Redox Reactions – Full Chapter Notes

Redox Reactions (Chapter 7)

Introduction: Chemistry is basically the study of change. “Redox” is a major category where Oxidation and Reduction occur simultaneously.

Ek haath se dena, doosre haath se lena!

1. Classical Idea (Old Definitions)

A. Oxidation

Addition of Oxygen: $2Mg + O_2 \rightarrow 2MgO$
Addition of Electronegative element: $Mg + F_2 \rightarrow MgF_2$
Removal of Hydrogen: $2H_2S + O_2 \rightarrow 2S + 2H_2O$
Removal of Electropositive element: $2K_4[Fe(CN)_6] + H_2O_2 \rightarrow 2K_3[Fe(CN)_6] + …$

B. Reduction

Removal of Oxygen: $2HgO \xrightarrow{\Delta} 2Hg + O_2$
Removal of Electronegative element: $2FeCl_3 + H_2 \rightarrow 2FeCl_2 + 2HCl$
Addition of Hydrogen: $CH_2=CH_2 + H_2 \rightarrow CH_3-CH_3$
Addition of Electropositive element: $2HgCl_2 + SnCl_2 \rightarrow Hg_2Cl_2 + SnCl_4$

REDOX = REDuction + OXidation

2. Electron Transfer Concept

This is the modern definition. Modern zamana, electrons ka aana jana!

Oxidation: Loss of electron(s) by any species.
$Na \rightarrow Na^+ + e^-$
Reduction: Gain of electron(s) by any species.
$Cl_2 + 2e^- \rightarrow 2Cl^-$

Oxidizing Agent (Oxidant)
Accepts electrons.
(Gets reduced itself)

Reducing Agent (Reductant)
Donates electrons.
(Gets oxidized itself)

Competitive Electron Transfer

Example: Metallic Zinc in Copper Nitrate solution.

$Zn(s) + Cu^{2+}(aq) \rightarrow Zn^{2+}(aq) + Cu(s)$

Blue colour fades because $Cu^{2+}$ is gone!

Conclusion: Zn releases electrons to Cu.
Order of tendency to lose electrons: $Zn > Cu > Ag$

3. Oxidation Number (O.N.)

The residual charge an atom has if all shared electrons belong to the more electronegative atom.

The Rules (Rat lo!)

Free State: Elements in uncombined state have O.N. = 0.
(e.g., $H_2, O_2, Cl_2, O_3, P_4, S_8, Na, Mg \rightarrow 0$)
Monatomic Ions: O.N. = Charge on ion.
($Na^+ = +1$, $Mg^{2+} = +2$, $Cl^- = -1$)
Oxygen: Generally -2.
- Exception 1 (Peroxides): $H_2O_2, Na_2O_2 \rightarrow$ -1
- Exception 2 (Superoxides): $KO_2 \rightarrow$ -½
- Exception 3 (with Fluorine): $OF_2 \rightarrow$ +2 (Fluorine is boss!)
Hydrogen: Generally +1.
- Exception (Metal Hydrides): $LiH, NaH \rightarrow$ -1 (Metals donate e-)
Halogens: F is always -1. Others (Cl, Br, I) usually -1 but positive with O or F.
Sum Rule:
- Neutral Molecule: Sum = 0
- Polyatomic Ion: Sum = Charge on ion

Example Calculation: $K_2Cr_2O_7$
Let Cr be $x$.
$2(+1) + 2(x) + 7(-2) = 0$
$2 + 2x – 14 = 0$
$2x = 12 \Rightarrow x = +6$ Easy peasy!

4. Stock Notation

Representing Oxidation state of metal using Roman Numerals in parenthesis.

$AuCl \rightarrow Au(I)Cl$ (Aurous)
$AuCl_3 \rightarrow Au(III)Cl_3$ (Auric)
$SnCl_2 \rightarrow Sn(II)Cl_2$
$MnO_2 \rightarrow Mn(IV)O_2$

5. Redox in terms of O.N.

Oxidation: Increase in Oxidation Number.
Reduction: Decrease in Oxidation Number.
Oxidizing Agent: Increases O.N. of others (its own decreases).
Reducing Agent: Decreases O.N. of others (its own increases).

