# $H_2O_2$ can be prepared by successive reactions

$2NH_4HSO_4\rightarrow H_2+(NH_4)_2S_2O_8$

$(NH_4)_2S_2O_8+2H_2O\rightarrow 2NH_4HSO_4+H_2O_2$.

The first is an electrolytic reaction and the second a steam distillation.What current needs to be passed in first reaction to produced enough intermediate to yield 100gm of pure $H_2O_2$ per hour? ( Assume 50% current efficiency )

$\begin{array}{1 1}(a)\;342A\\(b)\;315A\\(c)\;300A\\(d)\;320A\end{array}$

100gm of pure $H_2O_2$ will be $\large\frac{100}{34}$ moles of $H_2O_2$
$\Rightarrow$Moles of $(NH_4)_2S_2O_8$ that would have reacted =$\large\frac{100}{34}$
$\Rightarrow$Moles of $(NH_4)_2S_2O_8$ that needs to be formed by first reaction=$\large\frac{100}{34}$
$\Rightarrow$Moles of $(NH_4)_2S_2O_8$ that needs to react=$\large\frac{100}{34}$$\times 2=\large\frac{100}{7} When 1 mole of NH_4HSO_4 converts to (NH_4)_2S_2O_8,2 mole of electrons would be liberated.This is because in NH_4HSO_4 all the oxygen atoms have an oxidation state of -2.In (NH_4)_2S_2O_8 two oxygen atoms are linked to each other because which their oxidation state becomes -1. \therefore Two oxygen atoms have got oxidized from -2 to -1 state.So when 2 moles of NH_4HSO_4 reacts,2 moles of electrons are liberated. Moles of electrons given by \large\frac{100}{17} of NH_4HSO_4=\large\frac{100}{17}moles Charge=\large\frac{100}{27}$$\times 96500=567647.05$coulombs
Current with 50% efficiency $=\large\frac{567647.05}{60\times 60}\times \frac{100}{50}$
$\Rightarrow 315.36A$
Hence (b) is the correct answer.