# How do you balance NO_3^(-) + 4H^+ + Pb -> Pb^(2+) + NO_2 + 2H_2O?

Mar 28, 2017

2NO_3^(-)+4H^(+)+Pb→Pb^(2+)+2NO_2+2H_2O

#### Explanation:

2 units of nitrate ions balanced with 2 units of nitrogen dioxide will provide 6 units of oxygen on both sides of the reaction.

Mar 29, 2017

This is a redox equation, and we may separate the oxidation and reduction reactions SEPARATELY in order to balance the equation.

#### Explanation:

Oxidation: lead metal is oxidized to plumbic ion.

$P b \left(s\right) \rightarrow P {b}^{2 +} + 2 {e}^{-}$ $\left(i\right)$

Reduction: nitrate, $N \left(V\right)$, is REDUCED to nitrogen dioxide, $N \left(I V\right)$.

$N {O}_{3}^{-} + 2 {H}^{+} + {e}^{-} \rightarrow N {O}_{2} + {H}_{2} O$ $\left(i i\right)$

CHARGE and MASS are balanced as is absolutely required.

The final redox equation simply eliminates the electrons: we take $\left(i\right) + 2 \times \left(i i\right)$ to give:

$P b \left(s\right) + 2 N {O}_{3}^{-} + 4 {H}^{+} + \cancel{2 {e}^{-}} \rightarrow P {b}^{2 +} + \cancel{2 {e}^{-}} + 2 N {O}_{2} + 2 {H}_{2} O$

$P b \left(s\right) + 2 N {O}_{3}^{-} + 4 {H}^{+} \rightarrow P {b}^{2 +} + 2 N {O}_{2} + 2 {H}_{2} O$

Are charge and mass balanced here? How do you know?