
Potassium dichromate is a good oxidizing agent, in the acidic medium the oxidation state of chromium changes by:
A. 2
B. 3
C. 4
D. 5
Answer
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Hint: Potassium dichromate has a chemical formula $\left( {{\text{K}}_{2}}\text{C}{{\text{r}}_{2}}{{\text{O}}_{7}} \right)$. It is an inorganic chemical reagent which is commonly used as an oxidizing agent in laboratory and industrial applications. It shows reactions in all the mediums, acidic, basic and neutral medium. The change in oxidation state is the oxidation state of an element in the reactant side minus the oxidation state of an element in the product side.
Complete step by step answer:
Let us find the change in oxidation state step by step:
(1) The reaction shown by potassium dichromate in acidic medium is
In ionic form: $\text{C}{{\text{r}}_{2}}\text{O}_{7}^{2-}+14{{\text{H}}^{+}}+6{{\text{e}}^{-}}\to \text{2C}{{\text{r}}^{+3}}+7{{\text{H}}_{2}}\text{O}$.
(2) Write the oxidation state of chromium $\left( \text{Cr} \right)$ on the product side: The charge of chromium is $+3$.
As there are two chromium, so the charge becomes $+\left( 2\times 3 \right)$ or $+6$.
The oxidation state of chromium $\left( \text{Cr} \right)$ in product side is $+6$.
(3) Find the oxidation state of chromium in reactant side:
The element chromium $\left( \text{Cr} \right)$ is present in the compound $\text{C}{{\text{r}}_{2}}\text{O}_{7}^{2-}$.
The overall charge on the compound is $-2$.
Let the oxidation state of $\left( \text{Cr} \right)$ in the compound be ‘x’.
The oxidation state of oxygen is $-2$.
The number of atoms of chromium are 2 and the number of atoms of oxygen are 7.
The equation will be $\left[ \left( \text{x}\times \text{2} \right)+\left( -\text{2}\times \text{7} \right) \right]=-2$.
The charge on $\text{Cr}$ is $\text{x =}\dfrac{-2+14}{2}$ or $\text{x = }+\text{6}$.
The oxidation state of chromium $\left( \text{Cr} \right)$ in reactant side is $+6$.
The change in oxidation state of chromium will be oxidation state of chromium in reactant side minus with oxidation state of chromium in product side, which is equal to $+6-\left( +3 \right)$ or $+3$.
Potassium dichromate is a good oxidizing agent, in the acidic medium the oxidation state of chromium changes by $+3$. So, the correct answer is “Option B”.
Additional Information:
Uses of potassium dichromate are:
(1) Potassium dichromate is a common reagent in ‘wet tests’ in analytical chemistry.
(2) Potassium dichromate is used to prepare ‘chromic acid’ which is used for cleaning glassware and because of safety concerns associated with the compound, this practice is now avoided.
(3) It is also used as an ingredient in cement in which it improves its density and texture and retards the setting of the mixture. Moreover, its usage causes contact dermatitis in construction workers.
Note: In basic and neutral medium, the oxidation state of chromium changes from $+6$ to $+3$ and $+6$ to $+12$ respectively. The reaction in basic medium is
${{\text{K}}_{2}}\text{C}{{\text{r}}_{2}}{{\text{O}}_{7}}+\text{2NaOH}\to {{\text{K}}_{2}}\text{Cr}{{\text{O}}_{4}}+\text{N}{{\text{a}}_{2}}\text{Cr}{{\text{O}}_{4}}+{{\text{H}}_{2}}\text{O}$ and the reaction in neutral medium is ${{\text{K}}_{2}}\text{C}{{\text{r}}_{2}}{{\text{O}}_{7}}+{{\text{H}}_{2}}\text{O}\to \text{2KCrO}_{4}^{2-}+2{{\text{H}}^{+}}$.
Complete step by step answer:
Let us find the change in oxidation state step by step:
(1) The reaction shown by potassium dichromate in acidic medium is
In ionic form: $\text{C}{{\text{r}}_{2}}\text{O}_{7}^{2-}+14{{\text{H}}^{+}}+6{{\text{e}}^{-}}\to \text{2C}{{\text{r}}^{+3}}+7{{\text{H}}_{2}}\text{O}$.
(2) Write the oxidation state of chromium $\left( \text{Cr} \right)$ on the product side: The charge of chromium is $+3$.
As there are two chromium, so the charge becomes $+\left( 2\times 3 \right)$ or $+6$.
The oxidation state of chromium $\left( \text{Cr} \right)$ in product side is $+6$.
(3) Find the oxidation state of chromium in reactant side:
The element chromium $\left( \text{Cr} \right)$ is present in the compound $\text{C}{{\text{r}}_{2}}\text{O}_{7}^{2-}$.
The overall charge on the compound is $-2$.
Let the oxidation state of $\left( \text{Cr} \right)$ in the compound be ‘x’.
The oxidation state of oxygen is $-2$.
The number of atoms of chromium are 2 and the number of atoms of oxygen are 7.
The equation will be $\left[ \left( \text{x}\times \text{2} \right)+\left( -\text{2}\times \text{7} \right) \right]=-2$.
The charge on $\text{Cr}$ is $\text{x =}\dfrac{-2+14}{2}$ or $\text{x = }+\text{6}$.
The oxidation state of chromium $\left( \text{Cr} \right)$ in reactant side is $+6$.
The change in oxidation state of chromium will be oxidation state of chromium in reactant side minus with oxidation state of chromium in product side, which is equal to $+6-\left( +3 \right)$ or $+3$.
Potassium dichromate is a good oxidizing agent, in the acidic medium the oxidation state of chromium changes by $+3$. So, the correct answer is “Option B”.
Additional Information:
Uses of potassium dichromate are:
(1) Potassium dichromate is a common reagent in ‘wet tests’ in analytical chemistry.
(2) Potassium dichromate is used to prepare ‘chromic acid’ which is used for cleaning glassware and because of safety concerns associated with the compound, this practice is now avoided.
(3) It is also used as an ingredient in cement in which it improves its density and texture and retards the setting of the mixture. Moreover, its usage causes contact dermatitis in construction workers.
Note: In basic and neutral medium, the oxidation state of chromium changes from $+6$ to $+3$ and $+6$ to $+12$ respectively. The reaction in basic medium is
${{\text{K}}_{2}}\text{C}{{\text{r}}_{2}}{{\text{O}}_{7}}+\text{2NaOH}\to {{\text{K}}_{2}}\text{Cr}{{\text{O}}_{4}}+\text{N}{{\text{a}}_{2}}\text{Cr}{{\text{O}}_{4}}+{{\text{H}}_{2}}\text{O}$ and the reaction in neutral medium is ${{\text{K}}_{2}}\text{C}{{\text{r}}_{2}}{{\text{O}}_{7}}+{{\text{H}}_{2}}\text{O}\to \text{2KCrO}_{4}^{2-}+2{{\text{H}}^{+}}$.
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