Question

A volume of 10.0ml of 1M $Se{O_2}$ reacted with exactly 20ml of 2M $CrS{O_4}$. In the reaction, $C{r^{ + 2}}$ was oxidized to $C{r^{ + 3}}$.To what oxidation state was selenium converted by the reaction?

Answer 1

Hint: Firstly, we should write the chemical reaction and try to analyze which element is oxidized and reduced. In order to know the number of electrons gained or lost can be known easily by writing half a reaction.

Complete answer:
Firstly, we should write the chemical equation involved in the given question:
Let x be the unknown oxidation number of selenium in the product side.
\[Se{O_2} + CrS{O_4} \to C{r^{ + 3}} + S{e^{ + x}}\]
The reaction is a redox reaction:
- \[C{r^{ + 2}}\xrightarrow{{}}C{r^{^{ + 3}}} + {e^ - }\] (oxidation)
- \[S{e^{ + 4}} + ({x^{ - 4}}){e^ - }\xrightarrow{{}}S{e^{x + }}\] (reduction)
Therefore, it is a redox reaction.
The formula to calculate the equivalent of oxidizing or reducing agents.
Equivalent of reducing /oxidizing agent = Molarity $ \times $ volume used $ \times $ (No. electrons lost/gained)
We can solve this with the help of the mole concept. The equivalent of reducing agent is equal to the equivalent of the oxidizing agent.
\[Equivalent{\text{ }}of\;\;Se{O_2} = Equivalent{\text{ }}of{\text{ }}CrS{O_4}\]
\[Molarity \times volume{\text{ }}used\; \times No.{\text{ }}electrons{\text{ }}gained = Molarity \times volume{\text{ }}used \times No.{\text{ }}electrons{\text{ }}gained\]\[1 \times \dfrac{{10}}{{1000}} \times (4 - x) = 2 \times \dfrac{{20}}{{1000}} \times 1\]
\[x = 0\]
Thus, the oxidation state of selenium is 0 in the product side.

Note:
Always be careful with calculation. Mostly in these kinds of questions the volume will be given in terms of milliliter. So, make sure that the volume is converted to in terms of liters. Write the half reactions in order to not make mistakes in calculating the number of electrons gained or lost.