Problem 7.4 (NCERT): Justify that reaction is redox:
$2Cu_2O(s) + Cu_2S(s) \rightarrow 6Cu(s) + SO_2(g)$

Solution:
Check O.N.:
– In $Cu_2O$, Cu is +1.
– In $Cu_2S$, S is -2 (Cu is +1).
– In $Cu$, O.N. is 0 (Element).
– In $SO_2$, S is +4.

Cu (+1) $\rightarrow$ Cu (0) [Decrease $\rightarrow$ Reduction]
S (-2) $\rightarrow$ S (+4) [Increase $\rightarrow$ Oxidation]
Therefore, it is REDOX.

Sulphur ne jump maara -2 se +4!

6. Types of Redox Reactions

(i) Combination Reactions

$A + B \rightarrow C$ (One or both A/B must be elemental)

$C(s) + O_2(g) \rightarrow CO_2(g)$
$3Mg(s) + N_2(g) \rightarrow Mg_3N_2(s)$
$CH_4 + 2O_2 \rightarrow CO_2 + 2H_2O$ (Combustion is always redox!)

(ii) Decomposition Reactions

Breakdown into components. (Opposite of combination)

$2H_2O \xrightarrow{\Delta} 2H_2 + O_2$
$2KClO_3 \rightarrow 2KCl + 3O_2$ (O changes from -2 to 0, Cl from +5 to -1)

CAUTION!
Not all decomposition is redox.
$CaCO_3 \rightarrow CaO + CO_2$
(Ca is +2, C is +4, O is -2 everywhere. No change. NOT REDOX)

(iii) Displacement Reactions

$X + YZ \rightarrow XZ + Y$ (X replaces Y)

(a) Metal Displacement:

$CuSO_4 + Zn \rightarrow ZnSO_4 + Cu$
$V_2O_5 + 5Ca \rightarrow 2V + 5CaO$ (Metallurgy application)
$TiCl_4 + 2Mg \rightarrow Ti + 2MgCl_2$

(b) Non-Metal Displacement: (Usually H displacement)

Alkali metals + Cold water $\rightarrow H_2$ release
$2Na + 2H_2O \rightarrow 2NaOH + H_2$
Active metals + Acids $\rightarrow H_2$
$Zn + 2HCl \rightarrow ZnCl_2 + H_2$

Halogen Displacement:
$F_2 > Cl_2 > Br_2 > I_2$ (Oxidizing Power)
– $F_2$ can displace $Cl^-, Br^-, I^-$
– $Cl_2$ can displace $Br^-, I^-$ (e.g., $Cl_2 + 2KI \rightarrow 2KCl + I_2$)
– $Cl_2$ CANNOT displace $F^-$ (Fluorine sabka baap hai)

Layer Test uses this property (Violet color of Iodine in CCl4)

(iv) Disproportionation Reactions

A special redox where the SAME element is oxidized AND reduced simultaneously.

Condition: Element must exist in at least 3 oxidation states. The reactant must be in an intermediate state.

Peroxide decomposition:
$2H_2O_2^{-1} \rightarrow 2H_2O^{-2} + O_2^0$
Phosphorus in alkali:
$P_4^0 + 3OH^- + 3H_2O \rightarrow PH_3^{-3} + 3H_2PO_2^{+1}$
Chlorine + Alkali:
$Cl_2 + 2OH^- \rightarrow ClO^- + Cl^- + H_2O$ (Bleach formation)

7. Paradox of Fractional O.N.

Sometimes O.N. comes in fraction (e.g., $C_3O_2$ is 4/3). This is an average.

Carbon Suboxide ($C_3O_2$):
$O = C = C = C = O$
Structure shows ends are +2, middle is 0.
Avg = $(2+0+2)/3 = 4/3$.

