Question

Calculate concentration of $H_2{O}_2$ in g/litre in terms of percentage.

• A. $6.8\%$
• B. $3.4$%
• C. $68\%$
• D. $34\%$

Answer 1

The correct option is A $6.8\%$

The redox changes are as follows:
$\left(a\right)O_2^{-}+2e⟶2O^{2-}$
$2I^{-}⟶I_2^0+2e$
$\left(b\right)2(S^{2+}{)}_2⟶(S^{5/2+}{)}_4+2e$
$2e+I_2^0⟶2I^{-}$

The complete reaction is given below:
$H_2{O}_2\stackrel{KI}{\to }{I}_2+H_{2}O\stackrel{N{a}_2{S}_2{O}_3}{\to }N{a}_2{S}_4{O}_6+I^{-}$
Meq. of $H_2{O}_2=$ Meq. of $KI$ used $=$ Meq. of $I_2$ liberated $=$ Meq. of $N{a}_2{S}_2{O}_3$ used
Meq. of $H_2{O}_2=$ Meq. of $N{a}_2{S}_2{O}_3$ used
$N\times 50=20\times 0.1$
$\therefore N_{H_2{O}_2}=0.04$
Strength of $H_2{O}_2=N\times E=0.04\times \left(\frac{34}{2}\right)$ $=0.68$ g/L
Percentage of $H_2{O}_2=6.8\%$