Balance the following skeleton reaction and identify the oxidizing and reducing agents: CrO42- (aq) + N2O(g)+ H+(aq) + OH-(aq) + H2O(l) ⟶ Cr3+(aq) + NO(g) + H+(aq) + OH-(aq) + H2O(l)

oxidizing agent is (enter just the formula of the species, e. g. CrO42-,N2O, Cr3+,NO, H2O, H+, OH-)
reducing agent is (enter just the formula of the species, e. g. CrO42-,N2O, Cr3+,NO, H2O, H+, OH-)

The balanced net ionic equation is as follows:

CrO₄²- (aq) + 3 N₂O (g) + 2 H+ (aq) > Cr³+ (aq) + 6 NO (g) + H₂O (l)

The oxidizing agent is CrO₄²- as Cr⁶+ in CrO₄²- is reduced to Cr³+.

The reducing agent is N₂O as the nitrogen (i) is oxidized to nitrogen (ii).

Explanation:

The skeleton equation is given as follows:

CrO₄²- (aq) + N₂O(g)+ H+(aq) + OH-(aq) + H2O(l) > Cr³+(aq) + NO(g) + H+(aq) + OH-(aq) + H₂O(l)

The balanced net ionic equation with the spectator ions removed is as follows:

CrO₄²- (aq) + 3 N₂O (g) + 2 H+ (aq) > Cr³+ (aq) + 6 NO (g) + H₂O (l)

In a redox reaction, oxidizing agents are reduced while reducing agents are oxidized.

The oxidizing agent is CrO₄²- as Cr⁶+ in CrO₄²- is reduced to Cr³+.

One mole of CrO₄²- accepts three moles of electrons from three moles of N₂O to become reduced to 1 mole of Cr³+.

The reducing agent is N₂O as the nitrogen (i) is oxidized to nitrogen (ii).

Three moles of N₂O will each donate one mole of electrons to one mole of CrO₄²- to become oxidized to nitrogen (ii) oxide.

Hydrogen ions and a water molecule is added to the left-hand side and right-hand side of the equation respectively, in order to balance the number of oxygen atoms in the equation of the reaction.

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