Question

Balance by Ion electon method, $C{r}_2{O}_7^{-2}+F{e}^{+2}\to C{r}^{+3}+H_{2}O$ (in acidic medium).

• A. $C{r}_2{O}_7^{-2}+2F{e}^{+2}+14H^{+}\to C{r}^{+3}+2F{e}^{+3}+7H_{2}O$
• B. $C{r}_2{O}_7^{-2}+F{e}^{+2}+2H^{+}\to 2C{r}^{+3}+F{e}^{+3}+H_{2}O$
• C. $C{r}_2{O}_7^{-2}+3F{e}^{+2}+14H^{+}\to 2C{r}^{+3}+3F{e}^{+3}+H_{2}O$
• D. $C{r}_2{O}_7^{-2}+6F{e}^{+2}+14H^{+}\to 2C{r}^{+3}+6F{e}^{+3}+7H_{2}O$

Answer 1

The correct option is D $C{r}_2{O}_7^{-2}+6F{e}^{+2}+14H^{+}\to 2C{r}^{+3}+6F{e}^{+3}+7H_{2}O$

$C{r}_2{O}_7^{2-}+F{e}^{+2}\to C{r}^{+3}+F{e}^{3+}+H_{2}O$
Reduction half :
$\stackrel{+6}{C{r}_2}{O}_7^{2-}\to \stackrel{+3}{C{r}^{3+}}$
Oxidation half :
$F{e}^{+2}\to F{e}^{+3}$
By taking reduction half reaction balance the $Cr$ first.

$C{r}_2{O}_7^{2-}\to 2C{r}^{+3}$
Balance $O$ by adding $7H_{2}O$ to R.H.S. and balance $H$ by adding $14H^{+}$ to L.H. S.
$C{r}_2{O}_7^{2-}+14H^{+}\to 2C{r}^{+3}+7H_{2}O$

Above equation is balanced atom wise. Now balance both equations charge wise. To balance the charge, add electrons to the electrically positive side.
$C{r}_2{O}_7^{2-}+14H^{+}+6e^{-}\to 2C{r}^{+3}+7H_{2}O$ ----(1)

In oxidation half cell mass is balanced. So balance the charge :
$F{e}^{+2}\to F{e}^{+3}+e^{-}$----(2)

In equation (2) multiply by 6 and then add both equaiton (1) and (2).
$C{r}_2{O}_7^{-2}+6F{e}^{+2}+14H^{+}\to 2C{r}^{+3}+6F{e}^{+3}+7H_{2}O$