Question

When the equation Sn + HNO₃ → SnO₂ + NO₂ + H₂O is balanced in acidic solution, what is the smallest whole-number coefficient for Sn?

Answer 1

1.

Step-by-step explanation:

Hello,

In this case, for the given reaction we first assign the oxidation state for each species:

`Sn^0 + H^+N^(5+)O^(-2)_3 \rightarrow Sn^(4+)O_2 + N^(4+)O^(2-)_2 + H^+_2O^-`

Whereas the half reactions are:

`Sn^0+2H_2O \rightarrow Sn^(4+)O_2 +4H^++4e^-\\\\H^++H^+N^(5+)O^(-2)_3 +1e^-\rightarrow N^(4+)O^(2-)_2+H_2O`

Next, we exchange the transferred electrons:

`1*(Sn^0+2H_2O \rightarrow Sn^(4+)O_2 +4H^++4e^-)\\\\4* (H^++H^+N^(5+)O^(-2)_3 +1e^-\rightarrow N^(4+)O^(2-)_2+H_2O)\\\\\\Sn^0+2H_2O \rightarrow Sn^(4+)O_2 +4H^++4e^-\\\\4H^++4H^+N^(5+)O^(-2)_3 +4e^-\rightarrow 4N^(4+)O^(2-)_2+4H_2O`

Afterwards, we add them to obtain:

`Sn^0+2H_2O+4H^++4H^+N^(5+)O^(-2)_3 \rightarrow Sn^(4+)O_2 +4H^++4N^(4+)O^(2-)_2+4H_2O`

By adding and subtracting common terms we obtain:

`Sn^0+4H^+N^(5+)O^(-2)_3 \rightarrow Sn^(4+)O_2 +4N^(4+)O^(2-)_2+2H_2O`

Finally, by removing the oxidation states we have:

`Sn + 4HNO_3 \rightarrow SnO_2 + 4NO_2 + 2H_2O`

Therefore, the smallest whole-number coefficient for Sn is 1.

Regards.