Q16E - Chemistry (OpenStax)

Question

Balance the following reactions and write the reactions using cell notation. Ignore any inert electrodes, as they are never part of the half-reactions.

(a) \({\rm{Al}}(s) + {\rm{Z}}{{\rm{r}}^{4 + }}(aq) \to {\rm{A}}{{\rm{l}}^{3 + }}(aq) + {\rm{Zr}}(s)\)

(b) \({\rm{A}}{{\rm{g}}^ + }(aq) + {\rm{NO}}(g) \to {\rm{Ag}}(s) + {\rm{N}}{{\rm{O}}_3}^ - (aq)\)(acidic solution)

(c) \({\rm{Si}}{{\rm{O}}_3}^{2 - }(aq) + {\rm{Mg}}(s) \to {\rm{Si}}(s) + {\rm{Mg}}{({\rm{OH}})_2}(s)\)(basic solution)

(d) \({\rm{Cl}}{{\rm{O}}_3}^ - (aq) + {\rm{Mn}}{{\rm{O}}_2}(s) \to {\rm{C}}{{\rm{l}}^ - }(aq) + {\rm{Mn}}{{\rm{O}}_4}^ - (aq)\)(basic solution

Step-by-Step Solution

Answer

1. The balance reaction is \(4{\rm{Al}}({\rm{s}}) + 3{\rm{Z}}{{\rm{r}}^{4 + }}(aq) \to 4{\rm{A}}{{\rm{l}}^{3 + }}(aq) + 3{\rm{Zr}}({\rm{s}})\)
2. The overall reaction is \(3{\rm{A}}{{\rm{g}}^ + }({\rm{aq}}) + {\rm{NO}}({\rm{g}}) + 2{{\rm{H}}_2}{\rm{O}}({\rm{l}}) \to 3{\rm{Ag}}(s) + {\rm{NO}}_3^ - ({\rm{aq}}) + 4{{\rm{H}}^ + }(aq)\)
3. The overall reaction is \({\rm{Si}}{{\rm{O}}_3}^{2 - }(aq) + 3{{\rm{H}}_2}{\rm{O}}({\rm{l}}) + 2{\rm{Mg}}({\rm{s}}) \to {\rm{Si}}({\rm{s}}) + 2{\rm{Mg}}{({\rm{OH}})_2}(\;{\rm{s}}) + 2{\rm{O}}{{\rm{H}}^ - }\).
4. The overall reaction is\(2{\rm{Mn}}{{\rm{O}}_2}(\;{\rm{s}}) + {\rm{Cl}}{{\rm{O}}_{{3^ - }}}(aq) + 2{\rm{O}}{{\rm{H}}^ - }({\rm{aq}}) \to 2{\rm{MnO}}_4^ - (aq) + {\rm{C}}{{\rm{l}}^ - }({\rm{aq}}) + {{\rm{H}}_2}{\rm{O}}(l)\)

Step1: Define oxidation and reduction

• The oxidation-reduction reaction is also known as a redox reaction. In this reaction, one reactant is oxidized and other is reduced. In balancing an oxidation-reduction reaction, they must be first divided into two half reactions: one is oxidation reaction and other is reduction reaction.
• The balancing of redox reaction is complicated as compared to simple balancing. It is necessary to determine the half reactions of reactants undergoing oxidation and reduction. On adding the two half reactions, net total equation can be obtained. This method of balancing redox reaction is known as half equation method.

Step 2: a) Determine balanced reaction for each pair of half reactions in an acidic solution.

The balance reaction is \(4{\rm{Al}}({\rm{s}}) + 3{\rm{Z}}{{\rm{r}}^{4 + }}(aq) \to 4{\rm{A}}{{\rm{l}}^{3 + }}(aq) + 3{\rm{Zr}}({\rm{s}})\)

Consider the following equations:

Anode (Oxidation): \(4 \times \left( {{\rm{Al}}({\rm{s}}) \to {\rm{A}}{{\rm{l}}^{3 + }}(aq) + 3{e^ - }} \right)\)

Cathode (Reduction) : \(3 \times \left( {{\rm{Z}}{{\rm{r}}^{4 + }}({\rm{aq}}) + 4{e^ - } \to {\rm{Zr}}({\rm{s}})} \right)\)

Overall reaction: \(4{\rm{Al}}({\rm{s}}) + 3{\rm{Z}}{{\rm{r}}^{4 + }}(aq) \to 4{\rm{A}}{{\rm{l}}^{3 + }}(aq) + 3{\rm{Zr}}({\rm{s}})\)

Cell notation: \(Al\left| {A{l^{3 + }}(aq)} \right|Z{r^{4 + }}(aq)\mid Zr\)

Step 3: b) Determine balanced reaction for each pair of half reactions in an acidic solution.

