balancing redox reactions

SUALEHA IQBAL

BALANCING REDOX

EQUATIONS

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REDOX REACTIONS: “ Transfer of electrons between two species.”

OXIDING AGENT:

A compound that reduced is refer to

as oxidizing agent.

EXAMPLES: Br2 ,I2 , H2O2, KMnO4.

REDUCING AGENT:

A compound that is oxidize is refer to

as reducing agent.

EXAMPLES:

I- ,H2S, Zn.

INVOLVE TWO TYPES OF AGENTS:

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OXIDIZE SPECIE

REDUCE SPECIE

WHICH GIVES

WHICH GAIN

Types Of Species Involve In Redox Reactions:

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IDENTIFICATION OF REDOX REACTIONS:If the reaction is redox than it involve transfer of electrons from one atom, ion or molecule to another i.e. change in oxidation state (can observe through change in color).

OXIDATION STATE“It is the way to describe the number of electrons that have been transferred or shared between atoms of different kind.”

If two elements in a reaction change in oxidation state, one increasing, the other decreasing,Then the reaction is redox

Hydrogen =1 (except hydrides).Oxygen=(-2).Metal (+).Compound = 0.

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For example

BALANCING REDOX REACTIONS:

MnO4- (aq) + I- (aq)  Mn2+ (aq) + I2(s)

•Identification of oxidize and reduce species:

Manganese (Mn) goes from a charge of +7 to a charge of +2. Since it gained 5 electrons, it is being reduced.

Next we see that Iodine (I) goes from a charge of -1 to 0. Thus it lost electrons, and is being oxidized.

•Divide equation in half reactions:

First half is the reduction half: MnO4- (aq)  Mn2+ (aq)

Next we have the oxidation half:  I-

(aq) I2 (s)

From ion electron method:

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BALANCING PART:

•Four water molecules for 4 oxygen in MnO4.

•Balance hydrogen by adding hydrogen to other side

Reduction half: 

MnO4- (aq) + 8H+ + 5e-  Mn2+ 

(aq) +4H2O •5 electrons on the side having +7 charge are added to balance charge. Reduction half is balance.• Now come to oxidation half equation:

•In The oxidation half is iodine, so we can balance it easily by adding anotheriodine to the left side.

OXIDATION HALF :2I- (aq) I 2(S)+2(e-)

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• Multiply reduction half cell by 2 and oxidation half equation by 5 for balancing complete equation.

2{MnO4-(aq)+8H +5e- Mn+2

(aq) +H2O}

5{2I- (aq) I2(s) + 2e-}

Reduction half becomes: 2MnO4

- (aq) + 16H+ 

(aq) + 10e-  2Mn2+ (aq) + 8H2O (l)

Oxidation half becomes: 10I- (aq)   5I2 (s) + 10e-.

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Adding both equations :

2MnO4- 

(aq) + 16H+ (aq) + 10e-  2Mn2+ 

(aq) + 8H2O (l)

10I- (aq)   5I2 (s) + 10e-

10I- (aq) + 2MnO4- 

(aq) + 16H+ (aq)  5I2 (s) + 2Mn2+ 

(aq) + 8H2O (l)

Final inspection :

Proton :

Charge:

Oxygen:

Balanced

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2nd ExampleThrough oxidation state method:

Cu+2(aq)+ NO (g) +H2O

EQUATION :Cu (S)+HNO3(aq) Cu+2

(aq)+ NO (g) +H2O•Redox or not: (write oxidation states under equation)

Cu (S)+HNO3(aq)

• Insert co-efficient so that the total decrease in oxidation state of one Element equals the total increase in oxidation state of other element.

0 +5 +2 +2

3Cu (S)+2HNO3(aq

Cu (S)+HNO3(aq) Cu+2(aq)+ NO (g) +H2O

+2 ×3=+6

-3×2= -6 3Cu+2

(aq)+ 2NO (g)

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•Balance hydrogen by adding 6protons on left side.

3Cu (S)+2HNO3(aq)

+6H+

3Cu+2(aq)+ 2NO (g)

+4H2O•Spectators ions: 6 NO3

- on each side of equation

3Cu (S)

+8HNO3(aq)

3Cu(NO3)2 (aq)+ 2NO (g)

+4H2O(l)

BALANCED

•Balance oxygen: 4H2O on right will balance oxygen.

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Balancing organic redox reactions:oxidation of n-butyl alcohol with K2Cr2O7

CH3CH2CH2CH2-OH + K2Cr2O7 , H+ CH3CH2CH2COOH

half-reactions for the oxidation and the reduction involved:

oxidation: CH3CH2CH2CH2-OH CH3CH2CH2COOH

reduction: Cr2O72- Cr3+

Mass balance:

oxidation: CH3CH2CH2CH2-OH + H2O CH3CH2CH2COOH + 4 H+

reduction: Cr2O72- + 14H+ 2 Cr3+ + 7 H2O

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Charge balance

oxidation: CH3CH2CH2CH2-OH + H2O CH3CH2CH2COOH + 4H+ + 4e-

reduction: 6 e- + Cr2O72- + 14

H+

2 Cr3+ + 7 H2O

No. of electron in oxidation half = No. of electron in reduction half

• 3 x{( CH3CH2CH2CH2-OH + H2O CH3CH2CH2COOH + 4H+ + 4 e- )}

•3 CH3CH2CH2CH2-OH + 3 H2O 3 CH3CH2CH2COOH + 12H+ + 12 e-

•2× { (6 e- + Cr2O72- +

14 H+

2 Cr3+ + 7 H2O)}

•12 e- + 2 Cr2O72- + 28

H+

4 Cr3+ + 14 H2O

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Add the two half-reactions together

3 CH3CH2CH2CH2-OH + 3 H2O + 12e- + 2Cr2O72- +

28 H+

3 CH3CH2CH2COOH + 12 H+ + 12 e- + 4 Cr3+ + 14 H2O

Canceling out the electrons and extra waters, etc.

3 CH3CH2CH2CH2-OH + 2 Cr2O72- + 16 H+

3 CH3CH2CH2COOH + 4 Cr3+ + 11 H2O

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Spectator ions

3 CH3CH2CH2CH2-OH + 2K2Cr2O7 + 8H2SO4

3 CH3CH2CH2COOH + 2 Cr2(SO4)3 + 11 H2O + 2 K2SO4

Final equation:

3 CH3CH2CH2CH2-OH + 2 K2Cr2O7 + 8 H2SO4

3 CH3CH2CH2COOH + 2 Cr2(SO4)3 + 11 H2O + 2K2SO4

BALANCED

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THANK YOUreduced oxidize

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SUALEHA IQBAL..