Question

N2 and SO4²- reacting to produce NO2 and sulfate SO3²- ions in acidic solution.

Answer 1

Answer:- $N_2+4SO_4^-^2\rightarrow 2NO_2+4SO_3^-^2$

Solution:- The given reaction is:

$N_2+SO_4^-^2\rightarrow NO_2+SO_3^-^2$

This reaction is taking place in acidic medium and in acidic medium we use $H^+$ to balance hydrogen.

First of all we write two half equations, one for oxidation and another for reduction.

For this reaction, the oxidation number of N is changing from 0 to +4 which is oxidation. Note that in elemental form the oxidation number is zero and the oxidation number of O in its compounds is -2. This is how nitrogen is zero on reactant side and +4 on product side.

Oxidation number of S is changing from +6 to +4. the calculations for oxidation numbers are easy. We know that oxidation number of O in it's compounds is -2. The sum of oxidation numbers of an ion equals to the charge on it.

So, if we do the calculations for sulfate ion then:

$x+4(-2)=-2$

here we have multiplied the oxidation number of O(-2) by 4 as there are four oxygens in sulfate ion.

$x-8=-2$

$x=6$

The oxidation number of S for sulfite ion can also be calculated in the same way.

So, the two half equations are:

$N_2\rightarrow NO_2$

and

$SO_4^-^2\rightarrow SO_3^-^2$

Let's balance these one by one.

For the first half equation we multiply right side by 2 to balance nitrogen. Then there becomes four oxygen and so to balance oxygen we add four water molecules to the reactant side. On ding this, the reactant side has eight hydrogens and so to balance hydrogen we add eight $H^+$ .

$N_2+4H_2O\rightarrow 2NO_2+8H^+$

Charge is balanced by adding electrons. This equation has +8 charge on product side and zero charge on reactant side so we need to add 8 electrons to the right side.

$N_2+4H_2O\rightarrow 2NO_2+8H^++8e^-$

Let's balance the second half equation in the same way. For balancing oxygen we add one water molecule to the right side and then to balance hydrogen we need to add two $H^+$ to the left side.

$2H^++SO_4^-^2\rightarrow SO_3^-^2+H_2O$

For balancing the charge, we need to add two electrons to the left side.

$2e^-+2H^++SO_4^-^2\rightarrow SO_3^-^2+H_2O$

Next step is to makes the electrons equal and for this we need to multiply the second half equation by 4.

$8e^-+8H^++4SO_4^-^2\rightarrow 4SO_3^-^2+4H_2O$

Add the two half equations:

$N_2+4H_2O+8H^++4SO_4^-^2\rightarrow 2NO_2+8H^++4SO_3^-^2+4H_2O$

In above equation, water and hydrogen ion are common and also they are equal on both sides so they are canceled and the final equation we get is:

$N_2+4SO_4^-^2\rightarrow 2NO_2+4SO_3^-^2$