Question

250 ml of 1M $I_2$ is separately reacted with 0.5 M $C{u}_{2}S$ solution, 0.5 M CuS solution and 0.5 M $S_2{O}_3^{2-}$ solution, causing production of $C{u}^{+2}$ and $S{O}_4^{2-}$ in the first two and $S_4{O}_6^{2-}$ in the last case along with iodide ions. Which of the following option(s) is/are correct assuming 100% completion of the reaction?

• A. 100 ml of $C{u}_{2}S$ solution will be consumed.
• B. 125 ml of CuS solution will be consumed.
• C. 500 ml of $S_4{O}_6^2-$ solution will be consumed.
• D. Equivalent weight of $I_2$ in each of the reactions will be 127.

Answer 1

Answer: (A), (B), (D)

The correct options are
A

100 ml of $C{u}_{2}S$ solution will be consumed.

B

125 ml of CuS solution will be consumed.

D

Equivalent weight of $I_2$ in each of the reactions will be 127.

$\underset{(n_f=10)}{\stackrel{+1}{C{u}_2}\stackrel{-2}{S}}+\underset{(n_f=2)}{I_2}\to \stackrel{+2}{C{u}^{2+}}+\stackrel{+6}{S{O}_4^{2-}}+I^{-}$

Change in oxidation number of copper (I) sulphide is 10! 2 from two atoms going from Cu +1 to +2, +8 when sulphur goes from -2 to +6.

Similarly, two atoms of I go from 0 to -1, so that's a -2 there.
$\underset{(n_f=8)}{CuS}+\underset{(n_f=2)}{I_2}\to {\stackrel{+2}{Cu}}^{2+}+S{O}_4^{2-}+I^{-}$

Here only sulphur is getting oxide in copper(II) sulphide. So the change in oxidation number is + 8 for this compound. Iodine again goes from 0 to -1, two atoms in the iodine molecule, so that is a change of -2.
$\underset{(nf=2)}{I_2}+S_2\stackrel{+6}{O_3^{2-}}\to I^{-}\underset{(nf=2)}{S_4{O}_6^{2-}}$

Here iodine changes by -2 and sulphur by +2

In reaction (1):
Meq of $C{u}_{2}S$ = Meq of $I_2$
$0.5\times 10\times V=250\times 1\times 2$
V= 100 ml (of $C{u}_{2}S)$
In reaction (2):
Meq of CuS = Meq of $I_2$
$0.5\times 8\times V_{CuS}=1\times 2\times 250$; $V_{CuS}=125ml$
Eq.wt of $I_2=\frac{mol.wt}{n_f}=\frac{254}{2}=127$