Question

When 1 mol of $C{r^{3 + }}$ is oxidised to form 1 mole of $Cr{O_4}^{2 - }$. How many moles of electrons are transferred?
A.3 mol
B.4 mol
C.5 mol
D.6 mol
E.7 mol

Answer 1

Since, $C{r^{3 + }}$ is oxidised to $Cr{O_4}^{2 - }$. So, $C{r^{3 + }} \to Cr{O_4}^{2 - }$
First, we will find the oxidation state of Cr in $C{r^{3 + }}$ and Cr in $Cr{O_4}^{2 - }$. After finding the oxidation state, we need to balance out the equation by adding $O{H^ - }$ and ${H_2}O$ in order to find the number of electrons that are transferred.

Complete step by step answer:
The oxidation of 1 mole of $C{r^{3 + }}$ to 1 mol of $Cr{O_4}^{2 - }$ is given by:
$C{r^{3 + }} \to Cr{O_4}^{2 - }$
First let go through the oxidation states:
Oxidation state of Chromium in $C{r^{3 + }} = + 3$
Oxidation state of Chromium in $Cr{O_4}^{2 - }$
$\Rightarrow x + 4 \times ( - 2) = - 2$
$\Rightarrow x + ( - 8) = - 2$
$\Rightarrow x = + 6$
The oxidation state of Cr in $C{r^{3 + }}$ is +3 and that of $Cr{O_4}^{2 - }$ in which oxidation state of Cr is +6. This means that oxidation takes place and $C{r^{3 + }}$ is oxidised to $Cr{O_4}^{2 - }$.
This above equation is not balanced, so I need to add $O{H^ - }$ and ${H_2}O$ to balance it out.
$C{r^{3 + }} + 8O{H^ - } \to Cr{O_4}^{2 - } + 4{H_2}O + 3{e^ - }$
Hence, 3 moles of electrons are transferred when 1 mol of $C{r^{3 + }}$ is oxidised to form 1 mole of $Cr{O_4}^{2 - }$.

Note: Chromate $(Cr{O_4}^{2 - })$ is the oxyanion which results from the removal of protons from the chromic acid. It is also known as Chromium oxo anion or divalent inorganic anions. Potassium and sodium salts are the most commonly used chromate salts. They act as a strong oxidising agent and can become a predominant ion in acid solution. And in the alkaline solution, chromate ions act as predominant species $(Cr{O_4}^{2 - })$ combines with water to form chromium (III) hydroxide. Molecular weight of $Cr{O_4}^{2 - }$ is 194.1896 g / mol and its density is $2.73\dfrac{g}{{c{m^3}}}$.