Cell Voltages

To compare cells compare voltages of cells in their standard state.

Standard StatesFor solids and liquids: the state of the pure solid or liquid at 1

atm and at a specified temperatureFor gases: the gaseous phase at 1 atm and at a specified

temperatureFor solutions : Concentrations of 1 mol/L under a pressure of

1 atm and a specified temperature.

(temperature typically specified at 298.15 K)

Eo is the standard cell voltage or potential of a standard cell - when all reactants and products are in their standard state.

Gro = - n F Eo

Eo = - Gro

n F

Eo > 0 ; spontaneous cell reaction under standard conditions

A standard Cr3+(aq)|Cr(s) and a standard Co2+(aq)|Co(s) half cell are connected to make a galvanic cell. The voltage of the cell equals 0.464 V at 25oC. Write an equation to represent the reaction taking place in the cell and calculate its Go

r.

Cr(s) Cr3+(aq) + 3e- anode

Co2+(aq) + 2e- Co(s) cathode

Overall cell reaction

2Cr(s) + 3 Co2+(aq) 2Cr3+(aq) + 3 Co(s) Eo = 0.464 V

Gor= - n F Eo

= - (6 moles) (9.64853 x 104 coulomb/mole) (0.464 V)

= - 2.69 x 105 J or - 269 kJ

Standard Reduction PotentialsUnder standard conditions:

Eo = Eo (right half cell) - Eo (left half cell)

For a galvanic cell:Eo = Eo (cathode) - Eo (anode)where the Eo are the standard reduction potentials of the

electrodes.

For Eo > 0; spontaneous cell reaction

To determine which of two half cells will be the anode and which the cathode compare standard reduction potentials

Cr3+(aq) + 3e- Cr(s) Eo(Cr3+|Cr) = - 0.744 V

Co2+(aq) + 2e- Co(s) Eo(Co2+|Co) = - 0.28 V

Eo(Co2+|Co) > Eo(Cr3+|Cr)

Define the following half reaction to be the reference2 H+(aq, 1 M) + 2e- -> H2 (g, P = 1 atm) Eo = 0 V

All standard reduction potentials are determined relative to this reference.

If the standard reduction potential of a half reaction is > 0 => greater tendency to be reduced relative to H+(aq, 1 M)

If standard reduction potential of a half reaction < 0 => lower tendency to be reduced relative to H+(aq, 1 M)

In general, the more positive the standard reduction potential, the greater the electron-pulling power of the reduction half reaction, and therefore the more oxidizing the species

The more negative the standard reduction potential, the greater the electron-donating power of the oxidation half reaction, and therefore the more reducing the species

Variation of standard reduction potentials. The most negative values occur in the s block and the most positive values occur close to fluorine.

The standard potential of an electrode can be determined by setting up a standard cell in which one electrode has a known standard potential and measuring the resulting cell voltage.

For example, the standard potential of a zinc electrode is -0.76 V, and the standard emf of the cell

Zn(s) | Zn2+ (aq) || Sn4+ (aq), Sn2+ (aq) | Pt (s) is + 0.91 V

Eo = Eo (cathode) - Eo (anode)+ 0.91 V = Eo (Sn4+ (aq), Sn2+ (aq) ) - Eo (Zn(s) | Zn2+ (aq) )

Eo (Sn4+ (aq), Sn2+ (aq) ) = + 0.91 V + Eo (Zn(s) | Zn2+ (aq) ) = 0.91V - 0.76 V = + 0.15 V

Using standard reduction potentials

Strong oxidizing agents - have large positive standard reduction potentials

Examples: F2, MnO4-, H2O2

O2 (in acidic medium) is a fairly strong oxidizing agent.

Strong reducing agents - have large negative standard reduction potentials

Examples: Na, Li

Electrochemical series: list of relative strengths of oxidizing and reducing agents.

The strongest oxidizing agents are at the top of the table; the strongest reducing agents are at the bottom

M(s) + 2H+(aq) M2+(aq) + H2(g)

spontaneous if Eo(M2+|M) < 0

Disproportionation: a single species is both reduced and oxidized.

Must be able to both give up and accept electrons

Half reaction in which the species is reduced must have a larger reduction potential than the half reaction in which it is oxidized.

Is Fe2+(aq) in its standard state unstable with respect to disproportionation at 25oC?

Fe3+(aq) + e- Fe2+(aq) Eo = 0.771 V

Fe2+(aq) + 2e- Fe(s) Eo = -0.477 V

Overall disproportionation reaction

3 Fe2+(aq) 2 Fe3+(aq) + Fe(s) Eo = -0.477 - 0.771 = -1.218 V

No disproportionation

Effect of Concentration on EGr = Gr

o + RT ln Q

Gr = - n F EGr

o = - n F Eo

- n F E = - n F Eo + RT ln Q

E = Eo - (RT/ n F ) ln Q Nernst Equation

relates cell voltage with concentrations of reactants and products (through Q)

At 25.00oC (298.15 K), R T / F = 0.025693 V

E = Eo - (0.025693 / n ) ln Q

The reduction potential of a non-standard half cell is:

E = Eo - (RT/ nhc F ) ln Qhc

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

E = Eo - (RT/ 2 F ) ln (1 / [Zn2+(aq)]

Ion-Selective ElectrodespH or concentration of ions can be measured by using an

electrode that responds selectively to only one species of ion.

In a pH meter, one electrode is sensitive to the H3O+(aq) concentration, and the other electrode serves as a reference.

A calomel electrode has a reduction half reaction Hg2Cl2 (s) + 2 e- -> 2 Hg(l) + 2 Cl- (aq) Eo = +0.27 V

When combined with the H+(aq)/H2(g) electrode, the overall cell reaction is:Hg2Cl2 (s) + H2 (g) -> 2 H+ (aq) + 2 Hg(l) + 2 Cl- (aq)

Q = [H+(aq)]2 [Cl- (aq)]2 / PH2

If PH2 is held at 1 atm then Q = [H+(aq)]2 [Cl- (aq)]2

E = Eo - (RT/ n F ) ln [H+(aq)]2 [Cl- (aq)]2

The [Cl- (aq)] is held constant since the calomel electrode consists of a saturated solution of KCl.

E depends only on [H+(aq)].

Other electrodes are selectively sensitive to ions such as Ca2+, NH4

+, Na+, S2-.