Question

Calculate the equivalent weight of $C{r_2}{O_7}^{2 - }$ in an acidic medium.

Answer 1

Try to recall that $C{r_2}{O_7}^{2 - }$ 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 step by step solution:
It is known to you that for a redox reaction, $Equivalent{\text{ }}weight = \dfrac{{{\text{Molecular weight}}}}{{number{\text{ }}of{\text{ }}electrons{\text{ }}lost{\text{ }}or{\text{ }}gained{\text{ }}in{\text{ }}redox{\text{ }}reaction}}$.
Dichromate ion ($C{r_2}{O_7}^{2 - }$) in an 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:
$C{r_2}{O_7}^{2 - } + 14{H^ + } + 6{e^ - } \to 2C{r^{3 + }} + 7{H_2}O$.
In the above reaction, you can see that 1 molecule of $C{r_2}{O_7}^{2 - }$ is releasing 6 electrons and molecular weight of $C{r_2}{O_7}^{2 - }$= $(2 \times 52) + (7 \times 16)$
= 216 g/mol.
Calculation of equivalent weight of $C{r_2}{O_7}^{2 - }$:
Molecular weight of = 216 g/mol
Number of electrons gained in acidic medium=6
Now, putting these values in above formula of equivalent weight:
$Equivalent{\text{ }}weight = \dfrac{{216}}{6} = 36g/eq$.

Hence, the equivalent weight of $C{r_2}{O_7}^{2 - }$ in an acidic medium will be 36g/eq.

Note:
1. It should be remembered to you that dichromate ion ($C{r_2}{O_7}^{2 - }$) and the chromate ion ($Cr{O_4}^{2 - }$) exist in equilibrium at pH=4. These are interconvertible by changing the pH of the solution.
2. Also, you should remember that in alkaline solution, chromate ions are present while in acidic solution, dichromate ions are present.
3. Potassium dichromate has also important uses in photography and in photographic screen painting.