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4-5-2011

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Calculation of the standard emf of an electrochemical cell

The procedure is simple:

1. Arrange the two half reactions placing the one with the larger Eo up (the cathode).

2. The half reaction with the lower Eo is placed down (the anode).

3. Eocell = Eo

cathode - Eoanode

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A galvanic cell consists of a Mg electrode in a 1.0 M Mg(NO3)2 and a Ag electrode in a 1.0 M AgNO3 solution. Find the overall reaction and the standard cell emf.

From table of standard reduction potentials, we have:

Ag+ + e Ag(s) Eo = 0.80V

Mg2+ + 2e Mg(s) Eo = -2.37V

From diagonal rule Ag+ will react with Mg(s), therefore the reaction will be as follows:

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Multiply the first half reaction by 2 to account for the number of electrons in the second one and reverse the second half reaction:

2Ag+ + 2e 2Ag(s) Eo = 0.80V

Mg(s) Mg2+ + 2e Eo = +2.37V

2Ag+ + Mg(s) Mg2+ + 2Ag(s) Eo = 3.17V

Eorxn = Eo

cathode - Eoanode

Eorxn = 0.80 – (- 2.37) = 3.17V

Remember to use Eo values with sign as it appears in the table (reduction potential).

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What is the standard emf of an electrochemical cell made of a Cd electrode in a 1.0 M Cd(NO3)2 solution and a Cr electrode in a 1.0 M Cr(NO3)3 solution?

Cd2+ (aq) + 2e- Cd (s) E0 = -0.40 V

Cr3+ (aq) + 3e- Cr (s) E0 = -0.74 V

2e- + Cd2+ (1 M) Cd (s)

Cr (s) Cr3+ (1 M) + 3e-Anode (oxidation):

Cathode (reduction):

2Cr (s) + 3Cd2+ (1 M) 3Cd (s) + 2Cr3+ (1 M)

x 2

x 3

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Remember that multiplying a half reaction with any coefficient will not change the value of Eo.

E0 = Ecathode - Eanodecell

0 0

E0 = -0.40 – (-0.74) cell

E0 = 0.34 V cell

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Spontaneity of Redox Reactions

G = -nFEcell

G0 = -nFEcell0

n = number of moles of electrons in reaction

F = 96,500J

V • mol = 96,500 C/mol

G0 = -RT ln K = -nFEcell0

Ecell0 =

RTnF

ln K(8.314 J/K•mol)(298 K)

n (96,500 J/V•mol)ln K=

=0.0257 V

nln KEcell

0

=0.0592 V

nlog KEcell

0

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What is the equilibrium constant for the following reaction at 250C? Fe2+ (aq) + 2Ag (s) Fe (s) + 2Ag+ (aq)

=0.0257 V

nln KEcell

0

E0 = -0.44 – (0.80)

E0 = -1.24 V

e0.0257 V

x nE0 cell

K =

n = 2

= e0.0257 V

x 2-1.24 V

K = 1.23 x 10-42

E0 = EFe /Fe – EAg /Ag0 0

2+ +

From the reaction as written, Fe half cell is the cathode, and Ag half cell is the anode:

2Ag+ + 2e Ag(s) Eo = 0.80V

Fe2+ + 2e Fe(s) Eo = -0.44V

From diagonal rule, the rxn is non spontaneous

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Calculate the equilibrium constant for the following reaction at 25 oC:

Sn(s) + 2Cu2+ Sn2+ + 2Cu+

From the reaction as written, Cu half cell is the cathode, and Sn half cell is the anode:

2Cu2+ + 2e 2Cu+ Eo = 0.15V

Sn2+ + 2e Sn(s) Eo = -0.14V

According to the diagonal rule the reaction occurs spontaneously as written:

Eocell = Eo

cathode (Cu2+, Cu+) – Eoanode(Sn2+, Sn(s))

Eocell = 0.15 – (-0.14) = 0.29 V

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e 0.0257 V

x nE0 cell

K =

K = 6.3*109

K = e 0.0257

x 20.29

K = 6.3*109

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Calculate the standard free energy for the following reaction at 25 oC:

2Au(s) + 3Ca2+ 2Au3+ + 3Ca(s)First, we write the two half cells and calculate the cell potential as

they appear in the reaction. From the reaction as written, Ca half cell is the cathode, and Au half cell is the anode:

2Au3+ + 6e 2Au(s) Eo = +1.50V anode

3Ca2+ + 6e 3Ca(s) Eo = -2.87V cathode

According to the diagonal rule the reaction will not occur spontaneously as written:

Eocell = Eo

cathode– Eoanode

Eocell = -2.87 – 1.50 = -4.37 V

Go = - nFEocell

Go = - (6)(96500)(-4.37) = 2.53*106 JGo = 2.53*103 kJ

The large +ve Go suggests a non-spontaneous reaction at the standard states.

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The Effect of Concentration on Cell Emf

G = G0 + RT ln Q G = -nFE G0 = -nFE 0

-nFE = -nFE0 + RT ln Q

E = E0 - ln QRTnF

Nernst equation

At 298 K

-0.0257 V

nln QE0E = -

0.0592 Vn

log QE0E =

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Will the following reaction occur spontaneously at 250C if [Fe2+] = 0.60 M and [Cd2+] = 0.010 M? Fe2+ (aq) + Cd (s) Fe (s) + Cd2+ (aq)

E0 = -0.44 – (-0.40)

E0 = -0.04 V

E0 = EFe /Fe – ECd /Cd0 0

2+ 2+

-0.0257 V

nln QE0E =

-0.0257 V

2ln -0.04 VE =

0.0100.60

E = 0.013

E > 0 Spontaneous

From the reaction as written, Fe half cell is the cathode, and Cd half cell is the anode, use these to calculate the standard electrode potential:

We cannot use Eo to predict whether the rxn is spontaneous or not since we do not have standard conditions

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Will the following reaction occur spontaneously at 250C if [Fe2+] = 0.68 M and [Co2+] = 0.15 M? Fe2+ (aq) + Co (s) Fe (s) + Co2+ (aq)

E0 = -0.44 – (-0.28)

E0 = -0.16 V

E0 = EFe /Fe – ECo /Co0 0

2+ 2+

-0.0257 V

nln ([Co2+]/[Fe2+])E0E =

From the reaction as written, Fe half cell is the cathode, and Co half cell is the anode, use these to calculate the standard electrode potential:

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-0.0257 V

2ln -0.16 VE =

0.150.68

E = - 0.14

E < 0 Non-spontaneous

Now think at what ratio of [Co2+]/[Fe2+] will the reaction be spontaneous? To find such a ratio we must pass by the point where Eo

cell = 0

-0.0257 V

nln ([Co2+]/[Fe2+])- 0.16 0 =

[Co2+]/[Fe2+] = 4*10-6, the ratio must be less than 4*10-6

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Concentration CellsGalvanic cell from two half-cells composed of the same material but differing in ion concentrations.

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Selected Problems

2, 3, 6, 11-13, 15, 16, 17, 18, 21, 22, 24, 25, 29, 32, 34.