Question

In the ionic equation, $Bi{O}_3^{-}+6H^{+}+x{e}^{-}⟶B{i}^{3+}+3H_{2}O$, the value of $x$ is

• A. $6$
• B. $2$
• C. $4$
• D. $3$

Answer 1

Answer: (B)

$Bi{O}_3^{-}+6H^{+}+x{e}^{-}⟶B{i}^{3+}+3H_{2}O$
The half cell reaction can be written as :
$B{i}^{5+}\stackrel{+2e^{-}}{\to }B{i}^{3+}$
$\stackrel{+5}{\mathrm{Bi}}\stackrel{-2}{O_3^{-}}+6H^{+}+2e^{-}⟶B{i}^{3+}+3H_{2}O$
Thus, the value of $x=2$ in the above ionic equation. Also, by another way
$B{O}_3^{-}+6H^{+}+x{e}^{-}⟶B^{3+}+3H_{2}O$
Total charge on $LHS=$ Total charge on $RHS$
$-1+6+x(-1)=+3$
$-x=+3-5$
$-x=-2\Rightarrow x=2$