Question

Complete and balance the reaction in acidic solution. equation: ZnS + NO_{3}^{-} -> Zn^{2+} + S + NO ZnS+NO−3⟶Zn2++S+NO Which element is oxidized? S N Zn Which element is reduced? S N Zn Which species is the oxidizing agent? ZnS NO−3 Zn2+ Which species is the reducing agent? ZnS NO−3 Zn2+

Answer 1

Answer:

Balanced reaction: $3ZnS+2NO_{3}^{-}+8H^{+}\rightarrow 3Zn^{2+}+3S+2NO+4H_{2}O$

S is oxidized and N is reduced.

$NO_{3}^{-}$ is the oxidizing agent and ZnS is the reducing agent.

Explanation:

Reaction: $ZnS+NO_{3}^{-}\rightarrow Zn^{2+}+S+NO$

$\Rightarrow$ Oxidation: $ZnS\rightarrow Zn^{2+}+S$

Balance charge: $ZnS-2e^{-}\rightarrow Zn^{2+}+S$  ...............(1)

$\Rightarrow$ Reduction: $NO_{3}^{-}\rightarrow NO$

Balance H and O in acidic medium : $NO_{3}^{-}+4H^{+}\rightarrow NO+2H_{2}O$

Balance charge: $NO_{3}^{-}+4H^{+}+3e^{-}\rightarrow NO+2H_{2}O$ ...............(2)

[$3\times$Equation-(1)] + [ $2\times$Equation-(2)]:

$3ZnS+2NO_{3}^{-}+8H^{+}\rightarrow 3Zn^{2+}+3S+2NO+4H_{2}O$

Oxidation number of S increases from (-2) to (0) for the conversion of ZnS to S. Therefore S is oxidized.

Oxidation number of N decreases from (+5) to (+2) for the conversion of $NO_{3}^{-}$ to NO. Therefore N is reduced.

$NO_{3}^{-}$ consumes electron from ZnS. Therefore $NO_{3}^{-}$ is the oxidizing agent and ZnS is the reducing agent.