### Electrochemistry Part I: Redox Review & How to Balance Complex Redox Equations

Dr. C. Yau Fall 2013

Jespersen 6/e Chap. 6 Sec 1 & 2

## Redox Rxn

Cu

(s)

+ AgNO 3

(aq)

?

Net ionic equation?

## Review of Redox Rxns

Oxidation

is an

increase

in oxidation number.

Reduction

is a

decrease

in oxidation number.

Cu

(s)

+

2

AgNO 3

(aq)

Cu(NO 3 ) 2

(aq)

+

2

Ag

(s)

Cu

o

+

2

Ag + Cu 2+ +

2

Ag

o

0 +2 +1 0 Cu is oxidized. Ag is reduced.

Cu loses 2 e -

while

Ag gains 1 e each

.

A total of 2 e is transferred

from Cu to Ag + Cu = reducing agent ; Ag + = oxidizing agent .

LEO

### the lion says

GER

Loss of Electrons is Oxidation.

Gain of Electrons is Reduction.

Cu + Ag + Cu 2+ Cu Cu 2+ + 2 e + Ag

Oxidation (Loss of e )

Ag + + e Ag

Reduction (Gain of e )

### Recognition of Redox

Redox always involve a change in oxidation number.

Reduction must be accompanied by oxidation, and vice versa.

Oxidation numbers are

hypothetical charges

assigned to

each

atom.

Each atom, even in molecular substances that have no ions, is assigned a charge.

Note: The charge is

hypothetical. Oxidation numbers is only for "book keeping" to keep track of electrons.

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### Assigning Oxidation Numbers

LEARN THESE RULES IN ORDER!

1. Oxidation # of any free element is zero. e.g. H in H 2 , P in P 4 are assigned zero.

2. Oxid # of simple mon atomic ions is the charge of the ion.

e.g. In MgCl 2 , Mg is +2. Cl is -1.

In SnS, Sn is +2 and S is -2.

### Assigning Oxidation Numbers

3. In its cmpd, F is assigned -1.

4. In its cmpds, H is +1 unless it is bonded to a metal where it is -1.

e.g. In HCl, H is +1 and Cl is -1.

In MgH 2 , Mg is +2 and H is -1.

5. In its cmpds, O is -2 unless it is a peroxide. eg. in MgO, Mg is +2, O is -2.

In hydrogen peroxide (H 2 O 2 ), H is +1 and O is -1.

### Assigning Oxidation Numbers

6. Oxidation # of others are generally calculated from knowing that the sum of all charges must add up to charge of the particle. For example, In HClO, sum of charges = 0 +1 ? -2 H is assigned +1, O is assigned -2. Cl is calculated to be +1 in order for the sum to be zero.

In the chlorite ion, ClO 2 , sum of charges = -1 ? -2 Oxygen is assigned -2, what must Cl be in order for net charge to be -1?

Ans. Cl = +3

Give the oxidation number: 1. N in NO 3 2. Cr in CrSO 4 3. S in S 2 O 4 2 Identify which is the oxidizing agent: Cr 2 O 7 2− + 14H + + 6 Cl −  3Cl 2 + 2 Cr 3+ + 7 H 2 O 9

Answer to questions on previous slide: Give the oxidation number: 1.

N in NO 3 Ans. +5 2.

Cr in CrSO 4 Ans. +2 (Be sure to ask if you don’t know how you can tell.) 3.

S in S 2 O 4 2 Ans. +3 Identify which is the oxidizing agent: Cr 2 O 7 2− + 14H + + 6 Cl −  3Cl 2 + 2 Cr 3+ + 7 H 2 O Ans. Cl is going from -1 to 0, so it is being oxidized and cannot be the oxidizing agent.

Cr is going from +6 to +3, so it is being reduced.

The oxidizing agent is Cr 2 O 7 2−

name it as

Cr 2 O 7 2− (

Note: By convention, you and not just Cr.)

### Balancing Redox Under Acidic Conditions

1. Verify that rxn is redox.

2. Divide skeleton eqns into half-reactions.

3. Balance atoms other than H and O.

4. Balance O by adding H 2 O to each side of eqn.

5. Balance H by adding H + to side that needs H.

6. Balance charge by adding electrons.

7. Make the # of electrons gained equal to number lost 8. Add two half-reactions and cancel anything that is the same on both sides of eqn.

Balance the following reactions which are under

acidic

conditions.

Cr 2 O 7 2− + Cl −  Cl 2 + Cr 3+ Mg + VO 4 3−  Mg 2+ + V 2+ Sn 2+ + IO 4 −  Sn 4+ + I −

Practice Example 6.6, p.225, Pract Exer 12, 13, 14, 15 on p.226; p.245 #6.37.

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## Conditions

Follow the same steps as for acidic conditions but add on the following steps at the end. It would save you steps if you do this AFTER you have already combined the half-reactions into one equation: 9. Add to BOTH sides, the same number of OH as there are H + (Under basic condition, there cannot be any H + ).

10. 11. Combine H + Cancel any H and OH 2 O to form H 2 O .

that you can.

Balance the following reactions which are under

basic

conditions.

1. SO 3 2 – + MnO 4 –  SO 4 2 – + MnO 2 Which is the reducing agent?

2. CN – + MnO 4 –  CNO – + MnO 2 SO 3 2 Which is the oxidizing agent?

Practice Exer 6.16, 17, 18, 19on p. 227 and p.245 #6.39

MnO 4 14