Question

Balance the following equation by ion electron method.
$C{u}_{3}P+C{r}_2{O}_7^{2-}\to C{u}^{2+}+H_{3}P{O}_4+C{r}^{3+}$ (acid medium).

Answer 1

Write the reaction in two half reactions,oxidation half and reduction half as:

Oxidation:$C{u}_{3}P\to C{u}^{2+}+H_{3}P{O}_4$

Reduction:${C{r}_2{O}_7}^{2-}\to C{r}^{3+}$

Balance all other atoms except $O$ and $H$.

$C{u}_{3}P\to 3C{u}^{2+}+H_{3}P{O}_4$

${C{r}_2{O}_7}^{2-}\to 2C{r}^{3+}$

Now balance the oxygen atoms by adding $H_{2}O$ molecules.

$C{u}_{3}P+4H_{2}O\to 3C{u}^{2+}+H_{3}P{O}_4$

${C{r}_2{O}_7}^{2-}\to 2C{r}^{3+}+7H_{2}O$

Now balance hydrogen atoms by adding $H^{+}$ ions.

$C{u}_{3}P+4H_{2}O\to 3C{u}^{2+}+5H^{+}+H_{3}P{O}_4$

${C{r}_2{O}_7}^{2-}+14H^{+}\to 2C{r}^{3+}+7H_{2}O$

To balance the charge,add electrons to more positive side to equal the less positive side of the half reaction.

$C{u}_{3}P+4H_{2}O\to 3C{u}^{2+}+5H^{+}+H_{3}P{O}_4+11e^{-}$

${C{r}_2{O}_7}^{2-}+14H^{+}+6e^{-}\to 2C{r}^{3+}+7H_{2}O$

Now multiply oxidation half reaction by 6 and reduction half reaction by 11 and add both the reactions we will get

$6C{u}_{3}P+11C{r}_2{O}_7^{2-}+124H^{+}\to 18C{u}^{2+}+6H_{3}P{O}_4+53H_{2}O+22C{r}^{+3}$