Problem

Complete and balance the following acid-base equa…

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Problem 20 Medium Difficulty

Classify the following as acid-base reactions or oxidation-reduction reactions:
(a) $\mathrm{Na}_{2} \mathrm{S}(a q)+2 \mathrm{HCl}(a q) \longrightarrow 2 \mathrm{NaCl}(a q)+\mathrm{H}_{2} \mathrm{S}(g)$
(b) $\mathrm{PCl}_{3}(l)+\mathrm{Cl}_{2}(g) \longrightarrow \mathrm{PCl}_{5}(s)$
(c) $\mathrm{C}_{2} \mathrm{H}_{4}(g)+3 \mathrm{O}_{2}(g) \longrightarrow 2 \mathrm{CO}_{2}(g)+2 \mathrm{H}_{2} \mathrm{O}(g)$
(d) $\mathrm{Zn}(s)+\mathrm{H}_{2} \mathrm{SO}_{4}(a q) \longrightarrow \mathrm{ZnSO}_{4}(a q)+\mathrm{H}_{2}(g)$
(e) $2 \mathrm{K}_{2} \mathrm{S}_{2} \mathrm{O}_{3}(s)+\mathrm{I}_{2}(s) \longrightarrow \mathrm{K}_{2} \mathrm{S}_{4} \mathrm{O}_{6}(s)+2 \mathrm{KI}(s)$
(f) $3 \mathrm{Cu}(s)+8 \mathrm{HNO}_{3}(a q) \longrightarrow 3 \mathrm{Cu}\left(\mathrm{NO}_{3}\right)_{2}(a q)+2 \mathrm{NO}(g)+4 \mathrm{H}_{2} \mathrm{O}(l)$

Answer

(a) atom oxidized: $\operatorname{Mg}(0 \rightarrow+2)$
atom reduced: $\mathrm{Ni}(+2 \rightarrow 0)$
oxidizing agent (species reduced): $\mathrm{NiCl}_{2}(a q)$
reducing agent (species oxidized) : $\operatorname{Mg}(s)$.
(b) atom oxidized: $\mathrm{P}(+3 \rightarrow+5)$
atom reduced: $\mathrm{Cl}(0 \rightarrow-1)$
oxidizing agent (species reduced): $\mathrm{Cl}_{2}(g)$
reducing agent (species oxidized) : $\mathrm{PCl}_{3}(l)$
(c) atom oxidized: $\mathrm{C}(+2 \rightarrow+4)$
atom reduced: $\mathrm{O}(0 \rightarrow-2)$
oxidizing agent (species reduced): $\mathrm{O}_{2}(g)$
reducing agent (species oxidized): $\mathrm{C}_{2} \mathrm{H}_{4}(g)$
(d) atom oxidized: $\mathrm{Zn}(0 \rightarrow+2)$
atom reduced: $\mathrm{H}(+1 \rightarrow 0)$
oxidizing agent (species reduced): $\mathrm{H}_{2} \mathrm{SO}_{4}(a q)$
reducing agent (species oxidized): $\mathrm{Zn}(s)$
(e) atom oxidized: $\mathrm{S}(2 \rightarrow 2.5)$
atom reduced: I $(0 \rightarrow-1)$
oxidizing agent (species reduced): $\mathrm{I}_{2}(s)$
reducing agent (species oxidized) : $\mathrm{K}_{2} \mathrm{S}_{2} \mathrm{O}_{3}(s)$
(f) atom oxidized: $\mathrm{Cu}(0 \rightarrow+2)$
atom reduced: $\mathrm{N}(+5 \rightarrow+2)$
oxidizing agent (species reduced): HNO $_{3}(a q)$
reducing agent (species oxidized): $\mathrm{Cu}(s)$

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Video Transcript

for this question, you're asked to identify the reaction is an acid base reaction or an oxidation reduction reaction. Often you will have a tendency to identify an acid base reaction if you see an acid present as a reactant or a base present as a reactant. In addition, you may want to classify it as an acid base reaction if water is a product, but that's not always the case, it would only be an acid base reaction. If a hydrogen is transferred or a couple hydrogen czar transferred and there is no change in oxidation state. So for this first one, we do have a common acid hcl reacting with any to S to produce sodium chloride and hydrogen sulfide. This is an acid base reaction because there is no change in oxidation state of sodium, it's plus one, sulfur stays as minus two hydrogen stays as plus one chlorine states minus one. And we just have these hydrogen is being transferred to the sulfide. So it's an acid base reaction. In this case there's no acid or base that involved. We see a change in oxidation state of chlorine going from zero two minus one. So this would be an oxidation reduction reaction. For the next one, we have a combustion reaction, carbon hydrogen containing compound reacting with oxygen producing carbon dioxide and water. So water is a product, but it's not an acid base reaction. All combustion reactions are oxidation reduction reactions, all of them. Then for D, we have zinc with an oxidation state of zero reacting with an acid, sulfuric acid producing zinc sulfate and hydrogen gas. So we see hydrogen going from an oxidation state of plus 1 to 0 zinc going from zero to plus two. Because there is a change in oxidation state. This is an oxidation reduction reaction. For e. We have this potassium containing compound reacting with iodine producing these two products here, we do see a change in the oxidation state of iodine. It's zero here, It's minus one here. So definitely an oxidation reduction reaction. And for the last one f we have copper reacting with nitric acid producing copper to nitrate, nitrogen, monoxide and water. So again, you might be inclined to call this an oxygen or a acid base reaction. We have a common acid, water is a product, but there is a change in oxidation state. Copper goes from zero two plus two. So if you can identify anything having to change in oxidation state, then it's an oxidation reduction reaction.

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Paul Flowers, Klaus Theopold, Richard Langley, William R. Robinson

Chemistry

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