8.18 Balance the following redox reactions by ion – electron method :

(a) MnO4 (aq) + I (aq) MnO2 (s) + I2(s) (in basic medium)

(b) MnO4 (aq) + SO2 (g) Mn2+ (aq) + HSO4 (aq) (in acidic solution)

(c) H2O2 (aq) + Fe2+ (aq) Fe3+ (aq) + H2O (l) (in acidic solution)

(d) Cr2O7 2– + SO2(g) Cr3+ (aq) + SO42– (aq) (in acidic solution)

(a) Step 1: The two half-reactions involved in the given reaction are:
Oxidation                half  ${\stackrel{-\mathrm{I}}{\mathrm{I}}}_{\left(aq\right)}\to {\stackrel{0}{\mathrm{I}}}_{2\left(\mathrm{s}\right)}$
reaction:                           $\stackrel{+7}{M}n{{{O}_{4}}^{-}}_{\left(aq\right)}\to \stackrel{+4}{M}n{O}_{2\left(aq\right)}$
Reduction half reaction:
Step 2:
Balancing I in the oxidation half reaction, we have:
$2{{I}^{-}}_{\left(aq\right)}\to {I}_{2\left(s\right)}$
Now, to balance the charge, we add 2 ${e}^{-}$ to the RHS of the reaction.
$2{{I}^{-}}_{\left(aq\right)}\to {I}_{2\left(s\right)}+2{e}^{-}$
Step 3:
In the reduction half reaction, the oxidation state of Mn has reduced from +7 to +4.
Thus, 3 electrons are added to the LHS of the reaction.
${{{\mathrm{MnO}}_{4}}^{-}}_{\left(\mathrm{aq}\right)}+3{\mathrm{e}}^{-}\to {{\mathrm{MnO}}_{2}}_{\left(\mathrm{aq}\right)}$
Now, to balance the charge, we add 4 ${\mathrm{OH}}^{-}$ ions to the RHS of the reaction as the reaction is taking place in a basic medium.
${{\mathrm{MnO}}^{-}}_{4\left(\mathrm{aq}\right)}+3{\mathrm{e}}^{-}\to {\mathrm{MnO}}_{2\left(\mathrm{aq}\right)}+4{\mathrm{OH}}^{-}$
Step 4:
In this equation, there are 6 O atoms on the RHS and 4 O atoms on the LHS. Therefore, two water molecules are added to the LHS.
${{{\mathrm{MnO}}_{4}}^{-}}_{\left(\mathrm{aq}\right)}+2{\mathrm{H}}_{2}\mathrm{O}+3{\mathrm{e}}^{-}\to {\mathrm{MnO}}_{2\left(\mathrm{aq}\right)}+4{\mathrm{OH}}^{-}$
Step 5:
Equalizing the number of electrons by multiplying the oxidation half-reaction by 3 and the reduction half-reaction by 2, we have:
$6{{\mathrm{I}}^{-}}_{\left(\mathrm{aq}\right)}\to 3{\mathrm{I}}_{2\left(\mathrm{s}\right)}+6{\mathrm{e}}^{-}\phantom{\rule{0ex}{0ex}}2{{\mathrm{MnO}}^{-}}_{4\left(\mathrm{aq}\right)}+4{\mathrm{H}}_{2}\mathrm{O}+6{\mathrm{e}}^{-}\to 2{\mathrm{MnO}}_{2\left(\mathrm{s}\right)}+8{{\mathrm{OH}}^{-}}_{\left(\mathrm{aq}\right)}$
Step 6:
Adding the two half-reactions, we have the net balanced redox reaction as:
$6{{\mathrm{I}}^{-}}_{\left(\mathrm{aq}\right)}+2{{\mathrm{MnO}}^{-}}_{4\left(\mathrm{aq}\right)}+4{\mathrm{H}}_{2}{\mathrm{O}}_{\left(\mathrm{I}\right)}\to 3{\mathrm{I}}_{2\left(\mathrm{s}\right)}+2{\mathrm{MnO}}_{2\left(\mathrm{s}\right)}+8{{\mathrm{OH}}^{-}}_{\left(\mathrm{aq}\right)}$
(b)Following the steps as in part (a), we have the oxidation half-reaction as:
