Question

Which of the following statement does not hold true for the following reaction?
$xC{u}_{3}P+yC{r}_2{O}_7^{2-}+z{H}^{+}\to C{u}^{2+}+H_{3}P{O}_4+C{r}^{3+}$

• A. $Cu$ in $C{u}_{3}P$ is oxidized to $C{u}^{2+}$ whereas $P$ in $C{u}_{3}P$ is also oxidised to $P{O}_4^{3-}$.
• B. $Cu$ in $C{u}_{3}P$ is oxidized to $C{u}^{2+}$ whereas $P$ in $C{u}_{3}P$ is reduced to $H_{3}P{O}_4$
• C. In the conversion of $C{u}_{3}P$ to $C{u}^{2+}$ and $H_{3}P{O}_4$, 11 electrons are involved.
• D. The value of x is 6.

Answer 1

The correct option is B

$Cu$ in $C{u}_{3}P$ is oxidized to $C{u}^{2+}$ whereas $P$ in $C{u}_{3}P$ is reduced to $H_{3}P{O}_4$

Let us start by balancing the equation:

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

The oxidation state of $P$ is getting changed from -3 to +5 during this conversion. That is, P in $C{u}_{3}P$ is getting oxidised to $H_{3}P{O}_4$. Also, copper gets oxidized from $C{u}^{+}$ to $C{u}^{2+}$ oxidation state.