
Complete the following equation \[Pb{\text{ }} + {\text{ }}NaOH{\text{ }} \to \]
A. \[Pb{\text{ }} + {\text{ }}2NaOH{\text{ }} \to N{a_2}Pb{O_2}{\text{ }} + {\text{ }}{H_2}\]
B. \[Pb{\text{ }} + {\text{ }}NaOH{\text{ }} \to 2N{a_2}Pb{O_2}{\text{ }} + {\text{ }}{H_2}\]
C. \[Pb{\text{ }} + {\text{ }}NaOH{\text{ }} \to N{a_2}Pb{O_2}{\text{ }} + {\text{ }}{H_2}\]
D. \[Pb{\text{ }} + {\text{ 2}}NaOH{\text{ }} \to 2N{a_2}Pb{O_2}{\text{ }} + {\text{ }}{H_2}\]
Answer
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Hint: Many metals such as copper, zinc, aluminum, tin, lead, and beryllium form amphoteric in nature, that is they can react, with both acid and base. Ongoing down the group of the periodic table, the atomic radius increases so ionization energy decreases, which means the tendency of losing electrons increases which implies an increase in metallic character. The oxides and hydroxides of metals are basic in nature and thus basic character increases down the group.
Complete step by step answer:
When lead reacts with sodium hydroxide oxidation and reduction reaction happens. Where lead gets oxidized and hydrogen gets reduced. In this case, lead acts as a reducing agent and NaOH acts as an oxidizing agent.
The reactions are shown below.
oxidation reaction,
\[P{b^0}{\text{ }} \to P{b^{ + 2}} + 2{e^ - }\]
Reduction reaction
\[2{H^ + }{\text{ + }}2{e^ - } \to {H_2}\]
Here the reaction shown above is just to explain the change of oxidation state. This is not the actual reaction.
The actual reaction is as follows,
\[Pb{\text{ }} + {\text{ }}NaOH{\text{ }} \to 2N{a_2}Pb{O_2}{\text{ }} + {\text{ }}{H_2}\]
Therefore, the correct option is B.
Additional information:
Aluminum reacts with sodium hydroxide solution to produce hydrogen gas.
Further, the equation is,
\[{\text{2Al}} + 2NaOH{\text{ + 2}}{{\text{H}}_{\text{2}}}{\text{O}} \to 2NaAl{O_2} + {\text{3}}{H_2}\]
Copper is an amphoteric metal, this also reacts with NaOH.
The equations are as shown below,
\[Cu(s) + 2NaOH(aq) \to Cu{\left( {OH} \right)_2}(aq) + \,2Na(s)\]
Note:
Amphoteric metals form amphoteric oxides, amphoteric metals react with both acid and base. In general, the electropositive character of the oxide’s central atom will determine whether the oxide will be acidic or basic. The more electropositive the central atom, the more basic the oxide. The more electronegative the central atom, the more acidic the oxide.
Complete step by step answer:
When lead reacts with sodium hydroxide oxidation and reduction reaction happens. Where lead gets oxidized and hydrogen gets reduced. In this case, lead acts as a reducing agent and NaOH acts as an oxidizing agent.
The reactions are shown below.
oxidation reaction,
\[P{b^0}{\text{ }} \to P{b^{ + 2}} + 2{e^ - }\]
Reduction reaction
\[2{H^ + }{\text{ + }}2{e^ - } \to {H_2}\]
Here the reaction shown above is just to explain the change of oxidation state. This is not the actual reaction.
The actual reaction is as follows,
\[Pb{\text{ }} + {\text{ }}NaOH{\text{ }} \to 2N{a_2}Pb{O_2}{\text{ }} + {\text{ }}{H_2}\]
Therefore, the correct option is B.
Additional information:
Aluminum reacts with sodium hydroxide solution to produce hydrogen gas.
Further, the equation is,
\[{\text{2Al}} + 2NaOH{\text{ + 2}}{{\text{H}}_{\text{2}}}{\text{O}} \to 2NaAl{O_2} + {\text{3}}{H_2}\]
Copper is an amphoteric metal, this also reacts with NaOH.
The equations are as shown below,
\[Cu(s) + 2NaOH(aq) \to Cu{\left( {OH} \right)_2}(aq) + \,2Na(s)\]
Note:
Amphoteric metals form amphoteric oxides, amphoteric metals react with both acid and base. In general, the electropositive character of the oxide’s central atom will determine whether the oxide will be acidic or basic. The more electropositive the central atom, the more basic the oxide. The more electronegative the central atom, the more acidic the oxide.
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