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$H_2O_2$ 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 )

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

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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.
answered Feb 18, 2014 by sreemathi.v
 

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