23sec2

23.2 Half-cells and Cell Potentials

Electrode potentials

A measure of how willing a species is to gain or lose electrons.

Standard potentialsStandard potentials

Potential of a cell acting as a cathodecompared to a standard hydrogen electrode.

Values also require other standard conditions.

Standard hydrogen electrode

Hydrogen electrode (SHE)Hydrogen electrode (SHE)

The ultimate reference electrode.H2 is constantly bubbled into a 1 M HCl

solution

Eo = 0.000 000 V

All other standard potentials are

then reported relative to SHE.

H2

1 M HCl

Pt blackplate


Standard potentials are defined using specific concentrations.

All soluble species are at 1 M

Slightly soluble species must be at saturation.

Any gas is constantly introduced at 1 atm

Any metal must be in electrical contact

Other solids must also be present and in contact.


The standard potential for:

Cu2+ + 2e- Cu (s) is +0.337V.

This means that:This means that: If a sample of copper metal is placed in a

1 M Cu2+ solution, we’ll measure a value of 0.337V if compared to:

2H+ + 2e- H2 (g)

(1 M) (1atm)

Half reactions

A common approach for listing species that undergo REDOX is as half-reactions.

For 2Fe2Fe3+3+ + Zn + Znoo(s)(s) = 2Fe = 2Fe2+2+ + Zn + Zn2+2+

Fe3+ + e- Fe2+ (reduction)

Zno(s) Zn2+ + 2e- (oxidation)

You’ll find this approach useful for a number of reasons.

Half reactions

Tables are available which list half reactions as either oxidations or reductions.( Pg 688)

Will provideWill provide

Standard Eo values to help predict reactions and equilibria.

Other species that participate in the reaction.

Show the relative ability to gain or loss electrons.

Cell potentials

One thing that we would like to know is the spontaneous direction for a reaction.

This requires that we determine the Ecell.

Since our standard potentials (E o) are commonly listed as reductions, we’ll base our definitions on that.

Ecell = Ehalf-cell of reduction - Ehalf-cell of oxidation

Eocell = Eo

half-cell of reduction - Eohalf-cell of oxidation

Cell potentials

You know that both an oxidation and a reduction must occur.

One of your half reactions must be reversed.

The spontaneous or galvanic direction for a reaction is the one where EEcellcell is a positive value.

The half reaction with the largest E value will proceed as a reduction.

The other will be reversed - oxidation.oxidation.

Cell potentials

For our copper - zinc cell at standard conditions:

Cu2+ + 2e- Cu (s) +0.34 V

Zn2+ + 2e- Zn (s) -0.763 V

Ecell 1.03 V

Spontaneous reaction at standard conditions:

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

Concentration dependency of E

Eo values are based on standard conditions.

The E value will vary if any of the concentrations vary from standard conditions.

This effect can be experimentally determined by measuring E versus a standard (indicator) electrode.

Theoretically, the electrode potential can be determined by the Nernst equationNernst equation.

Calculation of cell potentials

To determine the Voltaic Ecell at standard conditions using reduction potentials:

Ecell = E ohalf-cell of reduction - E o

half-cell of oxidation

Where

EEhalf-cell of reductionhalf-cell of reduction - half reaction with the larger , or least negative E o value.

EEhalf-cell of oxidationhalf-cell of oxidation - half reaction with the smaller or more negative E o value.


Steps in determining the spontaneous direction and Steps in determining the spontaneous direction and EE of a cell at standard conditions of a cell at standard conditions

Calculate the E for each half reaction.

The half reaction with the largest or least negative E value will proceed as a reduction.

Calculate Ecell


ExampleExample Determine the spontaneous direction and Ecell for the following system.

Pb | Pb2+ (1M) || Sn2+ (1M) | Sn

Half reactionHalf reaction E Eoo

Pb2+ + 2e- Pb -0.13 V

Sn2+ + 2e- Sn -0.14 V Note: The above cell notation may or

may not be correct.


anode: Pb2+ + 2e- Pb -0.13 V

cathode:Sn2+ + 2e- Sn -0.14 V

Ecell = E ohalf-cell of reduction - E o

half-cell of oxidation

(-0.14) – (-0.13)= -0.1 V so not spontaneous


Calculate the E0cell to determine if the

following cell is spontaneous as written

Ni(s) + Fe2+(aq) -----Ni2+

(aq) + Fe(s)


Write each half reaction and look up the reduction cell potential

Oxidation: Ni----Ni2+ + 2e- -0.25 VReduction:Fe2+ +2e-------Fe -0.44 V

E0cell= (-0.44)-(-0.25)= -0.19V

Because the potential is negative, it is not spontaneous


A voltaic cell is constructed using the following half-reactions:

Ag+ + 1e------Ag +0.80

Cu2++2e------Cu +0.34

Determine the cell reaction and the standard cell potential


First determine which will undergo reduction and which will oxidize

Ag+ + 1e------Ag +0.80 reduction (larger)Cu2++2e------Cu +0.34 oxide (smaller)

E0cell= +0.80 -(+0.34)= +0.46 V

2Ag+ + Cu ------2Ag + Cu2+

Education

23sec2