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_2Oon 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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