Oxidation-Reduction

Dr. Ron Rusay

Balancing Oxidation-Reduction Reactions

Oxidation-Reduction

Oxidation is the loss of electrons.Reduction is the gain of electrons.The reactions occur together. One does

not occur without the other.The terms are used relative to the

change in the oxidation state or oxidation number of the reactant(s).

Aqueous Reactions:Oxidation - Reduction

In the following reaction, identify what is being oxidized and what is being reduced. What is the total number of electrons involved in the process?

Oxidation Reduction Reactions

QUESTIONIn a redox reaction, oxidation and reduction must both occur. Which statement provides an accurate premise of redox chemistry?

A.The substance that is oxidized must be the oxidizing agent.B.The substance that is oxidized must gain electrons.C.The substance that is oxidized must have a higher

oxidation number afterwards.D.The substance that is oxidized must combine with oxygen.

QUESTION

What is the oxidation number of chromium in ammonium dichromate?

A) +3 B) +4 C) +5 D) +6

QUESTION

Zinc

Reactivity Tables (usually reducing) show relative reactivities:In the examples from the previous slide, the acid solution (H+) will react with anything below it in the Table but not above. Nickel and Zinc,

….but not Copper.

Select all redox reactions by looking for a change in oxidation number as reactants are converted to products.

I) Ca + 2 H2O → Ca(OH)2 + H2

II) CaO + H2O → Ca(OH)2

III) Ca(OH)2 + H3PO4 → Ca3(PO4)2 + H2O

IV) Cl2 + 2 KBr → Br2 + 2 KCl

A) I and II B) II and III C) I and IV D) III and IV

QUESTION

QUESTION

Cu 2+ Cu (s)

H2 (g) 2 H +

0

0

+2 e-

- 2 e-

Use oxidation numbers to determine what is oxidized and what is reduced.

Number of electrons gained must equal the number of electrons lost.

- 2 e-

+2 e-

Refer to BalancingOxidation-Reduction Reactions

QUESTION

Balancing Redox Equationsin acidic solutions

1) Determine the oxidation numbers of atoms in both reactants and products.

2) Identify and select out those which change oxidation number (“redox” atoms) into separate “half reactions”.

3) Balance the “redox” atoms and charges (electron gain and loss must equal!).

4) In acidic reactions balance oxygen with water then hydrogen from water with acid proton(s).

Fe+2(aq)+ Cr2O7

2-(aq) +H+

(aq) Fe3+(aq)

+ Cr3+(aq) + H2O(l)

Fe 2+(aq)+ Cr2O7

2-(aq) +H+

(aq)

Fe 3+(aq) + Cr 3+

(aq) + H2O(l)

x = ? Cr ; 2x+7(-2) = -2; x = +6

? Cr oxidation number?

Balancing Redox Equationsin acidic solutions

Fe 2+(aq) Fe 3+

(aq)

Cr2O72-

(aq) + Cr 3+(aq)

Cr = (6+)

6 (Fe 2+(aq) -e -Fe3+

(aq))

6 Fe 2+(aq) 6 Fe3+

(aq) + 6 e -

Cr2O72-

(aq) + 6 e - 2 Cr3+(aq)

6 e -

-e -

2

Balancing Redox Equationsin acidic solutions

6 Fe2+(aq) 6 Fe3+

(aq) + 6 e -

Cr2O72-

(aq) + 6 e - 2 Cr3+(aq)

6 Fe2+(aq)+ Cr2O7

2-(aq) + ? 2nd H+

(aq) 6 Fe3+

(aq) + 2 Cr3+(aq)+ ? 1st Oxygen H2O(l)

Oxygen = 7 2nd (Hydrogen) = 14

Balancing Redox Equationsin acidic solutions

Completely Balanced Equation:

6 Fe2+(aq)+ Cr2O7

2-(aq) + 14 H+

(aq)

6 Fe3+(aq) + 2 Cr3+

(aq)+ 7 H2O(l)

Balancing Redox Equationsin acidic solutions

Dichromate ion in acidic medium converts ethanol, C2H5OH, to CO2 according to the unbalanced equation:

Cr2O72−(aq) + C2H5OH(aq) → Cr3+(aq) + CO2(g) + H2O(l)

The coefficient for H+ in the balanced equation using smallest integer coefficients is:

A) 8 B) 10 C) 13 D) 16 

QUESTION

Balancing Redox Equationsin basic solutions

1) Determine oxidation numbers of atoms in Reactants and Products

2) Identify and select out those which change oxidation number into separate “half reactions”

3) Balance redox atoms and charges (electron gain and loss must equal!)

4) In basic reactions balance the Oxygen with hydroxide then Hydrogen from hydroxide with water

MnO2 (aq)+ ClO31-

(aq) + OH 1-aq)

MnO41-

(aq)+ Cl 1-(aq) + H2O(l)

Mn4+ (MnO2) Mn7+ (MnO4 ) 1-

Cl+5 (ClO3 ) 1-+ 6 e- Cl 1-

Balancing Redox Equationsin basic solutions

Electronically Balanced Equation:

2 MnO2 (aq)+ ClO31-

(aq) + 6 e - 2 MnO4

1- + Cl 1- + 6 e-

Balancing Redox Equationsin basic solutions

Completely Balanced Equation:

2 MnO2 (aq)+ ClO31-

(aq) + 2 OH 1- (aq)

2 MnO4 (aq)1- + Cl 1-

(aq)+ 1 H2O (l)

9 O in product

Balancing Redox Equationsin basic solutions

QUESTIONOxalate ion can be found in rhubarb and spinach (among other green leafy plants). The following unbalanced equation carried out in a basic solution, shows how MnO4

– could be used to analyze samples for oxalate.

MnO4– + C2O4

2– MnO2 + CO32– (basic solution)

When properly balanced, how many OH– are present?

A.1B.2C.3D.4