Question

when the following equation for a reaction,which occurs in acidic solution, is correctly balanced using the smallest possible whole numbers, the coefficient of H+ is:
Sn + No2-_________ Sno2 + no

Answer 1

Answer:

Sn + HNO3 = SnO2 + NO2 + H2O

BALANCED:

Sn + 4HNO3 → SnO2 + 4NO2 + 2H2O

Tin + Nitric Acid = Tin(IV) Oxide + Nitrogen Dioxide + Water

Answer 2

In the balanced reaction using the smallest possible whole numbers, the coefficient of H+ is 4.

This reaction is a redox reaction.

Let's balance it by solving the half-reactions:

Tin (Sn) oxidized to Stannic oxide :

$Sn(s)+2H_2O(l)\longrightarrow SnO_2(s)+4H^{+}+4e^{-}$

Now, Nitrite ion is reduced to Nitrite oxide :

$NO_2^{-} + 2H^{+} +e^{-} \longrightarrow NO(g)+H_2O(l)$

After sum up two half-reactions if we multiply 4 to the latter half-reaction the full reaction will be:

$Sn(s)+4NO_2^{-}+2H_2O(l)+8H^{+}+4e^{-}\longrightarrow SnO_2(s)+4NO(g)+4H_2O+4H^{+}+4e^{-}$$\Longrightarrow Sn(s)+4NO_2^{-}+4H^{+}\longrightarrow SnO_2(s)+4NO(g)+2H_2O$

So, in the balanced reaction using the smallest possible whole numbers, the coefficient of H+ is 4.

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