Question

The number of coefficients of $BaCr{O}_4$, $KI$, $CrC{l}_3$ and $KCl$ in the balanced equation $BaCr{O}_4+KI+HCl\to BaC{l}_2+I_2+KCl+CrC{l}_3+H_{2}O$, are respectively :

• A. $2,6,2,6$
• B. $6,2,6,6$
• C. $2,2,6,6$
• D. $6,6,2,2$

Answer 1

The correct option is A $2,6,2,6$

The oxidation half reaction is $2I^{-}\to I_2+2e^{-}$...........(eq. 1)

The reduction half reaction is $C{r}^{+6}+3e^{-}\to C{r}^{3+}$............(eq. 2)

The oxidation half-reaction is multiplied by 3 and the reduction half-reaction is multiplied by 2 to balance the number of electrons. These two half-reactions (eq. 1 and eq. 2) are then added to get:

$2C{r}^{+6}+6I^{-}\to 2C{r}^{3+}+3I_2$ ...........(eq. 3)

Adding cations and anions in the above ionic equation (eq. 3) as given in the question, we get

$2BaCr{O}_4+6KI+xHCl\to 2CrC{l}_3+3I_2+yKCl+zBaC{l}_2+n{H}_{2}O$ (x, y, z and n are unbalanced coefficients)

The remaining atoms are now balanced to obtain the final equation:

$2BaCr{O}_4+6KI+16HCl\to 2CrC{l}_3+3I_2+6KCl+2BaC{l}_2+8H_{2}O$. This is the balanced equation.

The coefficients of $BaCr{O}_4,KI,CrC{l}_3$ and $KCl$ are $2,6,2$ and $6$ respectively.