How do you balance $N O_{3}^{-} + 4 H^{+} + P b \to P b^{2 +} + N O_{2} + 2 H_{2} O$ $NO_3^(-) + 4H^+ + Pb -> Pb^(2+) + NO_2 + 2H_2O$ ?

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Answer 1 · 2017-03-28T16:51:54

$2 N O_{3}^{-} + 4 H^{+} + P b \to P b^{2 +} + 2 N O_{2} + 2 H_{2} O$ $2NO_3^(-)+4H^(+)+Pb→Pb^(2+)+2NO_2+2H_2O$

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 · 2017-03-29T19:48:26

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

Oxidation: lead metal is oxidized to plumbic ion.

$P b (s) \to P b^{2 +} + 2 e^{-}$ $Pb(s) rarr Pb^(2+) + 2e^-$ $(i)$ $(i)$

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

$N O_{3}^{-} + 2 H^{+} + e^{-} \to N O_{2} + H_{2} O$ $NO_3^(-) + 2H^(+) + e^(-) rarr NO_2 + H_2O$ $(i i)$ $(ii)$

CHARGE and MASS are balanced as is absolutely required.

The final redox equation simply eliminates the electrons: we take $(i) + 2 \times (i i)$ $(i) + 2xx(ii)$ 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?

How do you balance NO_3^(-) + 4H^+ + Pb -> Pb^(2+) + NO_2 + 2H_2ONO−3+4H++Pb→Pb2++NO2+2H2ONO_3^(-) + 4H^+ + Pb -> Pb^(2+) + NO_2 + 2H_2O?