# How do we represent the oxidation of "bismuth metal" by the action of "nitric acid" to give "bismuth nitrate", and NO(g)?

Aug 25, 2017

#### Answer:

You are oxidizing $\text{bismuth metal}$ to $\text{bismuth(III) nitrate}$.......

#### Explanation:

Bismuth metal is oxidized to $B {i}^{3 +}$:

$B i \rightarrow B {i}^{3 +} + 3 {e}^{-}$ $\left(i\right)$

Nitrate, $\stackrel{+ V}{N} {O}_{3}^{-}$, is reduced to $\stackrel{+ I I}{N} O$:

$N {O}_{3}^{-} + 4 {H}^{+} + 3 {e}^{-} \rightarrow N O + 2 {H}_{2} O$ $\left(i i\right)$

Charge and mass are balanced in both equations, and we take $\left(i\right) + \left(i i\right)$

$B i + N {O}_{3}^{-} + 4 {H}^{+} \rightarrow B {i}^{3 +} + N O + 2 {H}_{2} O$

Is this balanced?

Else....we add $3 \times N {O}_{3}^{-}$ to each side to get......

$B i + 4 H N {O}_{3} \rightarrow 3 B i {\left(N {O}_{3}\right)}_{3} + N O \uparrow + 2 {H}_{2} O$