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 it is more appropriate to write these reactions as :

(a) $$6 \; CO_{ 2 \; (g) } \; + \; 12 \; H_{ 2 }O_{ (l) } \; \rightarrow \; C_{ 6 }H_{ 12 }O_{ 6 \; (aq) } \; + \; 6 \; H_{ 2 }O_{ (l) } \; + \; 6 \; O_{ 2 \; (g) }$$

Also suggest a technique to investigate the path of the above (a) and (b) redox reactions

Asked by Pragya Singh | 1 year ago |  56

##### Solution :-

The H2 produced in step 1 reduces CO2 , thereby producing glucose $$C_6 H_{12} O_6$$ and H2 O.

Now, the net reaction of the process is given as:

It is more appropriate to write the reaction as given above because water molecules are also produced in the process of photosynthesis. The path of this reaction can be investigated by using radioactive $$H_2 O_{18}$$ in place of $$H_2O$$

(b) O is produced from each of the two reactants Oand $$H_2 O_2$$ . For this reason, O2 is written twice. The given reaction involves two steps. First, O3 decomposes to form O2 and O. In the second step, H2O2 reacts with the O produced in the first step, thereby producing H2O and O2

The path can be found with the help of $$H_{ 2 }O_{ 2 }^{ 18 }$$​ or $$O_{ 3 }^{ 18 }$$​.

Answered by Abhisek | 1 year ago

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