$S{O}_{2\left(g\right)}+2{H}_{2}{O}_{\left(I\right)}\to HS{{O}^{-}}_{4\left(aq\right)}+3{{H}^{+}}_{\left(aq\right)}+2{{e}^{-}}_{\left(aq\right)}$
And the reduction half-reaction as:
${{{\mathrm{MnO}}_{4}}^{-}}_{\left(\mathrm{aq}\right)}+8{{\mathrm{H}}^{+}}_{\left(\mathrm{aq}\right)}+5{\mathrm{e}}^{-}\to {{\mathrm{Mn}}^{2+}}_{\left(\mathrm{aq}\right)}+4{\mathrm{H}}_{2}{\mathrm{O}}_{\left(\mathrm{I}\right)}$
Multiplying the oxidation half reaction by 5 and the reduction half reaction by 2, and then by adding them, we have the net balanced redox reaction as:
$2{{{\mathrm{MnO}}_{4}}^{-}}_{\left(\mathrm{aq}\right)}+5{\mathrm{SO}}_{2\left(\mathrm{g}\right)}+2{\mathrm{H}}_{2}{\mathrm{O}}_{\left(\mathrm{g}\right)}+2{\mathrm{H}}_{2}{\mathrm{O}}_{\left(\mathrm{I}\right)}+{{\mathrm{H}}^{+}}_{\left(\mathrm{aq}\right)}\to 2{{\mathrm{Mn}}^{2+}}_{\left(\mathrm{aq}\right)}+5{{{\mathrm{HSO}}_{4}}^{-}}_{\left(\mathrm{aq}\right)}$
(c) Following the steps as in part (a), we have the oxidation half reaction as:
${\mathrm{Fe}}_{\left(\mathrm{aq}\right)}^{2+}\to {{\mathrm{Fe}}^{3+}}_{\left(\mathrm{aq}\right)}+{\mathrm{e}}^{-}$
And the reduction half reaction as:
${\mathrm{H}}_{2}{\mathrm{O}}_{2\left(\mathrm{aq}\right)}+2{{\mathrm{H}}^{+}}_{\left(\mathrm{aq}\right)}+2{\mathrm{e}}^{-}\to 2{\mathrm{H}}_{2}{\mathrm{O}}_{\left(\mathrm{I}\right)}$
Multiplying the oxidation half reaction by 2 and then adding it to the reduction half-reaction, we have the net balanced redox reaction as:
${\mathrm{H}}_{2}{\mathrm{O}}_{2\left(\mathrm{aq}\right)}+2{{\mathrm{Fe}}^{2+}}_{\left(\mathrm{aq}\right)}+2{{\mathrm{H}}^{+}}_{\left(\mathrm{aq}\right)}\to 2{{\mathrm{Fe}}^{3+}}_{\left(\mathrm{aq}\right)}+2{\mathrm{H}}_{2}{\mathrm{O}}_{\left(\mathrm{I}\right)}$
(d) Following the steps as in part (a), we have the oxidation half reaction as:
${\mathrm{SO}}_{2\left(\mathrm{g}\right)}+2{\mathrm{H}}_{2}{\mathrm{O}}_{\left(\mathrm{l}\right)}\to {{\mathrm{SO}}_{4}^{2-}}_{\left(\mathrm{aq}\right)}+4{{\mathrm{H}}^{+}}_{\left(\mathrm{aq}\right)}+2{\mathrm{e}}^{-}$
And the reduction half reaction as:
${\mathrm{Cr}}_{2}{{\mathrm{O}}_{7}^{2-}}_{\left(\mathrm{aq}\right)}+14{{\mathrm{H}}^{+}}_{\left(\mathrm{aq}\right)}+6{\mathrm{e}}^{-}\to 2{{\mathrm{Cr}}^{3+}}_{\left(\mathrm{aq}\right)}+7{\mathrm{H}}_{2}{\mathrm{O}}_{\left(\mathrm{l}\right)}$
Multiplying the oxidation half reaction by 3 and then adding it to the reduction half reaction, we have the net balanced redox reaction as:
${\mathrm{Cr}}_{2}{{\mathrm{O}}_{7}^{2-}}_{\left(\mathrm{aq}\right)}+3{\mathrm{SO}}_{2\left(\mathrm{g}\right)}+2{{\mathrm{H}}^{+}}_{\left(\mathrm{aq}\right)}\to 2{{\mathrm{Cr}}^{3+}}_{\left(\mathrm{aq}\right)}+3{{\mathrm{SO}}_{4}^{2-}}_{\left(\mathrm{aq}\right)}+{\mathrm{H}}_{2}{\mathrm{O}}_{\left(\mathrm{l}\right)}$