrode.

Concentration 1 Molar

Pressure 1 atmosphere

Temperature 25°C

E°CellElectrodes have standard electrode potentials:

Red-ox potential

Standard hydrogen electrode

(Gas at 1 atm pressure and ions at 1M concentration)Assigned a potential of zero.

H2(g) 2H+ + 2e- EH = 0

Hydrogen can be used as an electrode in a copper reduction:

H2(g) 2H+ + 2e- EH = 0

Cu++(1M) + 2e- Cu(s) EºCu

H2(g) + Cu++(1M) 2H+(1M) + Cu(s) E = +0.337 VEºCu = EºCu

2+/Cu = +0.337 V

As per IUPAC, the standard electrode potential of an electrode (X) is the emf of a cell in which:

(i) The anode is a hydrogen electrode

(ii) The cathode is electrode X

Pt(s), H2(g) | H+(c) X║ + │X

½ H2(g) + X+ ( a = 1) → H+ (a = 1) + X

Zn2+ + 2e– → Zn ; Eº = +0.76 V

The example below shows some of the values for standard cell potentials.

Cathode (Reduction)Half-Reaction

Standard PotentialE° (volts)

Li+(aq) + e- -> Li(s) -3.04K+(aq) + e- -> K(s) -2.92Ca2+(aq) + 2e- -> Ca(s) -2.76Na+(aq) + e- -> Na(s) -2.71Zn2+(aq) + 2e- -> Zn(s) -0.76Cu2+(aq) + 2e- -> Cu(s) +0.34O3(g) + 2H+(aq) + 2e- -> O2(g) + H2O(l) +2.07F2(g) + 2e- -> 2F-(aq) +2.87

Strongest reducing agent.

Strongest oxidizing agent

Electrochemical series

Reference electrodes:

Calomel electrode: Hg in contact with Mercurous Chloride ( Hg2Cl2) and a solution of KCl

Hg2Cl2(s), KCl │Hg

½ Hg2Cl2(s) + e– → Hg(s) + Cl–

1. [KCl] = 1.0 N, ECal = 0.2800 V

2. [KCl] = 0.1 N, ECal = 0.3338 V

3. [KCl] = Saturated, ECal = 0.2415 V

Pt(s), H2(g) | H+(c)

(Anode) (Cathode)

Cathodic reaction

In an electrochemical cell n equivalents of reactants are converted into the products

n eqv. : Reactants Products

Then the quantity of electricity that flows through the cell = nF

F = 96486 C

If nF amount of charge is transported through the cell of emf E volts,

The amount of electrical work done is nFE

The decrease in Gibbs free energy is equal to the electrical work done by the cell.

Hence - ∆GP,T = nFE

∆G < 0 or E = +ve : Spontaneous reaction

∆G >0 or E = -ve : Non-Spontaneous reaction

∆G = 0 or E = 0 : Equilibrium

∆G = ∆ Go +RT lnQ

∆ G = -nFE

For standard state : ∆ Go = -nFEo

xA + yB pC + qD

acp. aD

q aA

x. aBy

Q =

R = 8.314 J /Kmol, F = 96486 C/mol, T = 298 K

R = gas constant, T = temperature in Kelvins, Q = thermodynamic reaction quotientF = Faraday's constant , n = number of electrons transferred

E = Eo - RT ln acp. aD

q aA

x. aBynF

E = Eo – 0.05915 log acp. aD

q aA

x. aByn

At equilibrium ∆ G = 0 and E = 0, Q= K,

At equilibrium:

Eo =(RT/nF) lnK

K = Equilibrium Constant

Mean activity coefficient of an electrolyte:

We assume a mean activity coefficient denoted as: γ±

and the assumption becomes: γ+ = γ- = γ± (mean activity coefficient)

So, mmasalt = (m+.m-)(γ+ .γ-) = (m+.m-)(γ±)2

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For NaCl (1:1) electrolyte,

Therefore.

(for 1:1 electrolyte)

It is possible to measure γ± experimentally!

If NaCl is of m molal, then m+ = m- = m, so asalt = (m)2 .(γ±)2

General expression for γ±

For any salt stoichiometry:

Now solve the following problem:

Solution of the problem:

Solution of the problem (continued):

H2 (g)H2(g) Pt (s)H+ (aq, a = 1.0) Pt(s) H+ (aq, a = ?)

