Question

Balance the following redox reaction by the ion electron method.
a) ${\text{Mn}}{{\text{O}}^{\text{ - }}}_{_{{\text{4(aq)}}}}{\text{ + }}{{\text{I}}^{\text{ - }}}_{{\text{(aq)}}} \to {\text{Mn}}{{\text{O}}_{{\text{2(s)}}}}{\text{ + }}{{\text{I}}_{{\text{2(s)}}}}$ (in basic medium)
b) ${\text{Mn}}{{\text{O}}^{\text{ - }}}_{{\text{4(aq)}}}{\text{ + S}}{{\text{O}}_{{\text{2(g)}}}} \to {\text{M}}{{\text{n}}^{{\text{ + 2}}}}_{{\text{(aq)}}}{\text{ + HS}}{{\text{O}}^{\text{ - }}}_{\text{4}}$ (in acidic medium)
c) ${{\text{H}}_{\text{2}}}{{\text{O}}_{{\text{2(aq)}}}}{\text{ + F}}{{\text{e}}^{{\text{2 + }}}}_{{\text{(aq)}}} \to {\text{F}}{{\text{e}}^{{\text{3 + }}}}_{{\text{(aq)}}}{\text{ + }}{{\text{H}}_{\text{2}}}{{\text{O}}_{{\text{(l)}}}}$ (in acidic medium)
d) ${\text{C}}{{\text{r}}_{\text{2}}}{{\text{O}}_{\text{7}}}{^{{\text{2 - }}}_{{\text{(aq)}}}}{\text{ + S}}{{\text{O}}_{{\text{2(g)}}}} \to {\text{C}}{{\text{r}}^{{\text{3 + }}}}_{{\text{(aq)}}}{\text{ + S}}{{\text{O}}_{\text{4}}}{^{{\text{2 - }}}_{{\text{(aq)}}}}$ (in acidic medium)
e) ${\text{F}}{{\text{e}}^{{\text{ + 2}}}}_{{\text{(aq)}}}{\text{ + C}}{{\text{r}}_{\text{2}}}{{\text{O}}_{\text{7}}}{^{{\text{ - 2}}}_{{\text{(aq)}}}} \to {\text{F}}{{\text{e}}^{{\text{ + 3}}}}_{{\text{(aq)}}}{\text{ + C}}{{\text{r}}^{{\text{ + 3}}}}_{{\text{(aq)}}}$

Answer 1

Hint: The ion electron method is also called the half-reaction method. In this method, the redox reaction is separated into two half-equations, one for oxidation and the other for reduction. Then each of these half reactions is balanced separately. Then they are combined to give the balanced redox equation.

