Question

Calculate the equivalent mass of $$HCl$$ in the given reaction: $$K_{2}C{r}_2{O}_7+14HCl\to 2KCl+2CrC{l}_3+3C{l}_2+H_{2}O$$.

Answer 1

Hint: The given reaction between hydrochloric acid and potassium dichromate is a redox reaction. The equivalent mass of hydrogen chloride can be calculated by knowing its molecular mass and its $$n-factor$$ in this particular reaction.

Complete answer:
The $$n-factor$$ is a number that measures the change in electrons taking place per molecule. The electrons in the process can be gained (reduction) as well as lost (oxidation). But in a balanced redox reaction we observe that the number of electrons being gained or lost as a result of oxidation or reduction are the same.
In the reaction between hydrochloric acid and potassium dichromate, the dichromate ion is being reduced to $$C{r}^{3+}$$ ions and the chloride ions are being oxidized into diatomic chlorine gas molecules with a zero oxidation state. The balanced redox reaction can be written as follows:

In order to find out the $$n-factor$$ of $$HCl$$, we need to break the reaction into two halves where one represents the oxidation half and the other represents the reduction half. Each half must be written with the number of electrons involved in the process in such a way that the oxidation numbers get balanced.
a.Oxidation
$$2C{l}^{-}\to C{l}_2+2e^{-}$$
b.Reduction
$$Cr(VI)\to C{r}^{3+}+3e^{-}$$
The electrons gained and lost in a balanced redox reaction must be equal. Hence the oxidation reaction is multiplied by three and the reduction reaction is multiplied by two.
$$6C{l}^{-}\to 3C{l}_2+6e^{-}$$
$$2Cr(VI)\to 2C{r}^{3+}+6e^{-}$$
Thus there are six electrons being involved in the redox reaction for fourteen $$HCl$$ molecules (the stoichiometric number of $$HCl$$ in the balanced reaction).
The formula of $$n-factor$$ calculation is given as follows:
$$n-factor={\displaystyle \frac{\text{number of electrons involved}}{\text{number of molecules involved}}}$$
Putting the number of involved electrons as six and number of involved molecules as fourteen in the above formula we get,
$$n-factor={\displaystyle \frac{6}{14}}$$
The equivalent mass can be calculated by finding out the ratio of the molar mass and $$n-factor$$ of hydrogen chloride
$$\text{equivalent mass}={\displaystyle \frac{\text{molecular mass of HCl}}{n-factor}}$$
$$\text{equivalent mass}={\displaystyle \frac{36.5}{{\displaystyle \frac{6}{14}}}}$$
Which when simplified gives,
$$\text{equivalent mass HCl}=85.17amu$$
Hence, the equivalent mass of $$HCl$$ in the given reaction is $$85.17amu$$.

Note:
 The oxidation numbers assigned to atoms in the chemical equation are hypothetical numbers that represent the number of electrons donated by or accepted by that atom in making bonds with the atoms present in the molecule.