Question

Determine the equivalent weight of ${{K}_{2}}C{{r}_{2}}{{O}_{7}}$ and $FeS{{O}_{4}}$ in the reaction.
     \[{{K}_{2}}C{{r}_{2}}{{O}_{7}}+7{{H}_{2}}S{{O}_{4}}+6FeS{{O}_{4}}\to C{{r}_{2}}{{(S{{O}_{4}})}_{3}}+{{K}_{2}}S{{O}_{4}}+3F{{e}_{2}}{{(S{{O}_{4}})}_{3}}+7{{H}_{2}}O\]

Answer 1

Hint: Equivalent weight of a compound refers to the mass of the substance that reacts exactly with an arbitrarily fixed amount of another compound. Equivalent weight of a compound is given as:
     \[\text{Equivalent weight = }\dfrac{\text{Molecular weight}}{\text{number of electrons lost or gained}}\]

Complete answer:
To determine the equivalent weight of potassium dichromate (${{K}_{2}}C{{r}_{2}}{{O}_{7}}$) and ferrous sulphate ($FeS{{O}_{4}}$), we first need to find out the number of gained and lost by ${{K}_{2}}C{{r}_{2}}{{O}_{7}}$ and $FeS{{O}_{4}}$, respectively.
The given chemical reaction:
     \[{{K}_{2}}C{{r}_{2}}{{O}_{7}}+7{{H}_{2}}S{{O}_{4}}+6FeS{{O}_{4}}\to C{{r}_{2}}{{(S{{O}_{4}})}_{3}}+{{K}_{2}}S{{O}_{4}}+3F{{e}_{2}}{{(S{{O}_{4}})}_{3}}+7{{H}_{2}}O\]
Let us first calculate the equivalent weight of ${{K}_{2}}C{{r}_{2}}{{O}_{7}}$.
Change in oxidation state of Cr in reactant and product side will give us the number of electrons gained by ${{K}_{2}}C{{r}_{2}}{{O}_{7}}$.
Oxidation state of Cr in ${{K}_{2}}C{{r}_{2}}{{O}_{7}}$ is +6.
Oxidation state of Cr in \[C{{r}_{2}}{{(S{{O}_{4}})}_{3}}\] is +3
Change in oxidation state of Cr will then be, 6-3 = 3
Therefore, the number of gained by one Cr = 3
One molecule of \[{{K}_{2}}C{{r}_{2}}{{O}_{7}}\] has two Cr atoms. Then, the total number of electrons gained by two Cr atoms = $2\times 3=6$
Molecular weight of ${{K}_{2}}C{{r}_{2}}{{O}_{7}}$ is calculated as: $39\times 2+52\times 2+16\times 7=294gmo{{l}^{-1}}$
Since, 6 electrons are gained per mole of ${{K}_{2}}C{{r}_{2}}{{O}_{7}}$, thus, the equivalent weight of ${{K}_{2}}C{{r}_{2}}{{O}_{7}}$ will now be calculated by dividing its molecular weight by 6.
     \[\text{Equivalent weight of }{{\text{K}}_{2}}\text{C}{{\text{r}}_{2}}{{\text{O}}_{7}}\text{= }\dfrac{294g\,mo{{l}^{-1}}}{6\,eq\,mo{{l}^{-1}}}=49g\,e{{q}^{-1}}\]
Similarly, let us calculate the change in oxidation number of Fe from $FeS{{O}_{4}}$ to \[F{{e}_{2}}{{(S{{O}_{4}})}_{3}}\]
Oxidation state of Fe in $FeS{{O}_{4}}$ is + 2.
Oxidation state of Fe in \[F{{e}_{2}}{{(S{{O}_{4}})}_{3}}\] is +3
Then, the change in oxidation state of Fe will then be, 3-2= 1
Therefore, the number of electrons lost by Fe atom = 1
Molecular weight of $FeS{{O}_{4}}$ is calculated as: \[55.845+32+16\times 4=151.845gmo{{l}^{-1}}\]
We can now find the equivalent weight of \[FeS{{O}_{4}}\] as:
     \[\text{Equivalent weight of FeS}{{\text{O}}_{4}}\text{= }\dfrac{151.845g\,mo{{l}^{-1}}}{1eq\,mo{{l}^{-1}}}=151.845g\,e{{q}^{-1}}\]
Hence, the equivalent weight of ${{K}_{2}}C{{r}_{2}}{{O}_{7}}$ and $FeS{{O}_{4}}$ in the reaction is \[49g\,e{{q}^{-1}}\] and\[151.845g\,e{{q}^{-1}}\] , respectively.

Note: I
n the given reaction oxidation state of Cr is decreasing from +6 to +3, that means ${{K}_{2}}C{{r}_{2}}{{O}_{7}}$ is undergoing reduction and thus, acting as an oxidizing agent. Conversely, the oxidation state of Fe is increasing from +2 to +3, so $FeS{{O}_{4}}$ is oxidized. We can say that one equivalent of ${{K}_{2}}C{{r}_{2}}{{O}_{7}}$ is oxidizing six equivalent of $FeS{{O}_{4}}$.