•Electrochemistry•Electrochemistry

Oxidation Reduction ReactionsOxidation Reduction Reactions• (1) Oxidation: Loss e- Increase in Oxidation Number Zn(s) Zn2+ + 2e-

• (2) Reduction: Acceptance of e-

Decrease in Oxidation Number Cl2(g) + 2e- 2Cl-

• (1) Oxidation: Loss e- Increase in Oxidation Number Zn(s) Zn2+ + 2e-

• (2) Reduction: Acceptance of e-

Decrease in Oxidation Number Cl2(g) + 2e- 2Cl-

Balancing Oxidation-Reduction Equations:Balancing Oxidation-Reduction Equations: Use Half-Reaction MethodUse Half-Reaction Method

The half-reactions forSn2+(aq) + 2Fe3+(aq) Sn4+(aq) + 2Fe3+(aq)

areSn2+(aq) Sn4+(aq) +2e-

2Fe3+(aq) + 2e- 2Fe2+(aq)

• Oxidation Half-Reaction: electrons are products.

• Reduction Half-Reaction: electrons are reactants.

Half-Reaction Method for Balancing Half-Reaction Method for Balancing Oxidation-Reduction EquationsOxidation-Reduction Equations

1. Separate the equation into the two half-reactions. Write down the two half reactions.

2. Balance each half reaction:a. First balance all elements other than H and O.b. Then balance O by adding water.c. Then balance H by adding H+ if have acidic

solution.d. Finish by balancing charge by adding electrons.

3. Multiply each half reaction to make the number of electrons equal.

4. Add the two half-reactions and simplify. To simplify, remove components common to both reactant and product sides.

5. Check!

Example

Balance: (acidic)MnO4

- (aq) + Fe2+

(aq) Mn2+ (aq) + Fe3+

(aq)

1. The two incomplete half reactions areMnO4

-(aq) Mn2+(aq)

Fe2+ (aq) Fe3+

(aq)

MnO4- Mn2+ + 4 H2O

2. Balance each half reaction:

Fe2+ (aq) Fe3+ (aq)

MnO4- + 5e- Mn2+ + 4 H2O

Fe2+ (aq) Fe3+ (aq) + e-

8H+ + MnO4- + 5e- Mn2+ + 4 H2O

Fe2+ (aq) Fe3+ (aq) + e-

8H+ + MnO4- + 5e- Mn2+ + 4 H2O

5 (Fe2+ (aq) Fe3+ (aq) + e-)

8H+ + MnO4- + 5e- Mn2+ + 4 H2O

5 (Fe2+ (aq) Fe3+ (aq) + e-)

8H+ + MnO4- + 5 Fe2+ (aq) Mn2+ + 4 H2O + 5Fe3+ (aq)

Example

Balance: (basic)MnO4

- (aq) + I- (aq) MnO2

+ I2

1. The two incomplete half reactions areMnO4

-(aq) MnO2

I- (aq) I2

2. Balance each half reaction:

MnO4-(aq) + 3e- MnO2

2I- (aq) I2 + 2e-

MnO4-(aq) + 3e- MnO2 + 4OH-

2I- (aq) I2 + 2e-

2H2O + MnO4-(aq) + 3e- MnO2 +

4OH-

2I- (aq) I2 + 2e-

2 (2H2O + MnO4-(aq) + 3e- MnO2 + 4OH-)

3 (2I- (aq) I2 + 2e-)

4H2O + 6I- (aq) + MnO4-(aq) 2MnO2 + 8OH- + 3I2

Problem

Complete and balance the following equations, and identify the oxidizing and reducing agents

Cr2O72- (aq) + I-(aq) Cr3+(aq) + IO3-(aq)(acidic solution)

Pb(OH)42- (aq) + ClO- (aq) PbO2(s) + Cl-(aq)(basic solution)

Zn Cu

salt bridge(K+) Cation flow →

(Cl-) flow ←

Brief Activity Series

Strong Reducing Agent.

Strong Oxidizing Agent

How many moles of Cu can be plated out of a solution containing Cu2+ ions if a 1.50 amp current is passed through for 300 s?

Cu2+ + 2e- → Cu(s)

Notes: 1 amp = C/s F = 9.65 x 104 c/ mole e-

Find: c → moles e-s → moles Cu

= moles Cu1.50 c

s

300 s mole e-

9.65 x 104 C

1 mole Cu

2 mole e-s

2.33 x 10-3

Effect of Concentration on EMF

A Battery going dead.

∆G = ∆Go + RT lnQ

-nFE = -nFEo + RT ln Q

Dividing by –nF gives Rise to the Nerst Equation.

The Nerst Equation:

QlognF

2.303RT - E E :log 10 basefor Or o

logQn

0.0592 - E E : K 298 @ o

QlnnF

RT - E E o

Q: aA + bB ↔ cC + dD

ba

d

BA

D

][][

][[C] Q

c

mol K

J 314.8

R

molV

JxF

41065.9

n = # of e- s transferred

The Nerst Equation allows us to find voltage under nonstandard conditions

Example:

Determine E (voltage) for: Fe(s) + Cd2+(aq) → Fe2+

(aq) + Cd(s)

When [Fe2+] = 0.10 M and [Cd2+] = 1.0 M @ 298K

Half RXNs: Fe(s) → Fe2+ + 2e- Eo = +0.44 V

Cd2+ + 2e- → Cd(s) Eo = -0.40 V

Eo = +0.04 V

][

][log

0592.02

2

Cd

Fe

nEE o

VVE 07.0]0.1[

]10.0[log

2

0592.004.0 + Value means spontaneous

Now try: [Fe2+] = 1.0 M and [Cd2+] = 0.01 M @ 298K

VVE 02.0]01.0[

]0.1[log

2

0592.004.0

- Value means non spontaneous and will go in opposite direction

What does an E = 0 value mean?

logQn

0.0592 - E E : K 298 @ o

logKn

0.0592 - E 0 E : K 298 @ o

0.0592

nElogK : K 298 @

o

For: Fe(s) + Cd2+(aq) → Fe2+(aq) + Cd(s) Eo = +0.04V

35.10.0592

0.04V))(e (2mollogK : K 298 @

-

K = 22

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My god preservation this country ( Iran ) from ENEMY , FALSE and ARID YEAR