Question

In following reaction which acts as reducing agent-$2FeC{l}_3+SnC{l}_2\to 2FeC{l}_2+SnC{l}_4$

• A. $S{n}_{2+}$
• B. $F{e}_{2+}$
• C. $S{n}_{4+}$
• D. $F{e}_{3+}$

Answer 1

Answer: (A)

$S{n}_{2+}$

$2\stackrel{+3}{Fe}{Cl}_3+\stackrel{+2}{Sn}{Cl}_2⟶\stackrel{+2}{Fe}{Cl}_2+\stackrel{+4}{Sn}{Cl}_4$

Here, $\stackrel{+3}{Fe}{Cl}_3$ is reduced to $\stackrel{+3}{Fe}{Cl}_2$

and $\stackrel{+2}{Sn}{Cl}_2$ is oxidized to $\stackrel{+4}{Sn}{Cl}_4$

Therefore, $\stackrel{+2}{Sn}$ acts as a reducing agent & itself gets oxidized to $\stackrel{+4}{Sn}$.