Complete step by step answer:
In the first option a) ${\text{Mn}}{{\text{O}}^{\text{ - }}}_{_{{\text{4(aq)}}}}{\text{ + }}{{\text{I}}^{\text{ - }}}_{{\text{(aq)}}} \to {\text{Mn}}{{\text{O}}_{{\text{2(s)}}}}{\text{ + }}{{\text{I}}_{{\text{2(s)}}}}$(in basic medium),
Step 1: The two half reactions are
Oxidation half reaction: ${{\text{I}}^{\text{ - }}}_{{\text{(aq)}}} \to {{\text{I}}_{\text{2}}}{\text{(s)}}$
Reduction half reaction: ${\text{Mn}}{{\text{O}}^{\text{ - }}}_{{\text{4(aq)}}} \to {\text{Mn}}{{\text{O}}_{{\text{2(aq)}}}}$
Step 2: Balancing ${\text{I}}$ in the oxidation half reaction, we have
${\text{2}}{{\text{I}}^{\text{ - }}}_{{\text{(aq)}}} \to {{\text{I}}_{{\text{2(s)}}}}$
Now, to balance the charge, we add ${\text{2}}{{\text{e}}^{\text{ - }}}$to the RHS of the reaction.
${\text{2}}{{\text{I}}^{\text{ - }}}_{{\text{(aq)}}} \to {{\text{I}}_{{\text{2(s)}}}}{\text{ + 2}}{{\text{e}}^{\text{ - }}}$
Step 3: In the reduction half reaction, the oxidation state of ${\text{Mn}}$ has reduced from ${\text{ + 7 to + 4}}$. Thus, ${\text{3}}{{\text{e}}^{\text{ - }}}$ are added to the LHS of the reaction.
${\text{Mn}}{{\text{O}}^{\text{ - }}}_{{\text{4(aq)}}}{\text{ + 3}}{{\text{e}}^{\text{ - }}} \to {\text{Mn}}{{\text{O}}_{{\text{2(aq)}}}}$
Now to balance the charge, we add ${\text{4 O}}{{\text{H}}^{\text{ - }}}$ ions on the RHS of the reaction as the reaction is taking place in basic medium.
${\text{Mn}}{{\text{O}}^{\text{ - }}}_{{\text{4(aq)}}}{\text{ + 3}}{{\text{e}}^{\text{ - }}} \to {\text{Mn}}{{\text{O}}_{{\text{2(aq)}}}}{\text{ + 4O}}{{\text{H}}^{\text{ - }}}$
Step 4: In this equation, there are ${\text{6 O}}$ atoms on the RHS and ${\text{4 O}}$ atoms on the LHS. Therefore, two water molecules are added to the LHS.
${\text{Mn}}{{\text{O}}^{\text{ - }}}_{{\text{4(aq)}}}{\text{ + 2}}{{\text{H}}_{\text{2}}}{\text{O + 3}}{{\text{e}}^{\text{ - }}} \to {\text{Mn}}{{\text{O}}_{{\text{2(aq)}}}}{\text{ + 4O}}{{\text{H}}^{\text{ - }}}$
Step 5: Equalising the number of electrons by multiplying the oxidation half reaction by ${\text{3}}$and the reduction half reaction by ${\text{2}}$, we have
${\text{6}}{{\text{I}}^{\text{ - }}}_{{\text{(aq)}}} \to {\text{3}}{{\text{I}}_{{\text{2(s)}}}}{\text{ + 2}}{{\text{e}}^{\text{ - }}}$
${\text{2Mn}}{{\text{O}}^{\text{ - }}}_{{\text{4(aq)}}}{\text{ + 4}}{{\text{H}}_{\text{2}}}{{\text{O}}_{{\text{(l)}}}}{\text{ + 6}}{{\text{e}}^{\text{ - }}} \to {\text{2Mn}}{{\text{O}}_{{\text{2(s)}}}}{\text{ + 8O}}{{\text{H}}^{\text{ - }}}_{{\text{(aq)}}}$
Step 6: Add the two half reactions and we get the net balanced redox reaction as:
${\text{6}}{{\text{I}}_{{\text{(aq)}}}}{\text{ + 2Mn}}{{\text{O}}^{\text{ - }}}_{{\text{4(aq)}}}{\text{ + 4}}{{\text{H}}_{\text{2}}}{{\text{O}}_{{\text{(l)}}}} \to {\text{3}}{{\text{I}}_{{\text{2(s)}}}}{\text{ + 2Mn}}{{\text{O}}_{{\text{2(s)}}}}{\text{ + 8O}}{{\text{H}}^{\text{ - }}}_{{\text{(aq)}}}$
Second option b)
Following the steps as in option a), we have the oxidation half reaction as ${\text{S}}{{\text{O}}_{{\text{2(g)}}}}{\text{ + 2}}{{\text{H}}_{\text{2}}}{{\text{O}}_{{\text{(l)}}}} \to {\text{HS}}{{\text{O}}^{\text{ - }}}_{{\text{4(aq)}}}{\text{ + 3}}{{\text{H}}^{\text{ + }}}_{{\text{(aq)}}}{\text{ + 2}}{{\text{e}}^{\text{ - }}}_{{\text{(aq)}}}$
And the reduction half reaction as ${\text{Mn}}{{\text{O}}^{\text{ - }}}_{{\text{4(aq)}}}{\text{ + 8}}{{\text{H}}^{\text{ + }}}_{{\text{(aq)}}}{\text{ + 5}}{{\text{e}}^{\text{ - }}} \to {\text{M}}{{\text{n}}^{{\text{2 + }}}}_{{\text{(aq)}}}{\text{ + 4}}{{\text{H}}_{\text{2}}}{{\text{O}}_{{\text{(l)}}}}$
Multiplying the oxidation half reaction by ${\text{5}}$ and the reduction half reaction by ${\text{2}}$ and then add them to get the net balanced equation
