which of the following is an oxidation - reduction reaction?\no zns(s) + 2o2(g) → znso4(s)\no cao(s) +…

Explanation:

Step1: Recall oxidation - reduction reaction concept

An oxidation - reduction reaction involves a change in oxidation numbers of elements.

Step2: Analyze first reaction

For $ZnS(s)+2O_2(g)\rightarrow ZnSO_4(s)$: In $ZnS$, $Zn$ has an oxidation number of + 2 and $S$ has - 2. In $O_2$, $O$ has an oxidation number of 0. In $ZnSO_4$, $Zn$ is + 2, $S$ is + 6 and $O$ is - 2. Sulfur is oxidized from - 2 to + 6 and oxygen is reduced from 0 to - 2.

Step3: Analyze second reaction

For $CaO(s)+H_2O(l)\rightarrow Ca(OH)_2(s)$: In $CaO$, $Ca$ is + 2 and $O$ is - 2. In $H_2O$, $H$ is + 1 and $O$ is - 2. In $Ca(OH)_2$, $Ca$ is + 2, $H$ is + 1 and $O$ is - 2. No change in oxidation numbers.

Step4: Analyze third reaction

For $6Li_2O(s)+P_4O_{10}(g)\rightarrow 4Li_3PO_4(s)$: In $Li_2O$, $Li$ is + 1 and $O$ is - 2. In $P_4O_{10}$, $P$ is + 5 and $O$ is - 2. In $Li_3PO_4$, $Li$ is + 1, $P$ is + 5 and $O$ is - 2. No change in oxidation numbers.

Step5: Analyze fourth reaction

For $SO_2(g)+H_2O(l)\rightarrow H_2SO_3(aq)$: In $SO_2$, $S$ is + 4 and $O$ is - 2. In $H_2O$, $H$ is + 1 and $O$ is - 2. In $H_2SO_3$, $S$ is + 4, $H$ is + 1 and $O$ is - 2. No change in oxidation numbers.

Answer:

$ZnS(s)+2O_2(g)\rightarrow ZnSO_4(s)$