Representing electrochemical cells

The electrochemical cell established by the following half cells:

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

Cu2+(aq) + 2 e- --> Cu(s)

san be represented as:

Anode is shown on the left, the cathode on the right, the single line separates phases, and the double dotted line indicates the salt bridge

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

anode cathode

The electrode need not always be involved in the redox reaction; it could be an inert substrate through which electrons flow into or out off.

A cell represented as:

Pt|Fe2+(aq), Fe3+(aq) Cl2(g)|Cl-(aq)|Pt

involves the following half cell reactions:

2Fe2+(aq) --> 2Fe3+ + 2e-

Cl2(g) +2e- --> 2Cl-(aq)

with two Pt strips acting as electrodes

What is the driving force for electron flow in the cell?

The affinity of the two electrodes for electrons is different; this sets up an electric potential difference E between the two electrodes.

The potential energy of the electrons is higher at the anode than at the cathode and the electrons flow spontaneously from anode to cathode.

The difference in potential energy, E, per electrical charge between the two electrodes is measured in units of volts.

One volt (V) is defined so that a charge of one coulomb (1C) falling through a potential difference of one volt (1 V) releases one joule (1 J) or energy.

1 V =1 J

1 C

The potential difference between the two electrodes of the galvanic cell provides the driving force that pushes electrons through the external circuit.

The potential difference of the cell, E, is also called the electromotive force, emf, or cell potential and has units of volts (V).

A galvanic or voltaic cell, takes advantage of the electron flow between the two electrodes due to the spontaneous redox reaction, converting chemical energy into electrical energy.

Under standard conditions (1 bar for gases, or 1M concentration for solutions) the emf is called the standard emf or standard cell potential, Eo.

For the Cu/Zn system the standard cell potential across the Cu and Zn electrodes is 1.10V; Eo = +1.10V

Differences in electric potential drives the flow of current.

This flow of current can do work - electrical work.

welec = - Q E

Q: amount of charge (in coulomb)

E : electric potential (in V)

Maximum amount of work done by the system during a process:

wmax = G (at constant T and P)

G = - Q E

If E is > 0, G < 0, cell reaction is spontaneous

It is also possible to drive an electrochemical cell in the reverse direction of the spontaneous reaction.

This can be done by applying an external voltage of magnitude greater than the potential difference established by the spontaneous redox reaction, and in the opposite direction

For example in the Cu/Zn system applying an external voltage >1.10 V forces the reverse reaction to occur, converting the applied electrical energy into chemical energy.

Such cells are called electrolytic cells, and force reactions to take place that do not occur spontaneously.

Batteries

In a battery a redox reaction occurs in the self-contained packaging of the battery.

The reaction produces a cell emf that can be used to drive electrons through an external circuit, producing electricity.

As the battery operates, reactants are consumed, the cell emf (which depends on the concentration of the reactants) drops till eventually it is zero.

Primary Cells: Galvanic cells; reactants sealed in the packaging. Cannot be recharged.

Alkaline Cell:

Anode: powdered Zn immobilized in a gel in contact with a concentrated solution of potassium hydroxide, KOH.

Cathode: mixture of MnO2(s) and graphite separated from the anode by a porous separator.

Cell Voltage: 1.5 V

Anode: Zn(s) + 2OH-(aq) --> ZnO(s) + H2O(l) + 2e-

Cathode: MnO2 (s) + 2 H2O(l) + 2e- -> Mn(OH)2(s) + 2 OH-(aq)

0 +2

+4 +2

Secondary Cells: Galvanic cells that must be charged before they can be used; rechargeable.

In the charging process, an external source of electricity reverses the spontaneous cell reaction and creates a non-equilibrium mixture.

After charging, the cell can again produce electricity as the reaction moves spontaneously in the direction of the equilibrium point.

Applications: batteries in laptops and automobiles.

Lead-Acid Battery - used in automobiles (12V battery, consists of six 2V batteries in series)

The cathode consists of lead dioxide, PbO2, packed on a metal grid.

The anode consists of Pb.

Both electrodes are immersed in sulfuric acid, H2SO4.

Anode: Pb(s) + HSO4-(aq) -> PbSO4(s) + H+(aq) + 2e-

Cathode: PbO2(s) + 3 H+(aq) + HSO4-(aq) + 2e- ->

PbSO4(s) + 2 H2O(l)

Cell Voltage: 2 V

0 +2

+4

+2

Fuel Cell: Designed for continuous operation; reactants supplied, and products removed continuously.

