Question

Identify the oxidant and reductant in the following reaction: $10H^{+}(aq)+4Zn(s)+NO_{3}^{-}(aq) \longrightarrow 4Zn^{2+}(aq)+NH_{4}^{+}(aq)+3H_{2}O(l)$

Answer 1

H+^ is the oxidant

Zn is the reductant

Answer 2

"$4Zn\quad +\quad { NO }_{ 3 }^{ - }\quad +\quad 10{ H }^{ + }\quad \rightarrow \quad { NH }_{ 4 }^{ + }\quad +\quad 4{ Zn }^{ 2+ }\quad +\quad 3{ H }_{ 2 }O$

In the above mentioned reaction, the oxidation part is  

${ Zn }^{ 0 }\quad \rightarrow \quad { Zn }^{ +22+ }\quad +\quad 2{ e }^{ - }$

The zinc is oxidized from 0 to +2 state

The reduction part is  the nitrogen is reduced from +5 to -3 state

${ N }^{ +5 }{ O }_{ 3 }^{ -2 }\quad +\quad 8{ e }^{ - }\quad \rightarrow \quad { N }^{ -3 }{ H }_{ 4 }^{ +1 }$

In the reaction the oxidant is ${ NO }_{ 3 }^{ - }$ and the reductant is Zn"