In acidic solution, nitrous oxide (N2O) reacts with potassium chromate (K2CrO4) to form Cr3+ (aq) and nitric oxide (NO), as products.


(i) Produce balanced half-equations to represent the oxidation and the reduction steps.
(ii) Generate a final balanced equation for the overall reaction.

If you want a chemistry tutor to respond, please type your School Subject into the appropriate box.

Balancing redox equations follows a set of rules. Surely you have those rules. If not, here is a site to help you get started. Otherwise, what is it you don't understand about the process.

http://www.chemteam.info/Redox/Redox.html

(i) Oxidation half-equation:

N2O -> NO + 2e-

Reduction half-equation:
CrO42- + 7H2O + 6e- -> Cr3+ + 14OH-

(ii) Overall balanced equation:
6N2O + 2K2CrO4 + 7H2O -> 6NO + 2Cr3+ + 4KOH

To answer these questions, we will go through the steps required to balance the half-equations and the overall equation.

(i) Balancing the half-equations:

1. Start by assigning oxidation numbers to each element in the reactants and products. The oxidation number is a hypothetical charge that an atom would have if all the bonds were completely ionic.

For the reaction between N2O and Cr3+, the oxidation number of N in N2O is +1. In NO, the oxidation number of N is +2.

Since N gains one electron, its oxidation state is reduced from +1 to +2. Therefore, the half-equation for the reduction step is:
N2O + 2e- → 2NO (reduction)

2. Next, consider the oxidation state of Cr in K2CrO4. The oxidation state of Cr is +6, and in Cr3+ it is +3.

Since Cr loses three electrons, its oxidation state is reduced from +6 to +3. Therefore, the half-equation for the oxidation step is:
Cr2O7^2- → 2Cr3+ + 3e- (oxidation)

(ii) Balancing the overall equation:

1. To balance the electrons in the half-equations, multiply the reduction half-equation by the coefficient of 3 and the oxidation half-equation by the coefficient of 2.

The balanced half-equations are:
2N2O + 4e- → 4NO (reduction)
Cr2O7^2- + 6e- → 2Cr3+ + 3O2 (oxidation)

2. In order to balance the number of nitrogen atoms, multiply the reduction half-equation by 2:
4N2O + 8e- → 8NO (reduction)

3. Now, sum up both equations to obtain the overall balanced equation:
4N2O + 8e- + Cr2O7^2- + 6e- → 8NO + 2Cr3+ + 3O2

Simplifying the equation gives the final balanced equation:
4N2O + Cr2O7^2- → 8NO + 2Cr3+ + 3O2

i don't know the answer