Ecat = 0.0

Ecell =

pH = 0.059 V Ecell

Ean0.059 V

1 log 10 1[aH+]

=

= 0.059 V pH

0.059 V pH

-

These days, we don’t use hydrogen gas electrode, but we use glass based ion selective electrode for pH reading.

pH electrodeConsider the following electrochemical cell:

Primary and secondary batteries

A primary cell, an ordinary flashlight battery, cannot be recharged with any efficiency, so the amount of energy it can deliver is limited to that obtainable from the reactants that were placed in it at the time of manufacture.

A secondary or storage battery is capable of being recharged; its electrode reactions can proceed in either direction. During charging, electrical work is done on the cell to provide the free energy needed to force the reaction in the non-spontaneous direction.

The modern dry cell is based on the one invented by Georges Leclanché in 1866.

The electrode reactions are

Zn → Zn2+ + 2e–

2 MnO2 + 2H+ + 2e– → Mn2O3 + H2O

The electrolyte is a wet paste containing NH4Cl to supply the hydrogen ions.

Alkaline Battery

A more modern version, introduced in 1949, KOH electrolyte zinc-powder anode MnO2 as cathode

Permits the cell to deliver higher currents and avoids the corrosive effects of the acidic ammonium ion on the zinc.

Zn + 2OH- --- Zn(OH)2 + 2e-

2MnO2 + H2O + 2e- ---- Mn2O3 + 2OH-

The lead-acid storage cell

Invented by Gaston Planté in 1859

The cell is represented by

Pb(s) | PbSO4(s) | H2SO4(aq) || PbSO4(s), PbO2(s) | Pb(s)

The reaction proceeds to the right during discharge and to the left during charging.

Cathode: PbO2 + 2e- + 4H+ + SO4

2- PbSO4 + 2H2O, Eº = 1.685 V

Anode: Pb + SO42- PbSO4 + 2e- , Eº = 0.356 V

Pb(s) + PbO2(s) + 2 H2SO4(aq) → 2 PbSO4(s) + 2 H2O Eº = 2.041 V

•The sulfuric acid electrolyte becomes quite viscous when the temperature is low, inhibiting the flow of ions between the plates and reducing the current that can be delivered.

•These batteries tend to slowly self-discharge, so a car left idle for several weeks might be unable to start.

•Over time, PbSO4 that does not get converted to PbO2 due to lack of complete discharge gradually changes to an inert form which limits the battery capacity.

Disadvantages:

The fuel cellThe principle was first demonstrated in 1839 by Sir William Grove, a Welsh lawyer and amateur chemist.

anode: H2(g) → 2 H+ + 2e– E° = 0 vcathode

:  ½ O2 + 2 H+ + 2e– → H2O(l) E° = +1.23 v

net:  H2(g) + ½ O2(g) → H2O(l) E° = +1.23 v

1959, the first working hydrogen-oxygen fuel cell was developed by Francis Thomas Bacon in England

Uses alkaline electrolyte

In acidic medium

Other fuels such as alcohols, hydrocarbon liquids, and even coal slurries have been used; methanol appears to be an especially promising fuel.

anode: H2(g) + 2OH- → 2 H2O + 2 e– E° = 0 v

cathode:  ½ O2 (g) + 2 H2O + 2 e– → 2OH- E° = +1.23 v

net:  H2(g) + ½ O2(g) → H2O E° = +1.23 v

Electrolytes are NaOH solution, phosphoric acid, or solid oxides.

Advantages:more efficient ( More than 70 %)Polution free ( Eco friendly)

Limitations:

The rates of the electrode reactions, especially the one in which oxygen is reduced, tend to be very small, and thus so is the output current per unit of electrode surface.

Coating the electrode with a suitable catalytic material is almost always necessary to obtain usable output currents, but good catalysts are mostly very expensive substances such as platinum, so that the resulting cells are too costly for most practical uses. There is no doubt that if an efficient, low-cost catalytic electrode surface is ever developed, the fuel cell would become a mainstay of the energy economy.

Button cell ( computers)

Silver oxide batteryZn as AnodeSiver Oxide as CathodeSodium hydroxide ( Electrolyte)

Zn + Ag2O ZnO + 2Ag

In pacemakers and hearing aid :

Lithium iodide battery

Lithium ion battery:

LiCoO2 and Carbon as electrode

Ni-Cd Battery:

Ni hydroxide and Cadmium

KOH ( electrolyte)

2NiO(OH) + Cd + 2H2O 2Ni(OH)2 + Cd(OH)2