L2 - Redox Reactions

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Deduction of redox reactions using half-equations in acidic or neutral solutions

starter activity

Complete Question 1 only

Copy of 10 - Exercise 1 - Oxidation and reduction.doc

07 - Exercise 1 - redox reactions.docx

OXIDATION AND REDUCTION IN TERMS OF ELECTRONS

Oxidation and reduction are not only defined as changes in O and H

REDOX When reduction and oxidation take place

OXIDATION Removal (loss) of electrons ‘OIL’

species will get less negative or more positive

REDUCTION Gain of electrons ‘RIG’

species will become more negative or less positive

REDUCTION in O.S. Species has been REDUCED

e.g. Cl is reduced to Cl¯ (0 to -1)

INCREASE in O.S. Species has been OXIDISED

e.g. Na is oxidised to Na⁺ (0 to +1)

State if the changes involve oxidation (O) or reduction (R) or neither (N)

Fe²⁺—> Fe³⁺

I₂ —> I¯

F₂ —> F₂O

C₂O₄^2- —> CO₂

H₂O₂ —> O₂

H₂O₂ —> H₂O

Cr₂O₇^2- —> Cr³⁺

Cr₂O₇^2- —> CrO₄^2-

SO₄^2- —> SO₂

Fe²⁺—> Fe³⁺O +2 to +3

I₂ —> I¯ R 0 to -1

F₂ —> F₂O R 0 to -1

C₂O₄^2- —> CO₂O +3 to +4

H₂O₂ —> O₂O -1 to 0

H₂O₂ —> H₂O R -1 to -2

Cr₂O₇^2- —> Cr³⁺R +6 to +3

Cr₂O₇^2- —> CrO₄^2-N +6 to +6

SO₄^2- —> SO₂R +6 to +4

Balancing Redox Equations

1 Work out formulae of the species before and after the change; balance if required

2 Work out oxidation state of the element before and after the change

3 Add electrons to one side of the equation so that the oxidation states balance

4 If the charges on the species (ions and electrons) on either side of the equation do not balance then add sufficient H⁺ ions to one of the sides to balance the charges

5 If equation still doesn’t balance, add sufficient water molecules to one side

Explanation

Cr₂O₇^2- being reduced to Cr³⁺ in acidic solution

Step 1 Cr₂O₇^2- ———> Cr³⁺there are two Cr’s on LHS

Cr₂O₇^2- ———> 2Cr³⁺both sides now have 2

Step 2 2 @ +6 2 @ +3 both Cr’s are reduced

Step 3 Cr₂O₇^2- + 6e¯ ——> 2Cr³⁺each Cr needs 3 electrons

Step 4 Cr₂O₇^2- + 6e¯ + 14H⁺ ——> 2Cr³⁺

Step 5 Cr₂O₇^2- + 6e¯ + 14H⁺ ——> 2Cr³⁺ + 7H₂O now balanced

Activity - Balancing Half Equations

Balance the Following Half Equations:

Na —> Na⁺

Fe²⁺ —> Fe³⁺

I₂ —> I¯

C₂O₄^2- —> CO₂

H₂O₂ —> O₂

H₂O₂—> H₂O

NO₃^- —> NO

NO₃^- —> NO₂

SO₄^2-—> SO₂

Na —> Na⁺ + e^-

Fe²⁺ —> Fe³⁺+ e^-

I₂ + 2e^- —> 2I¯

C₂O₄^2- —> 2CO₂+ 2e^-

H₂O₂ —> O₂+ 2H⁺ + 2e^-

H₂O₂ + 2H⁺ + 2e^- —> 2H₂O

NO₃^- + 4H⁺ + 3e^- —> NO+ 2H₂O

NO₃^- + 2H⁺ + e^- —> NO₂+ H₂O

SO₄^2-+ 4H⁺ + 2e^- —> SO₂+ 2H₂O

Combining Half Equations

A combination of two ionic half equations, one involving oxidation and the other reduction, produces a REDOX equation. The equations are balanced as follows...

Step 1 Write out the two half equations

Step 2 Multiply the equations so that the number of electrons in each is the same

Step 3 Add the two equations and cancel out the electrons on either side

Step 4 If necessary, cancel any other species which appear on both sides

Worked Example:

The reaction between manganate(VII) and iron(II)

Step 1 Write out the two half equations

Fe²⁺ ——> Fe³⁺ + e¯ Oxidation

MnO₄¯ + 5e¯ + 8H⁺ ——> Mn²⁺ + 4H₂O Reduction

Step 2 Multiply the equations so that the number of electrons in each is the same

5Fe²⁺ ——> 5Fe³⁺ + 5e¯ multiplied by 5

MnO₄¯ + 5e¯ + 8H⁺ ——> Mn²⁺ + 4H₂O multiplied by 1

Step 3 Add the two equations and cancel out the electrons on either side

MnO₄¯ + ~~5e¯~~ + 8H⁺ + 5Fe²⁺ ——> Mn²⁺ + 4H₂O + 5Fe³⁺ + ~~5e¯~~

Step 4 If necessary, cancel any other species which appear on both sides

MnO₄¯ + 8H⁺ + 5Fe²⁺ ——> Mn²⁺ + 4H₂O + 5Fe³⁺

Activity

Activity - Construct these Redox Equations:

Mg and H⁺
Cr₂O₇^2- and Fe²⁺
H₂O₂ and MnO₄¯
C₂O₄^2-and MnO₄¯
S₂O₃^2- and I₂
Cr₂O₇^2- and I¯

Mg ——> Mg²⁺ + 2e^¯(x1)

H⁺+ e^¯——> ½ H₂(x2)

Mg + 2H⁺——> Mg²⁺ + H₂

Cr₂O₇^2- + 14H⁺ + 6e^¯——> 2Cr³⁺ + 7H₂O (x1)

Fe²⁺——> Fe³⁺+ e^¯(x6)

Cr₂O₇^2- + 14H⁺ + 6Fe²⁺——> 2Cr³⁺ + 6Fe²⁺+ 7H₂O

MnO₄¯ + 5e¯ + 8H⁺ ——> Mn²⁺ + 4H₂O (x2)

H₂O₂ ——> O₂+ 2H⁺ + 2e^¯ (x5)

2MnO₄¯ + 5H₂O₂ + 6H⁺ ——> 2Mn²⁺ + 5O₂ + 8H₂O

MnO₄¯ + 5e¯ + 8H⁺ ——> Mn²⁺ + 4H₂O (x2)

C₂O₄^2-——> 2CO₂ + 2e^¯ (x5)

2MnO₄¯ + 5C₂O₄^2-+ 16H⁺ ——> 2Mn²⁺ + 10CO₂ + 8H₂O

2S₂O₃^2-——> S₄O₆^2- + 2e¯ (x1)

½ I₂ + e¯ ——> I^¯ (x2)

2S₂O₃^2-+ I₂——> S₄O₆^2- + 2I^¯