Question

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

Answer 1

`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.

Answer 2

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.

`Pb(s) rarr Pb^(2+) + 2e^-` `(i)`

Reduction: nitrate, `N(V)`, is REDUCED to nitrogen dioxide, `N(IV)`.

`NO_3^(-) + 2H^(+) + e^(-) rarr NO_2 + H_2O` `(ii)`

CHARGE and MASS are balanced as is absolutely required.

The final redox equation simply eliminates the electrons: we take `(i) + 2xx(ii)` to give:

`Pb(s) +2NO_3^(-) + 4H^(+) + cancel(2e^(-))rarr Pb^(2+) + cancel(2e^(-)) +2NO_2 + 2H_2O`

`Pb(s) +2NO_3^(-) + 4H^(+) rarr Pb^(2+) +2NO_2 + 2H_2O`

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