Question

In the following reaction, what is reactant being oxidized?

So confused, any help is very much appreciated!

Answer 1

The metal....

Explanation:

Tin metal is oxidized up to stannic ion...`Sn^(4+)`..

`Sn(s) rarr Sn^(4+) + 4e^(-)` `(i)`

Meanwhile the nitrogen in nitrate `N(V+)` is REDUCED to `N(IV+)`...

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

For both half equations, charge and mass are balanced...so we simply add `(i)+4xx(ii):`

`Sn(s)+4NO_3^(-) +8H^+ + cancel(4e^(-)) rarr Sn^(4+) +4NO_2(g) + 4H_2O+cancel(4e^(-))`

to give,..

`Sn(s)+4NO_3^(-) +8H^+ rarr Sn^(4+) +4NO_2(g) + 4H_2O`

Charge and mass are balanced as required....

Alternatively, you could assign oxidation numbers, and proceed from there...