Question

Balance the following redox equations by half reaction method:

(i) Cr₂O₇^2- + Fe²⁺ → Cr³⁺ + H₂O in acidic medium

(ii) Cr(OH)₃ + IO₃^- → I^- + CrO₄^2- in basic medium

Answer 1

(i) (Cr₂O₇)^2- + Fe²⁺ → Cr³⁺ + Fe³⁺ + H₂O

Step 1. Indication of O.N. of each atom

\(\overset{+6}{C}r_2\overset{-2}{O^{-2}}+Fe^{2+}\rightarrow Cr^{3+}+Fe^{3+}+\overset{+1}{H}_2\overset{-2}{O}\) 

Thus, Cr in Cr₂O₇^2- and Fe change their oxidation numbers.

Step 2. Writing the oxidation and reduction half reactions.

\(\overset{+6}{Cr}_2\overset{-2}{O^{2-}} \rightarrow 2Cr^{3+}\)  (Reduction half reaction)

\(Fe^{2+}\rightarrow Fe^{3+}\) (Oxidation half reaction)

Step 3. Addition of e^– to make up the difference in O.N.

\(\overset{+6}{C}_2\overset{2-}{O_7}+6e^-\rightarrow2Cr^{3+}\)

(Each Cr atom gains 3e^– . Thus, 2 Cr atom will gain 6e^–)

\(Fe^{2+}\rightarrow Fe^{3+}+6^-\)

Step 4. Balance ‘O’ by adding equal number of H₂O molecules to the side which is deficient in O atoms.

\(Cr_2O_7^{2-}+6e^- \rightarrow2Cr^{3+}+7H_2O\)

\(Fe^{2+}\rightarrow Fe^{3+}+e^-\)

Step 5. Balance H by adding H⁺ ions to the side which is deficient in H atoms.

\(Cr_2O_7^{2-}+6e^-+14H^+ \rightarrow2Cr^{3+}+7H_2O\)

\(Fe^{2+}\rightarrow Fe^{3+}+e^-\)

Step 6. Multiplying oxidation half reaction by 6 to equalize the electrons lost and gained and add the two.

\(Cr_2O_7^{2-}+6e^-+14H^+ \rightarrow2Cr^{3+}+7H_2O\)

[Fe²⁺ → Fe³⁺ e^-] x 6

(ii) Ce(OH)₃ + IO₃^- → I^- + CrO₄^2-

Step 1. Indication of oxidation number of each element.

Thus, we find that Cr in Cr(OH)₃ and iodine in IO₃^- undergo change in oxidation number.

Step 2. Writing oxidation and reduction half reactions.

\(\overset{+3}{Cr}(OH)_3\rightarrow \overset{+6}{Cr}O_4^{2-}\) (Oxidation half reaction)

\(\overset{+5}{I}O_3\rightarrow I^-\) (Reduction half reaction)

Step 3. Addition of e^- to make up the difference in O.N.

\(\overset{+3}{Cr}(OH)_3\rightarrow \overset{+6}{Cr}O_4^{2-} +3e^-\)

\(\overset{+5}{IO_3^-}+6e^-\rightarrow I^-\)

Step 4. Balance O atoms by adding H2O molecules to the side deficient in ‘O’ atoms.

Cr(OH)₃ + H₂O → CrO₄^2- + 3e^-

IO₃ + 6e^- → I^- + 3H₂O

Step 5. Balance H atoms. Since the medium is basic, therefore add proper number of H₂O molecules to the falling short of H atoms and equal number of OH^- ions to the other side.

Cr(OH)₃ + H₂O + 5OH^- → CrO₄^2- + 3e^- + 5H₂O

IO₃^- + 6e^- + 6H₂O → I^- + 3H₂O + 6OH^-

Step 6. Equalize the electrons lost and gained by multiplying the electron and gained by multiplying the oxidation half reaction with 2.