Question

For the following reaction, determine the number of electrons transferred. If the reaction is not a redox reaction, please enter "0" for your answer.3CuS(s)+8HNO3​(aq)3CuSO4​(aq)+8NO(g)+4H2​O(l)

Answer 1

Answer:

$24$.

Explanation:

$3\, {\rm CuS} + 8\, {\rm HNO_{3}} \to 3\, {\rm CuSO_{4}} + 8\, {\rm NO} + 4\, {\rm H_{2}O}$.

Start by finding the oxidation of each species in this reaction.

• The oxidation state of ${\rm O}$ in most compounds is mostly $(-2)$ (exceptions include peroxide and compounds with fluorine.) Thus, the oxidation state of ${\rm N}$ in ${\rm NO}$ would be $0 - (-2) = (+2)$.
• ${\rm S}^{2-}$, ${\rm NO_{3}}^{-}$, and ${\rm SO_{4}}^{2-}$ are polyatomic ions where the oxidation states are well-known. The oxidation state of ${\rm S}$ in ${\rm S^{2-}}$ is $(-2)$, the oxidation state of ${\rm N}$ in ${\rm NO_{3}}^{-}$ is $(-1) - 3 \times (-2) = (+5)$, and the oxidation state of ${\rm S}$ in ${\rm SO_{4}}^{2-}$ is $(-2) - 4 \times (-2) = (+6)$.

$3\, {\rm \stackrel{+2}{Cu}\stackrel{-2}{S}} + 8\, {\rm H\stackrel{+5}{N}O_{3}} \to 3\, {\rm \stackrel{+2}{Cu}\stackrel{+6}{S}O_{4}} + 8\, {\rm \stackrel{+2}{N}O} + 4\, {\rm H_{2}O}$.

Oxidation state changes in this reaction include:

• The oxidation state of three ${\rm S}$ atoms is increased by $8$ each, from $(-2)$ to $(+6)$.
• The oxidation state of eight ${\rm N}$ atoms is reduced by $3$ each, from $(+5)$ to $(+2)$.

Thus, three ${\rm S}$ atoms transferred a total of $3 \times 8 = 24$ electrons to eight ${\rm N}$ atoms in this reaction.