Question

Acidified potassium dichromate reacts with potassium iodide and oxidises it to $I_2$ . What is the oxidation state of chromium in the products of the reaction?
(A) +4
(B) +6
(C) +3
(D) +2

Answer 1

Hint: Chromium in the product of this reaction is existing as $\left({Cr}_2{\left({SO}_4\right)}_3\right)$ . Oxidation state of Sulphate ion is -2. Since, the compound here is a neutral molecule, the overall oxidation state of the compound is zero. Oxidation state of chromium can be found by a simple equation using the above given facts.

Complete step by step answer:
The reaction here is when acidified potassium dichromate is reacted with potassium iodide, so first of all, we can give the balanced chemical equation for this reaction.

$K_{2}C{r}_2{O}_7+7H_2{SO}_4+6KI\to {4K}_2{SO}_4+{Cr}_2{\left(S{O}_4\right)}_3+{7H}_{2}O+{3I}_2$

In the product form, Cr is existing as the compound $\left({Cr}_2{\left({SO}_4\right)}_3\right)$ .

Oxidation state of chromium can be found in the following method given below:
Consider the oxidation state of Cr as ‘x’.
Since the compound here is a neutral molecule, some of the oxidation states of all the elements in the compound will be zero.

We know that the oxidation state of sulphate ion is -2.
Therefore,

$$\begin{gathered} \Rightarrow 2x+(3\times -2)=0 \\ \Rightarrow 2x-6=0 \\ \Rightarrow 2x=6 \\ \Rightarrow x=+3 \end{gathered}$$

Hence, the oxidation state of Cr in the product is +3.

So, option (C) +3 is the correct answer.

Additional Information:
In the question, it is given that potassium iodide is oxidised into$I_2$. And this has been validated by the fact that the oxidation state of I in KI is -1 which is increased to 0 in $I_2$. Since there is an increase in oxidation state is happening here, and hence it is being referred to as iodine is being ‘oxidised’.

Oxidant: Oxidant or oxidising agent is referred to as a chemical compound which oxidises other compounds by itself getting reduced in a redox chemical reaction. In this question, when we consider the reaction , we can see that the oxidation state of Cr is getting reduced from +6 to +3 and so $K_2{Cr}_2{O}_7$ acts as an oxidising agent in this reaction.

Reductant: Reductant or reducing agent is referred to as a chemical compound which reduces other compounds by itself getting oxidised in a redox chemical reaction. In this question, when we consider the reaction , we can see that the oxidation state of I is getting reduced from -1 to 0 and so KI acts as a reducing agent in this reaction.

Increase in oxidation state is known as oxidation while a decrease in oxidation state is known as reduction.Increase in the oxidation state is actually referring to loss of electrons and decrease in oxidation state is actually the gain of electrons.

Note: Elements in their pure state have zero oxidation state whereas oxidation state of a pure ion is equal to its ionic charge. For example: ${Ca}^{2+}$ has an oxidation state of +2 , oxidation state of carbon in graphite and diamond is 0, oxidation state of $H_2$,$N_2$,${Cl}_2$ are 0.