# Write the complete stoichiometric equation for the reaction?

## The reaction between cerium(IV) ions and nitride ions can be expressed in the form of two half-equations: ${\text{Ce"^(4+) + e^- to "Ce}}^{3 +}$ ${\text{NO"_2^(-) + "H"_2"O" to "NO"_3^(-) + 2"H}}^{+} + 2 {e}^{-}$

Apr 17, 2018

$2 {\text{Ce"^(4+)+"NO"_2^(-)+"H"_2 "O" to 2"Ce"^(3+)+"NO"_3^(-)+2"H}}^{+}$

#### Explanation:

Half-reactions:

Reduction: ${\text{Ce"^(4+) + e^- to "Ce}}^{3 +}$
Oxidation: ${\text{NO"_2^(-) + "H"_2"O" to "NO"_3^(-) + 2"H}}^{+} + 2 {e}^{-}$

For each mole of the half-reaction:
the reduction half consumes one mole of electrons, whereas
the oxidation half produces two moles of electrons.

Adding both halves of the reaction shall give the complete equation for the redox reaction; however, since electrons are neither eliminated nor generated in any chemical process, a balanced equation shall contain no electrons.

Multiplying the first half-reaction with a coefficient of two shall eliminate ${e}^{-}$ from the net equation, giving

$\textcolor{b l u e}{2} {\text{Ce"^(4+) + color(green)(cancel[color(blue)(2)color(black)(e^(-))])+"NO"_2^(-) + "H"_2"O" to color(blue)(2)"Ce"^(3+)+"NO"_3^(-) + 2"H}}^{+} + \textcolor{g r e e n}{\cancel{\textcolor{b l a c k}{2} \textcolor{b l a c k}{{e}^{-}}}}$
$2 {\text{Ce"^(4+)+"NO"_2^(-)+"H"_2 "O" to 2"Ce"^(3+)+"NO"_3^(-)+2"H}}^{+}$

Double-check the equation:
Make sure that on the two sides of the equation

• electrons cancel out;
• charges balance;
• number of atoms balance.