Question

Determine the equivalent weight of $K_{2}C{r}_2{O}_7$ and $FeS{O}_4$ in the reaction.
     $$K_{2}C{r}_2{O}_7+7H_{2}S{O}_4+6FeS{O}_4\to C{r}_2{(S{O}_4)}_3+K_{2}S{O}_4+3F{e}_2{(S{O}_4)}_3+7H_{2}O$$

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:
     $$\text{Equivalent weight = }{\displaystyle \frac{\text{Molecular weight}}{\text{number of electrons lost or gained}}}$$

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:
     $$K_{2}C{r}_2{O}_7+7H_{2}S{O}_4+6FeS{O}_4\to C{r}_2{(S{O}_4)}_3+K_{2}S{O}_4+3F{e}_2{(S{O}_4)}_3+7H_{2}O$$
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.
     $$\text{Equivalent weight of }{\text{K}}_2\text{C}{\text{r}}_2{\text{O}}_7\text{= }{\displaystyle \frac{294gmo{l}^{-1}}{6eqmo{l}^{-1}}}=49ge{q}^{-1}$$
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:
     $$\text{Equivalent weight of FeS}{\text{O}}_4\text{= }{\displaystyle \frac{151.845gmo{l}^{-1}}{1eqmo{l}^{-1}}}=151.845ge{q}^{-1}$$
Hence, the equivalent weight of $K_{2}C{r}_2{O}_7$ and $FeS{O}_4$ in the reaction is $$49ge{q}^{-1}$$ and$$151.845ge{q}^{-1}$$ , 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 $FeS{O}_4$.