First slide

Faraday's Laws

Question

According to 2nd law of Faraday’s electrolysis the correct one is
$\large i)\,\frac{{wt.\,of\,{H_2}\,liberated}}{{wt.\,of\,C{l_2}\,liberated}}\, = \,\frac{{eq.\,wt.of\,{H_2}}}{{eq.\,wt.\,of\,C{l_2}}}$
$\large ii)\,\frac{{{m_{Ag}}}}{{{m_{Cu}}}}\, = \,\frac{{{E_{Ag}}}}{{{E_{Cu}}}}$
$\large iii)\,\frac{{{m_{Ag}}}}{{{m_{Cu}}}}\, = \,\frac{{{E_{Cu}}}}{{{E_{Ag}}}}$
$\large iv)\,\frac{{{m_{{H_2}}}}}{{{m_{Cu}}}}\, = \,\frac{{{E_{{H_2}}}}}{{{E_{Cu}}}}$
The correct combination is

Easy

A

only ii, iv
B

only i
C

only i, ii , iv
D

only ii, iii

Solution

According to 2nd law of Faraday:
Mass of substance liberated α Equivalent weight of substance Liberated.

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