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Q.
Equivalent weight of $C_{6}H_{5}CHO$ is equal to molar mass in
the following reaction.
Thus, species (A) is
Redox Reactions
Solution:
Gain of one O-atom $\equiv1\left(O\right)\equiv16 g\equiv8\times2 g\left(O\right) \frac{M}{2}\equiv8g\left(O\right)$
Thus, E (oxidation part) =$\frac{M}{2}$
Since, E (net) = M
Hence, E (reduction part) = $\frac{M}{2}$
$\therefore $ E (reduction) =$\frac{M}{2}$
Hence, A is $CH_{2}OH$