Question

A current of electricity is passed through copper voltmeter and a water voltmeter connected in series. If the copper of the copper voltmeter now weighs $16\,mg$ less, hydrogen liberated at the cathode of the water voltmeter measures about (at STP).
A. $4.0\,ml$
B. $5.6\,ml$
C. $6.4\,ml$
D. $8.4\,ml$

Answer 1

Hint:We know that voltmeter is a device capable of measuring the potential differences generated between two points in an electric circuit. In a circuit the voltmeter is always connected in parallel to the circuit. Here, recall the faraday law of electricity.

Complete step by step solution:
Given weight of copper displaced$ = 16mg$
Here, the amount of the hydrogen liberated at the cathode has to be calculated. The reaction will take place in the following manner;
\[Cu \to C{u^{2 + }} + 2{e^\_}\]
From the mechanism of the reaction it is clear that $2F$ of electricity is needed for $63.5\,gm$ of copper. This is because 1 mole of Cu is participating here and the molar mass of Cu is $63.5\,gm$. But it is given that the amount of copper gets reduced by $16\,mg$ i.e.\[16 \times {10^{3 - }}{\rm{ }}gm\].
Moles of $C{u^{2 + }}$used$ = \dfrac{{16 \times {{10}^{ - 3}}}}{{64}} = 0.25 \times {10^{ - 3}}$
Number of equivalents of $C{u^{2 + }}$=$2 \times 0.25 \times {10^{ - 3}} = 0.5 \times {10^{ - 3}}$
\[2{H^ + } + 2{e^ - } \to {H_2}\]
$2 \times $moles of hydrogen$ = 0.5 \times {10^{ - 3}}$
So, moles of hydrogen$ = \dfrac{{0.5}}{2} \times {10^{ - 3}}moles$
We know that 1 mole of hydrogen at STP has a volume of $22.4\,L$
So,$0.25 \times {10^{ - 3}}$moles have volume$ = 22.4 \times 0.25 \times {10^{ - 3}}L = 5.6\,ml$

Hence, options (B) $5.6\,ml$, is the correct option.

Note:This is due to faraday law of electricity which states that when the same amount of electricity is passed through, then the different substances are directly proportional to their equivalent weights.