Consider the reactions:
$(a)$ $6 CO_{2(g)} + 6 H_{2}O_{(l)} \rightarrow C_{6}H_{12}O_{6(aq)} + 6 O_{2(g)}$
$(b)$ $O_{3(g)} + H_{2}O_{2(l)} \rightarrow H_{2}O_{(l)} + 2 O_{2(g)}$
Why is it more appropriate to write these reactions as:
$(a)$ $6 CO_{2(g)} + 12 H_{2}O_{(l)} \rightarrow C_{6}H_{12}O_{6(aq)} + 6 H_{2}O_{(l)} + 6 O_{2(g)}$
$(b)$ $O_{3(g)} + H_{2}O_{2(l)} \rightarrow H_{2}O_{(l)} + O_{2(g)} + O_{2(g)}$
Also,suggest a technique to investigate the path of the above $(a)$ and $(b)$ redox reactions.

(N/A) Photosynthesis involves two steps. Step $1$: $H_{2}O$ decomposes to give $H_{2}$ and $O_{2}$ $(2 H_{2}O_{(l)} \rightarrow 2 H_{2(g)} + O_{2(g)})$. Step $2$: The $H_{2}$ produced reduces $CO_{2}$ to glucose $(6 CO_{2(g)} + 12 H_{2(g)} \rightarrow C_{6}H_{12}O_{6(s)} + 6 H_{2}O_{(l)})$. The net reaction is $6 CO_{2(g)} + 12 H_{2}O_{(l)} \rightarrow C_{6}H_{12}O_{6(s)} + 6 H_{2}O_{(l)} + 6 O_{2(g)}$. This is more appropriate because water is both a reactant and a product. The path can be investigated using radioactive isotope labeling,such as $H_{2}O^{18}$.
$(b)$ $O_{2}$ is produced from both $O_{3}$ and $H_{2}O_{2}$. The reaction occurs in two steps: $O_{3(g)} \rightarrow O_{2(g)} + O_{(g)}$ and $H_{2}O_{2(l)} + O_{(g)} \rightarrow H_{2}O_{(l)} + O_{2(g)}$. The net reaction $O_{3(g)} + H_{2}O_{2(l)} \rightarrow H_{2}O_{(l)} + O_{2(g)} + O_{2(g)}$ shows the origin of the oxygen molecules. The path can be investigated using $H_{2}O_{2}^{18}$ or $O_{3}^{18}$.