Q.Consider the reactions:

Concept: Oxidation-Reduction (Redox) — specifically, tracking the source of oxygen atoms to identify which species is oxidised and which is reduced.

Reasoning:

1. In reaction (a), the simple equation shows CO₂ being reduced to glucose and H₂O being oxidised to O₂. But the oxygen in the evolved O₂ comes entirely from H₂O, not from CO₂. Writing 12 H₂O on the left makes this clear: 6 H₂O are consumed in the reduction of CO₂, and the other 6 H₂O appear as a product, while the 6 O₂ molecules come from the remaining oxygen atoms of water.

2. In reaction (b), O₃ is reduced to H₂O and H₂O₂ is oxidised to O₂. However, the two O₂ molecules come from different sources: one from O₃ and one from H₂O₂. Writing O₂ + O₂ separately (or as 2O₂ with a note) emphasises that each originates from a different reactant.

3. The balanced forms make the electron transfer and atom source explicit — essential for understanding the redox mechanism. …

Why the original equations are misleading

Both reactions are redox reactions — electrons are transferred between species, and the oxidation states of key atoms change. The original equations (a) and (b) are chemically correct in terms of atom balance, but they obscure the actual mechanism of electron transfer. In redox chemistry, it is crucial to see which atoms lose electrons (are oxidised) and which gain electrons (are reduced). The rewritten forms make these half‑reactions visible.

1. Reaction (a) — Photosynthesis

The classic equation

suggests that the oxygen in $\mathrm{O_2}$ comes from $\mathrm{CO_2}$. But experiments (using isotopic oxygen‑18) show that all the $\mathrm{O_2}$ released comes from water, not from carbon dioxide.

The correct half‑reactions are:

• Oxidation (loss of electrons):

• Reduction (gain of electrons):

To see the full stoichiometry, multiply the oxidation half‑reaction by 6 (to produce 6 $\mathrm{O_2}$) and the reduction half‑reaction by 6 (to consume 6 $\mathrm{CO_2}$). Adding them gives:

The 12 water molecules on the left are the source of the 6 $\mathrm{O_2}$ molecules. The 6 water molecules on the right are a product of the reduction step. So water appears on both sides — it is not simply a reactant; it is both consumed and produced.

2. Reaction (b) — Ozone with hydrogen peroxide

The original equation

is balanced, but it does not show which oxygen atoms end up where. In reality, the reaction proceeds via two distinct steps:

• Step 1: Ozone oxidises hydrogen peroxide:

Here, one oxygen atom from $\mathrm{O_3}$ and one from $\mathrm{H_2O_2}$ combine to form one $\mathrm{O_2}$ molecule. The remaining two oxygen atoms from $\mathrm{O_3}$ form the second $\mathrm{O_2}$.

• Step 2 (if written separately): The extra $\mathrm{O_2}$ is just a product — no further reaction.

The rewritten form

makes it clear that two distinct dioxygen molecules are formed from different sources. The original equation lumps them together as $2\,\mathrm{O_2}$, which hides the fact that one $\mathrm{O_2}$ comes from the peroxide and the other from the ozone. …