- A$107.88$
- B$1.6$
- C$0.8$
- ✓$21.60$
At Anode: $2O{H^ - } \to {H_2}O + \frac{1}{2}{O_2} + 2{e^ - }$
${E_{Ag}} = \frac{{108}}{1} = 108;\,\,{E_{{O_2}}} = \frac{{\frac{1}{2} \times 32}}{2} = 8$
$\frac{{{W_{Ag}}}}{{{E_{Ag}}}} = \frac{{{W_{{O_2}}}}}{{{E_{{O_2}}}}}$; ${W_{Ag}} = \frac{{1.6 \times 108}}{8} = 21.6\,\,gm$.
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$C{H_3}CN\xrightarrow{{Na + {C_2}{H_5}OH}}X\xrightarrow{{HN{O_2}}}Y\mathop {\xrightarrow{{{K_2}C{r_2}{O_7}}}}\limits_{{H_2}S{O_4}} Z$

Assertion $A$ : A solution of the product obtained by heating a mole of glycine with a mole of chlorine in presence of red phosphorous generates chiral carbon atom.
Reason $R$ : A molecule with 2 chiral carbons is always optically active.
In the light of the above statements, choose the correct answer from the options given below: