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Question: What is the balanced redox reaction between lead and ammonia? \[PbO\left( s \right)+N{{H}_{3}}\lef...

What is the balanced redox reaction between lead and ammonia?
PbO(s)+NH3(g)N2(g)+H2O(I)+Pb(s)PbO\left( s \right)+N{{H}_{3}}\left( g \right)\to {{N}_{2}}\left( g \right)+{{H}_{2}}O\left( I \right)+Pb\left( s \right)
Which element is being oxidized and which is being reduced?

Explanation

Solution

The balance equation is
3PbO+2NH3N2+3H2O+3Pb3PbO+2N{{H}_{3}}\to N_{2}^{{}}+3{{H}_{2}}O+3Pb
From the reaction, we can identify that Lead oxide is reduced to lead and ammonia is being oxidized to nitrogen
Oxidation and reduction reaction occurs
Identification of loss or gain of electrons helps to detect which element is being oxidized or reduced

Complete answer:
PbO on the reactant side has been reduced to PbPb product side
PbO(+2)+2H++2ePb(0)+H2OPbO\left( +2 \right)+2{{H}^{+}}+2{{e}^{-}}\to Pb\left( 0 \right)+{{H}_{2}}Owhich is marked as equation 1
The oxidation number of Pb is (+2)(+2) on the reactant side which is reduced to (0)(0)on the product side.
The oxidation number is decreasing. Therefore, Pb element is reduced
Similarly, NH3 on the reactant side has been oxidized to N2{{N}_{2}}
2NH3(3)N2(0)+6H++6e2N{{H}_{3}}\left( -3 \right)\to {{N}_{2}}\left( 0 \right)+6{{H}^{+}}+6{{e}^{-}}which is marked as equation 2
The oxidation number of N is (3)(-3) on the reactant side which is oxidized to (0)(0) NH3N{{H}_{3}} (0)(0)on the product side.
The oxidation number is increasing. Therefore, NH3N{{H}_{3}} element is oxidized
From the following steps, it is clear that PbPbelement is reduced and NH3N{{H}_{3}} element is oxidized
Balanced equation:
Writing down the equations
equation 1
PbO(+2)+2H++2ePb(0)+H2OPbO\left( +2 \right)+2{{H}^{+}}+2{{e}^{-}}\to Pb\left( 0 \right)+{{H}_{2}}O
Multiply equation 1 by 33 we get
3PbO(+2)+6H++6e3Pb(0)+3H2O3PbO\left( +2 \right)+6{{H}^{+}}+6{{e}^{-}}\to 3Pb\left( 0 \right)+3{{H}_{2}}O which is equation 11*
equation 22
2NH3(3)N2(0)+6H++6e2N{{H}_{3}}\left( -3 \right)\to {{N}_{2}}\left( 0 \right)+6{{H}^{+}}+6{{e}^{-}}
By adding 11* and 22 we get,
3PbO+2NH3+6H++6eN2+6H++6e+3Pb+3H2O3PbO+2N{{H}_{3}}+6{{H}^{+}}+6{{e}^{-}}\to {{N}_{2}}+6{{H}^{+}}+6{{e}^{-}}+3Pb+3{{H}_{2}}O
Solving on both sides, we get the final equation
3PbO+2NH3N2+3H2O+3Pb3PbO+2N{{H}_{3}}\to N_{2}^{{}}+3{{H}_{2}}O+3Pb
Hence derived, the final equation by the oxidation and reduction reactions.

Note:
If an oxidation number decreases in a reaction, it is reduced.
Similarly, if the oxidation number increases in a reaction, it is oxidized.
Here, Lead oxide is reduced to lead and ammonia is being oxidized to nitrogen
Calculating reduction reaction of lead, oxidation reaction of ammonia and by solving those reaction we can get the final required equation