Question

In the standardization of $N{a_2}{S_2}{O_3}$ using ${K_2}C{r_2}{O_7}$ by iodometry, the equivalent weight of ${K_2}C{r_2}{O_7}$ is:
(A) \[\dfrac{{Molecular\,\,weight}}{2}\]
(B) $\dfrac{{Molecular\,\,weight}}{6}$
(C) $\dfrac{{Molecular\,\,weight}}{3}$
(D) $Molecular\,\,weight$

Answer 1

Hint: Equivalent weight of an oxidising or reducing agent is equal to its molecular weight divided by the number of electrons gained or lost by it. In the standardization of $N{a_2}{S_2}{O_3}$using ${K_2}C{r_2}{O_7}$ by iodometry taking place in acidic medium, ${K_2}C{r_2}{O_7}$ gains 6 electrons.

Formula used: $Equivalent\,\,weight = \dfrac{{Molecular\,\,weight}}{{No.\,\,of\,electrons\,\,gained\,\,or\,\,lost\,\,per\,\,mole}}$

Complete step by step solution:
Iodimetric titrations refers to the titrations involving iodine liberated with a standard solution of sodium thiosulfate in a chemical reaction.
Iodine is a weak oxidant and in the presence of excess iodide ion, iodine is converted to tri-iodide ion. So, when a strong oxidising agent like ${K_2}C{r_2}{O_7}$ is treated with an excess of iodide ion (${I^{\, - \,}}$) in acidic medium, the iodide ions react as reducing agent and it is quantitatively oxidised to iodine (${I_2}$) by the oxidant. Thus, an equivalent amount of iodine is liberated. It is then titrated with a standard solution of reducing agent such as sodium thiosulfate ($N{a_2}{S_2}{O_3}$), which quantitatively reduces iodine to iodide and itself gets oxidised to sodium tetrathionate ($N{a_2}{S_4}{O_6}$).
$
  C{r_2}O_7^{2 - } + 14{H^ + } + 6I{\,^ - } \to 2C{r^{3 + }} + 7{H_2}O + 3{I_2} \\
  {I_2} + 2{S_2}O_3^{2 - } \to 2I{\,^ - } + {S_4}O_6^{2 - } \\
  $
We know that, $Equivalent\,\,weight = \dfrac{{Molecular\,\,weight}}{{No.\,\,of\,electrons\,\,gained\,\,or\,\,lost\,\,per\,\,mole}}$
Now, from the above equations we can observe that $C{r_2}O_7^{2 - }$(in which $Cr$is in +6 oxidation state) is gaining 6 electrons and getting reduced to $C{r^{3 + }}$.
Thus, $Equivalent\,\,weight\,of\,\,{K_2}C{r_2}{O_7} = \dfrac{{Molecular\,\,weight}}{6}$

Hence, option (B) is the correct answer.

Note: ${K_2}C{r_2}{O_7}$ acts as a powerful oxidising agent in acidic medium only and not in basic medium.
So, in acidic medium, $Equivalent\,\,weight\,of\,\,{K_2}C{r_2}{O_7} = \dfrac{{Molecular\,\,weight}}{6} = \dfrac{{294}}{6} = 49$.
In iodometric titrations, strong reducing agents such as tin (II) chloride, sulphurous acid, hydrogen sulphide and sodium thiosulphate, etc., react completely with iodine. Weaker reducing agents such as arsenic (III) or antimony (III) ions react only in neutral or very faintly acidic conditions.