Tetrathionate ($S_4O_6^{2-}$):
$O_3S – S – S – SO_3$
Ends are +5, Middles are 0.
Avg = $(5+0+0+5)/4 = 2.5$.

8. Balancing Redox Reactions

Method A: Oxidation Number Method

Steps:

Write correct formula.
Identify atoms whose O.N. changes.
Calculate Increase/Decrease per atom. Multiply to make them equal.
Balance charge:
- Acidic Medium: Add $H^+$
- Basic Medium: Add $OH^-$
Balance H atoms by adding $H_2O$.

Example: $Cr_2O_7^{2-} + SO_3^{2-} \rightarrow Cr^{3+} + SO_4^{2-}$ (Acidic)
1. Cr: +6 $\rightarrow$ +3 (Change = 3 per atom, total 6 for $Cr_2$)
2. S: +4 $\rightarrow$ +6 (Change = 2)
3. Multiply S part by 3 to match Cr change (6 vs 2 $\rightarrow$ need $3 \times 2 = 6$).
4. $Cr_2O_7^{2-} + 3SO_3^{2-} \rightarrow 2Cr^{3+} + 3SO_4^{2-}$
5. Charge Left: -2 + 3(-2) = -8. Charge Right: 2(+3) + 3(-2) = 0.
6. Add $8H^+$ to left.
7. Final: $Cr_2O_7^{2-} + 3SO_3^{2-} + 8H^+ \rightarrow 2Cr^{3+} + 3SO_4^{2-} + 4H_2O$

Method B: Half-Reaction (Ion-Electron) Method

Yeh method zyada reliable hai exams ke liye!

Separate into Oxidation Half and Reduction Half.
Balance atoms other than O and H.
Balance O by adding $H_2O$.
Balance H by adding $H^+$ (acidic) or then $OH^-$ trick (basic).
Balance Charge by adding electrons ($e^-$).
Equalize electrons in both halves and ADD.

Example: $MnO_4^- + I^- \rightarrow MnO_2 + I_2$ (Basic)

Oxidation Half: $2I^- \rightarrow I_2 + 2e^-$

Reduction Half:
$MnO_4^- \rightarrow MnO_2$
$MnO_4^- + 4H^+ + 3e^- \rightarrow MnO_2 + 2H_2O$ (Charge balance)
(Convert to basic: add $4OH^-$ to both sides)
$MnO_4^- + 2H_2O + 3e^- \rightarrow MnO_2 + 4OH^-$

Combine: Multiply Ox by 3, Red by 2 to cancel $e^-$.
$6I^- + 2MnO_4^- + 4H_2O \rightarrow 3I_2 + 2MnO_2 + 8OH^-$

Redox Titrations

KMnO4: Self-indicator (Purple to Colorless/Pink).
K2Cr2O7: Needs indicator (Diphenylamine).
Iodometry: Uses Starch (Blue color with $I_2$).

9. Electrode Processes

Redox Couple: Together oxidized and reduced forms ($Zn^{2+}/Zn$).

The Daniell Cell (Galvanic)

Anode (Left): Oxidation ($Zn \rightarrow Zn^{2+} + 2e^-$) – Negative Pole
Cathode (Right): Reduction ($Cu^{2+} + 2e^- \rightarrow Cu$) – Positive Pole
Salt Bridge: Maintains electrical neutrality (KCl / agar-agar).

LOAN
Left Oxidation Anode Negative

Electrochemical Series

Arrangement based on Standard Electrode Potential ($E^\theta$).

High Negative Value $\rightarrow$ Strong Reducing Agent (Li is best).
High Positive Value $\rightarrow$ Strong Oxidizing Agent ($F_2$ is best).
$H^+/H_2$ is Reference ($0.00 V$).

Feasibility: Reaction works if $E_{cell}$ is positive.

$E_{cell} = E_{cathode} – E_{anode}$

~ THE END ~
(All the best for Exams!)