Question

The equivalent weight of ${{\text{K}}_{\text{2}}}{\text{C}}{{\text{r}}_{\text{2}}}{{\text{O}}_{\text{7}}}$ in acidic medium is equal to:
A ) molecular weight
B ) 1/2 molecular weight
C ) 1/6 molecular weight
D ) 1/5 molecular weight

Answer 1

The equivalent weight of an oxidising/reducing agent is the ratio of the molecular weight of the oxidising agent/reducing agent to the number of electrons that participate in the oxidation / reduction half reaction. Determine the number of electrons that participate in the reaction and the change in the oxidation number of chromium atoms during the reaction.

Complete answer:
In acidic medium, dichromate ion is reduced to ${\text{C}}{{\text{r}}^{3 + }}$ ion
Write the balanced chemical equation

${\text{C}}{{\text{r}}_{\text{2}}}{\text{O}}_7^{2 - }{\text{ + 14 }}{{\text{H}}^ + }{\text{ + 6 }}{{\text{e}}^ - }{\text{ }} \to {\text{ 2 C}}{{\text{r}}^{3 + }}{\text{ + 7 }}{{\text{H}}_2}{\text{O}}$

In the above reaction, six electrons participate.
Let X be the oxidation number of chromium in dichromate ion. The oxidation number of oxygen is -2. The sum of oxidation numbers of all the atoms in an ion is equal to the charge on an ion.

$${\text{2x + 7}}\left( { - 2} \right){\text{ = }} - {\text{2 }} \\\ {\text{2x}} - {\text{14 = }} - {\text{2}} \\\ {\text{2x = 12}} \\\ {\text{x = + 6}} \\\$$

The oxidation number of chromium atom in dichromate ion is +6. The oxidation number of chromium atom in ${\text{C}}{{\text{r}}^{3 + }}$ ion is +3.
During the reaction, the decrease in the oxidation number of one chromium atom is from +6 to +3. The decrease in the oxidation number of one chromium atom is $6 - 3 = 3$ . For two chromium atoms of dichromate ion, the total decrease in the oxidation number is $2 \times 3 = 6$ .
The equivalent weight of ${{\text{K}}_{\text{2}}}{\text{C}}{{\text{r}}_{\text{2}}}{{\text{O}}_{\text{7}}}$ in acidic medium is given by the following relationship.

$${\text{The equivalent weight =}}\dfrac{{{\text{Molecular weight}}}}{{{\text{Number of electrons participating in the reaction}}}} \\\ {\text{The equivalent weight =}}\dfrac{{{\text{Molecular weight}}}}{{{\text{Total change in the oxidation number for two Cr atoms}}}} \\\ {\text{The equivalent weight = }}\dfrac{{{\text{molecular weight}}}}{6} \\\$$

The equivalent weight of ${{\text{K}}_{\text{2}}}{\text{C}}{{\text{r}}_{\text{2}}}{{\text{O}}_{\text{7}}}$ in acidic medium is equal to 1/6 molecular weight.

Hence, the option C ) is the correct answer.

Additional information: The molecular weight of potassium dichromate is $294{\text{ }}g/mol$ . Hence, the equivalent weight of potassium dichromate in acidic medium is $\dfrac{{{\text{molecular weight}}}}{6} = \dfrac{{294}}{6} = 49$.

Note: During conversion of dichromate ion to chromium(III) ion, six electrons are transferred. The oxidation number of chromium decreases from +6 to +3. For one chromium atom, the decrease in the oxidation number is 3. For two chromium atoms, total decrease in the oxidation number is 6.