Does hydrogen undergoes reduction or oxidation on the basis of oxidation number in 2NH3 +H2SO4 which gives (NH4)SO4
neither. this is an acid/base neutralization reaction, and none of the species changes oxidation numbers. N is -3 on the left, and N is -3 on the right. S, O remain the same, as does H.
2NH3 + H2SO4 >> (NH4)2SO4
In order to determine whether hydrogen undergoes reduction or oxidation in the given reaction, we need to calculate the oxidation numbers of hydrogen in the reactants and products. Oxidation number is a measure of the number of electrons transferred between atoms in a compound or a reaction.
Let's start by assigning oxidation numbers to each element:
In ammonia (NH3), the oxidation number of hydrogen is +1. The oxidation number of nitrogen is -3 since it is in group 15 of the periodic table. The sum of the oxidation numbers in ammonia is 0 (since it is a neutral molecule).
In sulfuric acid (H2SO4), the oxidation number of hydrogen is again +1. The oxidation number of sulfur can be determined by the overall charge of the compound, which is 0. To find the oxidation number of sulfur, we can set up an equation:
(2 x +1) + x + (4 x -2) = 0 (where x is the oxidation number of sulfur).
Simplifying the equation, we get:
2 + x - 8 = 0
x - 6 = 0
x = +6
Therefore, the oxidation number of sulfur in H2SO4 is +6.
Now, let's examine the product, ammonium sulfate ((NH4)SO4). The oxidation numbers of nitrogen and sulfur in (NH4)SO4 can be determined in a similar way as explained above.
In ammonium ((NH4)+), the oxidation number of hydrogen is +1. The oxidation number of nitrogen can be calculated by setting up an equation:
(4 x +1) + x = +1 (since the overall charge of ammonium is +1)
Simplifying the equation, we get:
4 + x = 1
x = -3
Therefore, the oxidation number of nitrogen in ammonium is -3.
In the sulfate (SO42-), the oxidation number of sulfur can be determined by setting up an equation:
x + (4 x -2) = -2 (since the overall charge of sulfate is -2)
Simplifying the equation, we get:
x - 8 = -2
x = +6
Therefore, the oxidation number of sulfur in sulfate is +6.
Now, comparing the oxidation numbers of hydrogen in the reactants and products, we can see that the oxidation number of hydrogen remains the same (+1) throughout the reaction. Since there is no change in oxidation number, we can conclude that hydrogen neither undergoes oxidation nor reduction in this reaction.
To summarize, in the reaction 2NH3 + H2SO4 → (NH4)2SO4, hydrogen doesn't undergo any oxidation or reduction based on its oxidation number.