Question

Calculate equivalent weight of ${K_2}C{r_2}{O_7}$ in acidic medium.

Answer 1

Hint: Try to recall that ${K_2}C{r_2}{O_7}$ is an oxidizing agent in acidic medium and the equivalent weight of an oxidizing or reducing agent is equal to the molecular weight divided by the number of electrons lost or gained by one molecule of the substance in a redox reaction.
Complete answer:
It is known to you that for a redox reaction, \[Equivalent{\text{ }}weight = \dfrac{{{\text{Molecular weight}}}}{{number{\text{ }}of{\text{ }}electrons{\text{ }}lost{\text{ }}or{\text{ }}gained{\text{ }}in{\text{ }}redox{\text{ }}reaction}}\].
Potassium dichromate (${K_2}C{r_2}{O_7}$ in acidic medium is a versatile, powerful oxidizing agent. It means it gains electrons during a redox reaction. The reaction which takes place when potassium dichromate (${K_2}C{r_2}{O_7}$) is in acidic medium is as follows:
\[{K_2}C{r_2}{O_7} + 14{H^ + } + 6{e^ - } \to 2{K^ + } + 2C{r^{3 + }} + 7{H_2}O\]
In the above reaction, you can see that 1 molecule of ${K_2}C{r_2}{O_7}$ is releasing 6 electrons and molecular weight of ${K_2}C{r_2}{O_7}$= $(2 \times 39) + (2 \times 52) + (7 \times 16)$
                                                       = 294 g/mol.
Calculation of equivalent weight of ${K_2}C{r_2}{O_7}$:
Molecular weight of = 294 g/mol
Number of electrons gained in acidic medium=6
Now, putting these values in above formula of equivalent weight:
\[Equivalent{\text{ }}weight = \dfrac{{294}}{6} = 49g/eq\].

Hence, the equivalent weight of ${K_2}C{r_2}{O_7}$ in acidic medium will be 49g/eq.

Note: It should be remembered to you that potassium dichromate has a wide range of uses, including as an oxidizer in many chemical and industrial applications and in dying, staining and tanning of leather.
Also, you should remember that nowadays they are being replaced with processes using trivalent chromium because of the toxicity of hexavalent chrome.
Potassium dichromate has also important uses in photography and in photographic screen painting.