Question

How many electrons are involved in the following redox reaction?
$\text{C}{{\text{r}}_{2}}\text{O}_{7}^{-2}\text{ + F}{{\text{e}}^{+2}}\text{ + }{{\text{C}}_{2}}\text{O}_{4}^{-2}\text{ }\to \text{ C}{{\text{r}}^{+3}}\text{ + F}{{\text{e}}^{3+}}\text{ + C}{{\text{O}}_{2}}$ (Unbalanced)
A. 3
B. 4
C. 6
D. 5

Answer 1

For this problem, firstly we have to balance the reaction such that the number of moles of the reactant is equal to the number of moles of the product and then we will count the total number of electrons that will be involved in the reaction.

Complete step by step solution:
- In the given question, we have to calculate the total number of electrons which will be involved in the given redox reaction.
- As we know that redox reaction is the type of reaction in which both oxidation and reduction processes will take place.
- Oxidation is determined when there is an increase in the oxidation state of the element whereas reduction is determined when there is a decrease in the oxidation state of the element.
- So, as we can see that the chromium is reduced from +6 to +3 and the balanced chemical reaction will be:
$\text{C}{{\text{r}}_{2}}\text{O}_{7}^{-2}\,\text{+ 14}{{\text{H}}^{+}}\text{ + 6}{{\text{e}}^{-}}\text{ }\to \text{ 2C}{{\text{r}}^{3+}}\text{ + 7}{{\text{H}}_{2}}\text{O}$ …. (1)
- Now, in iron the oxidation state changes from +2 to +3 which shows it is oxidised as there is an increase in the oxidation state.
- Also, there is the removal of one electron and the balanced reaction will be:
$\text{F}{{\text{e}}^{2+}}\text{ }\to \text{ F}{{\text{e}}^{3+}}\text{ + 1}{{\text{e}}^{-}}$ … (2)
- At last in the oxalate ion, the change in the oxidation state of carbon is from +3 to +4 which shows that there is the oxidation of the carbon.
- The balanced chemical reaction will be:
${{\text{C}}_{2}}\text{O}_{4}^{-2}\text{ }\to \text{ 2C}{{\text{O}}_{2}}\text{ + 2}{{\text{e}}^{-}}$ … (3)
- Now, we will add equation 1, 2 and 3 after which we get the net reaction and the total number of electrons which are involved in the reaction.
$\text{C}{{\text{r}}_{2}}\text{O}_{7}^{-2}\text{ + F}{{\text{e}}^{+2}}\text{ + }{{\text{C}}_{2}}\text{O}_{4}^{-2}\text{ + 14}{{\text{H}}^{+}}\ \text{+ 3}{{\text{e}}^{-}}\to \text{ 2C}{{\text{r}}^{+3}}\text{ + F}{{\text{e}}^{3+}}\text{ + 2C}{{\text{O}}_{2}}\text{ + 7}{{\text{H}}_{2}}\text{O}$
- So, the total number of electrons involved is 3.

Therefore, A is the correct answer.

Note: The oxidation state of chromium in $\text{C}{{\text{r}}_{2}}\text{O}_{7}^{-2}$ and $\text{C}{{\text{r}}^{+3}}$ is $\text{2x + (7 }\times \text{ -2) = -2; x = +6}$ and $+3$ respectively. Whereas the oxidation state of carbon in ${{\text{C}}_{2}}\text{O}_{4}^{-2}$ and $\text{C}{{\text{O}}_{2}}$ is $\text{2x + (7 }\times \text{ -2) = -2; x = +3}$ and $\text{x + (2 }\times \text{ -2) = 0; x = +4}$ respectively.