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Electrochemistry Part IV: Spontaneity & Nernst Equation

Dr. C. Yau

Spring 2014

Jespersen Chap. 20 Sec 4 & 5

Skipping Sec 6 & 8

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Spontaneity of Reaction

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We know to have a spontaneous rxn…

E > 0

ΔG < 0

How are these two related?ΔG = - n FEcell where n = moles of e-

F = Faraday's constant9.65x104 C/mol e-

(remember 1 V = 1J/C)

ΔGo = - n FEocell

(under standard conditions)

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Example 20.7 p. 937

Calculate ΔGo for the reaction, given that its standard cell potential is 0.320 V at 25oC.

NiO2 (s) + 2Cl(aq) + 4H+ (aq)

Cl2 (g) + Ni2+ (aq) + 2H2O (l)

ΔGo = - n FEocell

F = 9.65x104 C/mol e-

(1 V = 1J/C, so Eo = 0.320 J/C)How do we figure out what n is? Ans. -61.8 kJ

Do Pract Exer 13 & 14 p. 938

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Calculating K from Cell Potential

We know ΔGo = - n FEocell

We also know ΔGo = - RT ln K (Chap. 19)

so - n FEo = - RT ln K

Example 20.8 p. 789

Calculate K for the reaction in Example 20.8.

NiO2 (s) + 2Cl(aq) + 4H+ (aq)

2Cl2 (g) + Ni2+ (aq) + 2H2O (l)

Collect all the constants we need.

Do Pract Exer 15 & 16 p. 939

Derivation of the Nernst Eqn

What happens when it is not under standard conditions?

Divide both sides of eqn by (-nF) we get...

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o

In the previous chapter on thermodynamics, we know...

G = G + RT lnQ

ΔG = - n FEcell

ΔGo = - n FEocell

0cell-n F Ecell = -n F E + RT ln Q

0cell cell

RTE = E ln Q

n F Nernst Equation

Common simplified version of the Nernst Equation for 25.0 oC:

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0cell cell

0cell cell

0cell cell

0cel

-1 -

l cell

0cell c

1

4

ell

-1

RT2.303

F

(8

RTE = E Q

n FRT

E = E

.314 J mol K )(298.15 K)(2.303)

9.6

Qn

5x10 C mol

F1

E = E Qn

1E = E Q

n

1E = E 0.0

ln

(2.303 log)

log

log

n

-1

0cell cell

0cell cell

592 J C Q

1E = E 0.0592

log

lo V Qn0.0592

g

VE = E Q

nlog

This version of Nernst Eqn will be given also, but remember it’s only for 25.0oC

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Nernst Equation

Eo is the cell potential under standard conditions (for aqueous soln, 1M)

What if it is not 1 M?

Example 20.9 p. 940

Suppose a galvanic cell employs the following:

Ni2+ + 2e- Ni Eo = - 0.25 V

Cr3+ + 3e- Cr Eo = - 0.74 V

Calculate the cell potential when [Ni2+] = 4.87x10-4M

and [Cr3+] = 2.48x10-3 M

0cell cell

0.0592 VE = E log Q

n

Ans. +0.44 V

This type of quest will be on your final exam.

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Example 20.10 p. 941The rxn of tin metal with acid can be written as

Sn (s) + 2H+ (aq) Sn2+ (aq) + H2 (g)

Calculate the cell potential(a) when the system is at standard state.(b) when the pH = 2.00(c) when the pH is 5.00.Assume that [Sn2+] = 1.00 M and the partial

pressure of H2 is also 1.00 atm.

Do Pract Exer 17, 18, 20 p. 942

Ans. +0.02 V

Ans. -0.16V

What we are skipping in Chap. 20:

pp. 943-951

Concentration from E Measurements

Sec 20.6 Electricity

Batteries: Lead Storage Batteries

Zinc-Manganese Dioxide Cells (LeClanche cell)

Nickel-Cadmium Rechargeable Batteries

Nickel-Metal Hydride Batteries

Lithium Batteries

Lithium Ion Cells

Fuel Cells

Photovoltaic Cells

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