Question

How many electrons are lost or gained by Cr in the given chemical reaction below:
$C{r}_2{O}_7^{2-}\to C{r}^{3+}$

• A. 5
• B. 4
• C. 2
• D. 6

Answer 1

The correct option is D 6

Let’s go step by step here.
First multiply $C{r}^{3+}\times 2$ to balance the number of Cr atoms.
$C{r}_2{O}_7^{2-}\to 2C{r}^{3+}$
Oxidation state of $Cr$ on the LHS:
$2\times Cr+7\times (-2)=-2,Cr=+6$
Oxidation state of $Cr$ on the RHS:
$Cr=+3$
Since there are two $Cr$ atoms, the total change is $-6$
So, we need to add 6 electrons to the LHS to balance this reaction.
$6e^{-}+C{r}_2{O}_7^{2-}\to 2C{r}^{3+}$