H2(g) + 1/2O2(g) -------> H2O(l) H = -286 kJ/mol

Anode: H2(g) -> 2 H+(aq) + 2e-

Cathode: 2 H+(aq) + 1/2 O2(g) + 2e- -> H2O(l)

Cell voltage: 0.615 V

-2

+1

0

0

Ecell ~ 0.8 V

Individual fuel cells can be stacked; the number of fuel cells determines the total voltage, the surface area of each determines the total current.

Voltage x total current = Electrical Power

200 kW UTC Power Plant in Central Park

Corrosion: an undesirable application of spontaneous redox reactions

When iron is exposed to damp air, with both O2 and H2O present, the half reaction

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

takes place

Fe(s) is first oxidized by O2 to Fe2+

The Fe2+ is then further oxidized to Fe3+

Fe3+ then forms Fe2O3.H2O - rust

O2-free water O2 dissolved In water

Electrolysis

Electrical energy to drive a non-spontaneous redox reaction.

For example, electricity can be used to decompose molten NaCl to Na and Cl2, a non-spontaneous process

2NaCl(l) --> 2Na(l) + Cl2(g)

Go > 0

(for the reaction of 2NaCl(s) -> 2 Na(s) + Cl2(g)

Go ~ 770 kJ/mol )

Electrolysis is a process driven by an external source of electrical energy and takes place in an electrolytic cell.

Cathode: 2Na+(l) + 2e- --> 2Na(l) Reduction

Anode: 2Cl-(l) --> Cl2(g) + 2e- Oxidation

In an electrolytic cell oxidation occurs at the anode and reduction at the cathode, just as in a voltaic cell, but the signs are reversed.

Quantitative Electrochemistry - Faraday’s laws

In redox reactions charge must be balanced; the number of electrons given up in the oxidation process must equal the number of electrons involved in the reduction process.

The amount of charge transferred between the two electrodes must be related to the amount of each species reacting at each electrode.

The stoichiometry of a half reaction indicates how many electrons are needed to achieve an electrolytic process.

For example, in the reduction of Na+ to Na(s), one mole of electrons reduces one mole of Na+ resulting in the deposition of one mole of Na(s) on the cathode

Similarly two moles of electrons produce one mole of Cu(s) from one mole of Cu2+

Cu2+ + 2e- --> Cu

Or 3 moles of electrons produce one mole of Al(s)

Al3+ + 3e- --> Al

Hence, for any half reaction, the amount of substance reduced or oxidized is directly proportional to the number of electrons passed into the cell.

This is the basis of Faraday’s laws

1) the quantities of substance produced and consumed at the electrodes are directly proportional to the amount of electric charge passing through the cell.

2) When a given amount of electric charge passes through the cell, the quantity of substance produced or consumed at an electrode is proportional to its molar mass divided by the number of moles of electrons required to produce or consume one mole of the substance.

A coulomb (C) is the quantity of charge passing a point in a circuit in 1 s when the current is 1 ampere (A)

If a current of I amperes flows for a period of t secs, the amount of charge transferred = I amperes x t sec

(coulomb = amperes x seconds)

Current is the rate of flow of charge

1 A = 1C/sec

The charge on a single electron = 1.6022 x 10-19 C

The charge associated with one mole of electrons

= 1.6022 x 10-19 C/e- x 6.0221 x 1023 e-/mole = 9.6485x104 C/mol

The charge associated with one mole of electrons, 9.6485x104

C/mol, is the FARADAY CONSTANT, F

To use Faraday’s law to calculate the amount of substance oxidized or reduced:

1) Need to know the amount of current passed through the cell and for how long

2) Calculate the quantity of charge transferred- coulombs

3) Using the Faraday constant to calculate the number of moles of electrons that corresponds to the quantity of charge determined in step 2

4) Relate the number of moles of electrons to the number of moles of substance oxidized or reduced

5) Convert moles of substance to grams of substance

A galvanic cell generates an average current of 0.121 A for 15.6 min. The cathode half-reaction in the cell is

Pb2+(aq) + 2e- -->Pb(s)

Determine the amount of lead, in grams, deposited at the cathode?

The amount of charge passed through the cell

= (0.121 A) (15.6 min x 60sec/min) = 113 C

Amount of electrons in moles = 113 C/(9.6485x104 C/mol) = 1.17 x 10-3 mol

Amount of Pb deposited in moles = 5.85x10-4 mol

Amount of Pb deposited in grams = 0.122 g