# Oxidation and Reduction Equation

Example 1. Balance the following equations: $(i) MnO^-_4 + Fe^{2+} \to Mn^{2+} + Fe^{3+} \\ (ii) Cr_2O^{2-}_7 + I^- + H^+ \to Cr^3 + I_2 + H_2O \\ (iii) KMnO_4 + KCl + H_2SO_4 \to K_2SO_4 +MnSO_4 + H_2O + Cl_2$

Solution:

(i) Oxidising agent is $MnO^-_4$ $\underset{+7}{MnO^-_4} \to \underset{+2}{Mn^{2+}}$ gain of electron = 5

Reducing agent is $Fe^{2+}$ $\underset{+2}{Fe^{2+}} \to Fe^{3+} \text{loss of electrons} = 1 \\ MnO_4 + 5Fe^{2+} \to 5Fe^{3+} + Mn^{2+}$

Now add on right hand side to balance oxygen atoms and ions on left hand side to balance hydrogen atoms, we get $MnO^-_4 + 5Fe^{2+} + 8H^+ \to 5Fe^{3+} + Mn^{2+} + 4H_2O$ $(ii) Cr_2O^{2-}_7 + I^- + H^+ \to Cr^{3+} + I_2 + H_2O \\[3mm] \text{Oxidising agent is} Cr_2O^{2-}_7 \\[3mm] Cr_2O^{2-}_7 \to 2Cr^{3+}- 6e^-$……..(I)

Reducing agent is $I^-$ $2I^- \to I_2 + 2e^-$……..(II)

Multiply eq. (II) by 3 and add in eq. (I) $Cr_2O_7^{2-} + 6I^- \to 2Cr^{3+} +3I_2$ ………(III)

Now balance the eq. (III) for oxygen by adding $7H_2O$on right hand side and then add ion on left hand side. We get $Cr_2O^{2-}_7 + 6I^- \to 2 Cr^{3+} + 3I_2 + 7H_2O$ $(iii) KMnO_4 + KCL + H_2SO_4 \to K_2SO_4 + MnSO_4 + H_2O + Cl_2$

Oxidising agent is $KMnO_4$ $KMnO_4 \to MnSO_4- 5e^-$ ……..(I)

Reducing agent is KCl $2KCl \to Cl_2 + 2e^-$ …………(II)

Multiply eq. (I) by 2 and eq. (II) by 5 and add: $2KMnO_4 + 10KCl \to 2MnSO_4 + 5 Cl_2$

Adjusting for $H_2O, H_2SO_4 \text{and} K_2SO_4 + 6K_2SO_4 + 5Cl_2 + 8H_2SO$

Example 2. Balance the following equations: $Hg^{2+} + Sn^{2+} \to Hg^+ + Sn^{4+}$

Solution: Two half reactions may be:

Oxidation: $Sn^{2+} \to Sn^{4+} + 2e^-$ ……..(i)

Reduction: $Hg^{2+} + e^- \to Hg^+$ ………….(ii)

Multiply eq. (ii) by 2 and on adding in eq. (i) we get: $Sn^{2+} + 2Hg^{2+} \to Sn^{4+} + 2Hg^+$

Example 3. Balance the following equations:

(i) $Cu_2O + H^+ + 2NO^-_3 \to Cu^{2+} + NO + H_2O$

(ii) $K_4Fe(CN)_6 + H_2S0_4 + H_2O \to K_2SO_4 + FeSO_4 + (NH_4)_2SO_4 + CO$

(iii) $C_2H_5OH+ I_2 + OH^- \to CHI_3 + HCO^-_2 + I^- + H_2O$

Solution:

(i) $Cu_2O + H^+ + 2NO^-_3 \to Cu^{2+} + NO + H_2O$ $(a) Cu_2O \to 2Cu^{2+} + 2e^- \hspace{10mm} \text{(Oxidation)} \\[3mm] (b) NO^-_3 + 3e^- \to NO \hspace{10mm} \text{(Reduction)}$

Multiplying eq. (a) by 3 and (b) by 2, and add $3Cu_2O + 2NO^-_3 \to 6Cu^{2+} + NO$

Charge      -2           +12

Now add $14H^+$ on left side to equalize the charge and $7H_2O$ on right side. We get $2Cu_2O + 14H^+ + 2NO^-_3 \to 6Cu^{2+} + 2NO + 7H_2O$

(ii) $K_4Fe(CN)_6 + H_2SO_4 + H_2O \to K_2SO_4 + FeSO_4 + (NH_4)_2SO_4 + CO$

In this case C of $CN^-$ ion is reduced to CO(change in Ox. No. +4 to +2) and N of $CN^-$ ion is oxidised to $NH^+_4$ ion  (change in Ox. N0. -5 to -3). Since both oxidation and reduction take place in the same compound hence it is written two times as: $K_4Fe(CN)_6 \to 6NH^+_4 \cdots(I) \\[3mm] \text{or} \hspace{10mm} K_4Fe(CN)_6 + 24H^+ \to 6NH^+_4 \\[3mm] K_4Fe(CN)_6 \to 6CO \\[3mm] \text{or} \hspace{10mm} K_4Fe(CN)_6 + 6H_2O \to 6CO + 12H^+ \cdots(II)$

On adding eq. (I) and (II), we get $2K_4[Fe(CN)_6] + 12H^+ + 6H_2O \to 6NH^+_4 + 6CO$

Or, $\hspace{10mm} K_4[Fe(CN)_6] + 6H^+ + 3H_2O \to 3NH^+_4 + 3CO$

Now balancing C and N atoms, $K_4[Fe(CN)_6] + 12H^+ + 6H_2O \to 6NH^+_4 + 6CO$

Now we can write, $K_4[Fe(CN)_6] + 6H_2SO_4 + H_2O \to 2K_2SO_4 + FeSO_4 + 3(NH_4)_2SO_4 + 6CO$

(iii) $C_2H_5OH+ I_2 + OH^- \to CHI_3 + HCO^-_2 + I^- + H_2O$

In this case $I_2$ is reduced to $I^-$, therefore $I_2 \to 2I^- 2e^-$ ……..(I)

In $C_2H_5OH$ carbon atom is oxidised to $CHI_3$ and $HCOO^-$ (Ox. No. from -2 to +2) $C_2H_5OH \to CHI_3 + HCOO^- + 8e^-$ ……..(II)

Now multiply eq. (I) by 4 and add in eq. (II) $C_2H_5OH + 4I_2 \to CHI_3 + HCOO^- + 8I^-$

Since three iodine atoms are in $CHI_3$ molecule hence on adjusting these atoms. $C_2H_5OH + 4I_2 \to CHI_3 + HCOO^- + 5I^-$

Since this reaction takes place in alkaline medium hence on balancing charge $6OH^-$ are added on left side then balance oxygen atoms by adding water molecules, we get $C_2H_5OH + 4I_2 + 6OH^- \to CHI_3 + HCOO^- + 5I^- + 5H_2O$

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