Question

which of the following is an oxidation - reduction reaction?
o so₂(g) + h₂o(l) → h₂so₃(aq)
o caco₃(s) → cao(s) + co₂(g)
o ca(oh)₂(s) + h₂co₃(l) → caco₃(aq) + 2h₂o(l)
o c₆h₁₂o₆(s) + 6o₂(g) → 6co₂(g) + 6h₂o(l)

Explanation:

Step1: Recall oxidation - reduction reaction concept

An oxidation - reduction reaction involves a transfer of electrons, which is indicated by a change in oxidation numbers of elements in the reactants and products.

Step2: Analyze first reaction

For $SO_2(g)+H_2O(l)\longrightarrow H_2SO_3(aq)$, the oxidation numbers of S, H and O do not change. S has an oxidation number of +4 in $SO_2$ and in $H_2SO_3$, H is + 1 and O is - 2 throughout, so it is not a redox reaction.

Step3: Analyze second reaction

For $CaCO_3(s)\longrightarrow CaO(s)+CO_2(g)$, the oxidation numbers of Ca, C and O remain the same (Ca = + 2, C = + 4, O=-2) in reactants and products, so it is not a redox reaction.

Step4: Analyze third reaction

For $Ca(OH)_2(s)+H_2CO_3(l)\longrightarrow CaCO_3(aq)+2H_2O(l)$, this is an acid - base reaction (neutralization). The oxidation numbers of Ca, H, C and O do not change, so it is not a redox reaction.

Step5: Analyze fourth reaction

For $C_6H_{12}O_6(s)+6O_2(g)\longrightarrow 6CO_2(g)+6H_2O(l)$, in $C_6H_{12}O_6$, the average oxidation number of C can be calculated. Let the oxidation number of C be $x$. Then $6x + 12\times(+1)+6\times(-2)=0$, so $x = 0$. In $CO_2$, the oxidation number of C is +4. Also, the oxidation number of O changes from 0 in $O_2$ to - 2 in $CO_2$ and $H_2O$. There is a change in oxidation numbers, so it is a redox reaction.

Answer:

$C_6H_{12}O_6(s)+6O_2(g)\longrightarrow 6CO_2(g)+6H_2O(l)$