Question

the presence of which reactant is the best indicator of an oxidation-reduction reaction?

kno₃

h₂so₄

naoh

h₂

Explanation:

1. Recall the definition of an oxidation - reduction (redox) reaction: A redox reaction involves the transfer of electrons, which is accompanied by a change in oxidation states of the reactants.
2. Analyze each option:

• For $\ce{KNO_{3}}$: In $\ce{KNO_{3}}$, the oxidation state of $\ce{K}$ is + 1, $\ce{N}$ is + 5, and $\ce{O}$ is - 2. Reactions involving $\ce{KNO_{3}}$ may or may not be redox reactions. For example, in a double - displacement reaction like $\ce{KNO_{3}+NaCl = KCl + NaNO_{3}}$, there is no change in oxidation states, so it is not a redox reaction.
• For $\ce{H_{2}SO_{4}}$: In $\ce{H_{2}SO_{4}}$, the oxidation state of $\ce{H}$ is + 1, $\ce{S}$ is + 6, and $\ce{O}$ is - 2. Reactions like $\ce{H_{2}SO_{4}+2NaOH=Na_{2}SO_{4}+2H_{2}O}$ (neutralization) are not redox reactions as there is no change in oxidation states.
• For $\ce{NaOH}$: In $\ce{NaOH}$, the oxidation state of $\ce{Na}$ is + 1, $\ce{O}$ is - 2, and $\ce{H}$ is + 1. Reactions involving $\ce{NaOH}$ in acid - base reactions (like the one with $\ce{H_{2}SO_{4}}$ above) are not redox reactions.
• For $\ce{H_{2}}$: In $\ce{H_{2}}$, the oxidation state of $\ce{H}$ is 0. When $\ce{H_{2}}$ participates in a reaction, for example, $\ce{2H_{2}+O_{2}\xlongequal{点燃}2H_{2}O}$, the oxidation state of $\ce{H}$ changes from 0 in $\ce{H_{2}}$ to + 1 in $\ce{H_{2}O}$, and the oxidation state of $\ce{O}$ changes from 0 in $\ce{O_{2}}$ to - 2[SSE Completed, Client Connection Error][SSE Completed, Client Connection Error][LLM SSE On Failure]

Answer:

D. $\ce{H_{2}}$