ChemistryElectrochemistry: E0, G0, Kp. 1

ELECTROCHEMISTRY – RELATIONSHIPS BETWEEN E0, G0, and K

Background

E0 is a measure of the direction of a chemical reaction that is thermodynamically favored. If E0 > 0, the forward direction is favored. You’ve probably already realized that E0, G0, and K are related in determining the direction of a chemical reaction. These are related by the following equations:

G0 = –RT lnK = –nFE0
where: / R / = 8.31 J/molV
n / = number of moles of electrons transferred in the reaction
F / = 9.648 x 104 J/molV (Faraday’s constant)

N.B.G0 and E0 have opposite signs. The direction of the reaction is thermodynamically favored when G0 is negative but E0 is positive.

The above equation can be rearranged to lnK

RT/F is a constant: =

Problem-1: Determining G0 and K.

3Ag(s) + NO3–(aq) + 4H+(aq) 3Ag+(aq) + NO(g) + 2H2O(l)

(Remember the answers for balancing redox reactions in acids and bases? This is one way to use it.)

CalculateG0

(1) Oxidation Numbers:Ag: 0  +1 (oxidation)N: +5  +2 (reduction)

(2) Half-Reactions:NO3–(aq) + 4H+(aq) + 3e– NO(g) + 2H2OE0red = + 0.964 V

Ag(s) + e– Ag+E0ox = – 0.799 V

E0 = (0.964 V) + (– 0.799 V) = 0.165 V

(3) nn = 3 moles of electrons (after scaling half-reactions, 3e–cancel out)

(4) G0G0 = nFE

= (3e– mole)(9.648 x 104 J/molV)(0.165 V)

G0 = –4.78 kJ

(problem is continued )

Calculate K

From above: E0 = 0.165 V ; n = 3 mol; G0 = –4.78 kJ

(1) Equation:lnK

E0 = ln K

0.165 V = ln K

ln K = 19.3

K = e19.3 => . K = 2.4 x 108 .

Problem-2: DetermineK for the Formation of Rust

Does iron rust at room temperature? Calculate the equilibrium constant for the reaction:

2 Fe(s) + 1.5 O2(g) Fe2O3(s) (298 K)

ΔH0 rxn = –824.2 kJ ΔS0 rxn = -270.5 J/K

ELECTROCHEMISTRY – RELATIONSHIPS BETWEEN E0, G0, and K KEY

Background

E0 is a measure of the direction of a chemical reaction that is thermodynamically favored. If E0 > 0, the forward direction is favored. You’ve probably already realized that E0, G0, and K are related in determining the direction of a chemical reaction. These are related by the following equations:

G0 = –RT lnK = –nFE0
where: / R / = 8.31 J/molV
n / = number of moles of electrons transferred in the reaction
F / = 9.648 x 104 J/molV (Faraday’s constant)

N.B.G0 and E0 have opposite signs. The direction of the reaction is thermodynamically favored when G0 is negative but E0 is positive.

The above equation can be rearranged to lnK

RT/F is a constant: =

Problem-1: Determining G0 and K.

3Ag(s) + NO3–(aq) + 4H+(aq) 3Ag+(aq) + NO(g) + 2H2O(l)

(Remember the answers for balancing redox reactions in acids and bases? This is one way to use it.)

CalculateG0

(1) Oxidation Numbers:Ag: 0  +1 (oxidation)N: +5  +2 (reduction)

(2) Half-Reactions:NO3–(aq) + 4H+(aq) + 3e– NO(g) + 2H2OE0red = + 0.964 V

Ag(s) + e– Ag+E0ox = – 0.799 V

E0 = (0.964 V) + (– 0.799 V) = 0.165 V

(3) nn = 3 moles of electrons (after scaling half-reactions, 3e– cancel out)

(4) G0G0 = nFE

= (3e– mole)(9.648 x 104 J/molV)(0.165 V)

G0 = –4.78 kJ

(problem is continued )

Calculate K

From above: E0 = 0.165 V ; n = 3 mol; G0 = –4.78 kJ

(1) Equation:lnK

E0 = ln K

0.165 V = ln K

ln K = 19.3

K = e19.3 => . K = 2.4 x 108 .

Problem-2: Determine K for the Formation of Rust

Does iron rust at room temperature? Calculate the equilibrium constant for the reaction:

2 Fe(s) + 1.5 O2(g) Fe2O3(s) (298 K)

ΔH0 rxn = –824.2 kJ ΔS0 rxn = -270.5 J/K

Strategy, Overall: The equation we want to use relates the thermodynamic values to K.