The overall reaction is \(3{\rm{A}}{{\rm{g}}^ + }({\rm{aq}}) + {\rm{NO}}({\rm{g}}) + 2{{\rm{H}}_2}{\rm{O}}({\rm{l}}) \to 3{\rm{Ag}}(s) + {\rm{NO}}_3^ - ({\rm{aq}}) + 4{{\rm{H}}^ + }(aq)\)

Consider the following equations:

Anode (Oxidation): \({\rm{NO}}({\rm{g}}) + 2{{\rm{H}}_2}{\rm{O}}({\rm{l}}) \to {\rm{NO}}_3^ - ({\rm{aq}}) + 4{{\rm{H}}^ + }({\rm{aq}}) + 3{e^ - }\))

Cathode (Reduction) : \(3 \times \left( {{\rm{A}}{{\rm{g}}^ + }(aq) + {e^ - } \to Ag(\;{\rm{s}})} \right)\)

Overall reaction: \(3{\rm{A}}{{\rm{g}}^ + }({\rm{aq}}) + {\rm{NO}}({\rm{g}}) + 2{{\rm{H}}_2}{\rm{O}}\) (l)

\( \to 3{\rm{Ag}}({\rm{s}}) + {\rm{N}}{{\rm{O}}_3}^ - ({\rm{aq}}) + 4{{\rm{H}}^ + }(aq)\)

Cell notation: \({\rm{NO}}\mid {\rm{NO}}_3^ - \)(aq) 

Step 3: c) Determine balanced reaction for each pair of half reactions in an acidic solution.

The overall reaction is \({\rm{Si}}{{\rm{O}}_3}^{2 - }(aq) + 3{{\rm{H}}_2}{\rm{O}}({\rm{l}}) + 2{\rm{Mg}}({\rm{s}}) \to {\rm{Si}}({\rm{s}}) + 2{\rm{Mg}}{({\rm{OH}})_2}(\;{\rm{s}}) + 2{\rm{O}}{{\rm{H}}^ - }\).

Anode (Oxidation): \(2{\rm{Mg}}({\rm{s}}) + 4{\rm{O}}{{\rm{H}}^ - }({\rm{aq}}) \to 2{\rm{Mg}}{({\rm{OH}})_2}(\;{\rm{s}}) + 4{e^ - }\)

Cathode (Reduction) \(:{{\mathop{\rm SiO}\nolimits} _3}^{2 - }(aq) + 3{{\rm{H}}_2}{\rm{O}}({\rm{l}}) + 4{e^ - } \to {\rm{Si}}({\rm{s}}) + 6{\rm{O}}{{\rm{H}}^ - }\)

Overall reaction: \({\rm{Si}}{{\rm{O}}_3}^{2 - }(aq) + 3{{\rm{H}}_2}{\rm{O}}({\rm{l}}) + 2{\rm{Mg}}({\rm{s}}) \to {\rm{Si}}({\rm{s}}) + 2{\rm{Mg}}{({\rm{OH}})_2}(\;{\rm{s}})\)

\( + 2{\rm{O}}{{\rm{H}}^ - }\)

Step 4: d) Determine balanced reaction for each pair of half reactions in an acidic solution.

The overall reaction is\(2{\rm{Mn}}{{\rm{O}}_2}(\;{\rm{s}}) + {\rm{Cl}}{{\rm{O}}_{{3^ - }}}(aq) + 2{\rm{O}}{{\rm{H}}^ - }({\rm{aq}}) \to 2{\rm{MnO}}_4^ - (aq) + {\rm{C}}{{\rm{l}}^ - }({\rm{aq}}) + {{\rm{H}}_2}{\rm{O}}(l)\)

Anode (Oxidation): \(2{\rm{Mn}}{{\rm{O}}_2}(\;{\rm{s}}) + 8{\rm{O}}{{\rm{H}}^ - }({\rm{aq}}) \to 2{\rm{MnO}}_4^ - ({\rm{aq}}) + 4{{\rm{H}}_2}{\rm{O}}({\rm{l}}) + 6{e^ - }\) 

Cathode (Reduction) : \({\rm{ClO}}_3^ - (aq) + 3{{\rm{H}}_2}{\rm{O}}({\rm{l}}) + 6{e^ - } \to {\rm{C}}{{\rm{l}}^ - }({\rm{s}}) + 6{\rm{O}}{{\rm{H}}^ - }(aq)\)

Overall reaction: \(2{\rm{Mn}}{{\rm{O}}_2}(\;{\rm{s}}) + {\rm{Cl}}{{\rm{O}}_3}^ - (aq) + 2{\rm{O}}{{\rm{H}}^ - }({\rm{aq}})\)

\( \to 2{\rm{MnO}}_4^ - (aq) + {\rm{C}}{{\rm{l}}^ - }({\rm{aq}}) + {{\rm{H}}_2}{\rm{O}}(l)\)

Cell notation: (aq)