${\text{2Mn}}{{\text{O}}^{\text{ - }}}_{{\text{4(aq)}}}{\text{ + 5S}}{{\text{O}}_{{\text{2(g)}}}}{\text{ + 2}}{{\text{H}}_{\text{2}}}{{\text{O}}_{{\text{(l)}}}}{\text{ + }}{{\text{H}}^{\text{ + }}}_{{\text{(aq)}}} \to {\text{M}}{{\text{n}}^{{\text{2 + }}}}_{{\text{(aq)}}}{\text{ + HS}}{{\text{O}}^{\text{ - }}}_{{\text{4(aq)}}}$
Option c) Following the steps as in option a, we get the oxidation reaction as ${\text{F}}{{\text{e}}^{{\text{2 + }}}}_{{\text{(aq)}}} \to {\text{F}}{{\text{e}}^{{\text{3 + }}}}_{{\text{(aq)}}}{\text{ + }}{{\text{e}}^{\text{ - }}}$
And the reduction half reaction as ${{\text{H}}_{\text{2}}}{{\text{O}}_{{\text{2(aq)}}}}{\text{ + 2}}{{\text{H}}^{\text{ + }}}_{{\text{(aq)}}}{\text{ + 2}}{{\text{e}}^{\text{ - }}} \to {\text{2}}{{\text{H}}_{\text{2}}}{{\text{O}}_{{\text{(l)}}}}$
Multiply the oxidation half reaction by ${\text{2}}$ and then add it to the reduction half reaction.
We get the net balanced redox reaction as ${{\text{H}}_{\text{2}}}{{\text{O}}_{{\text{2(aq)}}}}{\text{ + 2F}}{{\text{e}}^{{\text{2 + }}}}_{{\text{(aq)}}}{\text{ + 2}}{{\text{H}}^{\text{ + }}}_{{\text{(aq)}}} \to {\text{2F}}{{\text{e}}^{{\text{3 + }}}}_{{\text{(aq)}}}{\text{ + 2}}{{\text{H}}_{\text{2}}}{{\text{O}}_{{\text{(l)}}}}$
Option d) Following the same steps as in option a, oxidation half reaction after balancing:${\text{S}}{{\text{O}}_{{\text{2(g)}}}}{\text{ + 2}}{{\text{H}}_{\text{2}}}{{\text{O}}_{{\text{(l)}}}} \to {\text{S}}{{\text{O}}^{{\text{2 - }}}}_{{\text{4(aq)}}}{\text{ + 4}}{{\text{H}}^{\text{ + }}}_{{\text{(aq)}}}{\text{ + 2}}{{\text{e}}^{\text{ - }}}$
Reduction half reaction after balancing:${\text{C}}{{\text{r}}_{\text{2}}}{{\text{O}}^{{\text{2 - }}}}_{{\text{7(aq)}}}{\text{ + 14}}{{\text{H}}^{\text{ + }}}_{{\text{(aq)}}}{\text{ + 6}}{{\text{e}}^{\text{ - }}} \to {\text{2C}}{{\text{r}}^{{\text{3 + }}}}_{{\text{(aq)}}}{\text{ + 3S}}{{\text{O}}^{{\text{2 - }}}}_{{\text{4(aq)}}}{\text{ + }}{{\text{H}}_{\text{2}}}{{\text{O}}_{{\text{(l)}}}}$
Multiply oxidation half reaction by 3 and then add it to the reduction half reaction.
Net balanced redox reaction is ${\text{C}}{{\text{r}}_{\text{2}}}{{\text{O}}^{{\text{2 - }}}}_{{\text{7(aq)}}}{\text{ + 3S}}{{\text{O}}_{{\text{2(g)}}}}{\text{ + 2}}{{\text{H}}^{\text{ + }}}_{{\text{(aq)}}} \to {\text{2C}}{{\text{r}}^{{\text{3 + }}}}_{{\text{(aq)}}}{\text{ + 3S}}{{\text{O}}^{{\text{2 - }}}}_{{\text{4(aq)}}}{\text{ + }}{{\text{H}}_{\text{2}}}{{\text{O}}_{{\text{(l)}}}}$
Option e) Oxidation half reaction after balancing: ${\text{F}}{{\text{e}}^{{\text{2 + }}}} \to {\text{F}}{{\text{e}}^{{\text{3 + }}}}{\text{ + 1}}{{\text{e}}^{\text{ - }}}$
Reduction half reaction after balancing: ${\text{C}}{{\text{r}}_{\text{2}}}{{\text{O}}^{{\text{2 - }}}}_{\text{7}}{\text{ + 14}}{{\text{H}}^{\text{ + }}}{\text{ + 6}}{{\text{e}}^{\text{ - }}} \to {\text{2C}}{{\text{r}}^{{\text{3 + }}}}{\text{ + 7}}{{\text{H}}_{\text{2}}}{\text{O}}$
Multiply the oxidation half reaction by 6 and then add both the half reactions, we get the net balanced redox reaction as ${\text{6F}}{{\text{e}}^{{\text{2 + }}}}{\text{ + C}}{{\text{r}}_{\text{2}}}{{\text{O}}^{{\text{2 - }}}}_{\text{7}}{\text{ + 14}}{{\text{H}}^{\text{ + }}} \to {\text{6F}}{{\text{e}}^{{\text{3 + }}}}{\text{ + 2C}}{{\text{r}}^{{\text{3 + }}}}{\text{ + 7}}{{\text{H}}_{\text{2}}}{\text{O}}$

Note: The species which oxidizes a substance and gets reduced is called an oxidizing agent. It gains electrons. The species which reduces a substance and gets oxidized is called a reducing agent, it loses electrons.
In oxidation reaction, there will be loss of electrons and in reduction reaction, there will be gain of electrons.
Or we can also say that in the oxidation reaction, there’s an increase in the oxidation state and reduction reaction means there’s a decrease in the oxidation state.