(1) Calculate G0:ΔG0 rxn= ΔH0 – TΔS0

= -824,200 J/mol – (298 K * -270.5 J/K·mol)

= -743.59 kJ

(2) Calculate K:ΔG0= –RT lnK

lnK= –ΔG0 / (RT)

= 743,59 J / (8.314 J/K * 298 K)

= 743,050 J / 2,478 J

= 299.9

K= e299.9

K= 7.146 x 10129

K is a very large number. Thus, the formation of rust is extremely thermodynamically favored.

Strategy, Overall: The equation we want to use relates the thermodynamic values to K.

(1) Calculate G0:ΔG0 rxn= ΔH0 – TΔS0

= -824,200 J/mol – (298 K * -270.5 J/K·mol)

= -743.59 kJ

(2) Calculate K:ΔG0= –RT lnK

lnK= –ΔG0 / (RT)

= 743,59 J / (8.314 J/K * 298 K)

= 743,050 J / 2,478 J

= 299.9

K= e299.9

K= 7.146 x 10129

K is a very large number. Thus, the formation of rust is extremely thermodynamically favored.

Problem-3: E0 and Ksp

What is E0 at 25oC for the following reaction? Ksp for BaSO4 is 1.1 x 10–10.

Ba2+(aq) + SO42–(aq) BaSO4(s)

Final Useful Equation:ln K

E0 = ln K

(1)Determine K for the equation.

Ksp = = 1.10 x 10–10

But the above reaction is the reverse of the one for Ksp. Using the reciprocal law:

K = 1/Ksp(BaSO4) = 9.1 x 109

(2)Determine n.

This should be 2. Barium exists as only Ba2+ ions or Ba0: Ba2++ 2e–  Ba(s)

(3)G0 = –RT lnK = –nFE0

 E0 = = =

E0 = +0.29 V

ANSWERS

Problem-3: E0 and Ksp

What is E0 at 25oC for the following reaction? Ksp for BaSO4 is 1.1 x 10–10.

Ba2+(aq) + SO42–(aq) BaSO4(s)

Final Useful Equation:ln K

E0 = ln K

(4)Determine K for the equation.

Ksp = = 1.10 x 10–10

But the above reaction is the reverse of the one for Ksp. Using the reciprocal law:

K = 1/Ksp(BaSO4) = 9.1 x 109

(5)Determine n.

This should be 2. Barium exists as only Ba2+ ions or Ba0: Ba2+ + 2e– Ba(s)

(6)G0 = –RT lnK = –nFE0

 E0 = = =

E0 =+0.29 V

Problem-4: 2014 AP Released Question

A student is given a standard galvanic cell, represented above, that has a Cu electrode and a Sn electrode. As current flows through the cell, the student determines that the Cu electrode increases in mass and the Sn electrode decreases in mass.

(a)Identify the electrode at which oxidation is occurring. Explain your reasoning based on the student’s observations.

(b)As the mass of the Sn electrode decreases, where does the mass go?

(c)In the expanded view of the center portion of the salt bridge shown in the diagram below, draw and label a particle view of what occurs in the salt bridge as the cell begins to operate. Omit solvent molecules and use arrows to show the movement of particles.

(d)A nonstandard cell is made by replacing the 1.0 M solutions of Cu(NO3)2 and Sn(NO3)2 in the standard cell with 0.50 M solutions of Cu(NO3)2 and Sn(NO3)2. The volumes of solutions in the nonstandard cell are identical with those in the standard cell.

(i) Is the cell potential of the nonstandard cell greater than, less than, or equal to the cell potential of the standard cell? Justify your answer.

(ii)Both the standard and nonstandard cells can be used to power an electronic device. Would the nonstandard cell power the device for the same time, a longer time, or a shorter time as compared with the standard cell? Justify your answer.

(e)In another experiment, the student places a new Sn electrode into a fresh solution of 1.0 M Cu(NO3)2.

Half-Reaction / E0 (V)
Cu+ + e– Cu(s) / 0.52
Cu2+ + 2e– Cu(s) / 0.34
Sn4+ + 2e–Sn2+ / 0.15
Sn2+ + 2e–Sn(s) / –0.14

(i) Using information from the table above, write and net-ionic equation for the reaction between the Sn electrode and the Cu(NO3)2 solution that would be thermodynamically favorable. Justify that the reaction is thermodynamically favorable.

(ii)Calculate the value of DG0 for the reaction. Include units with your answer..