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.

it should be all of them because it asks if you included all of those in your answer

explanation:

when [oh-] is 10-5  m, and [h3o+] is 10-9  m, the poh is 5 and ph is 9. again in both cases the sum of ph and poh is 14. however, the hydrogen ion concentration is not always going to be equal to exactly 1 x 10 raised to a negative number. for example, we skipped over the value of 2.0 x 10-7.