Question

Complete the given equations:
(i) Cu + 8$HN{O}_3$ $\Rightarrow$ 3$Cu(N{O}_3{)}_2$ + ___W__ + 4$H_{2}O$
(ii) 4Zn +10$HN{O}_3$ $\Rightarrow$ 4$Zn(N{O}_3{)}_2$ + 5$H_{2}O$ + __X__
(iii) $I_2$ + 10$HN{O}_3$ $\Rightarrow$___Y__ + 10$N{O}_2$ + 4$H_{2}O$

• A. (W) 2$N{O}_2$ (X) NO (Y) 5$HI{O}_3$
• B. (W) 2NO (X)$N_{2}O$ (Y) 2$HI{O}_3$
• C. (W) $N_2$ (X)$N{O}_2$ (Y) HI
• D. (W) $N_{2}O$ (X)$N{O}_2$ (Y) 3HI

Answer 1

The correct option is B (W) 2NO (X)$N_{2}O$ (Y) 2$HI{O}_3$

(I) In this reaction, $C_4⟶C_4+2$ occurs in which $6$ electrons are released. Now out of $8\left({NO}_{\overline{3}}\right)$ got reduced . So $2NO$ should be formed using $6\overline{e}.$

(ii) $Zn$ released $8\overline{e}$ which are used by $2\left({NO}_3\right)$ to get reduced and hence they are reduced to $N_{2}O.$

(iii) $I_2$ must be oxidised to reduce $10$ moles of ${NO}_{\overline{3}}$ to $10$ moles of ${NO}_2$. So it will oxidise to its $2nd$ highest oxidisation state forming ${HIO}_3.$