Question

A copper voltameter, a silver voltameter and a water voltameter are connected in series and current is passed through them for some time. The ratio of the number of moles of copper, silver and hydrogen formed at the cathode is.
A)$2:1:1$
B)$1:1:1$
C) $1:2:1$
D)$1:2:2$

Answer 1

Hint: We know that the charge of the element is the same as the current produced in the voltameter.
The number of moles of a species in the voltameter is given by,
$Number\,of\,moles = \dfrac{1}{{Number\,of\,electrons}}$

Complete step by step answer:
To find out the quantity of substance either produced or consumed during electrolysis given the time a known current flowed:
Write the balanced half-reactions involved.
Calculate the number of moles of electrons that were transferred.
Calculate the number of moles of substance that was produced or consumed at the electrode.
1) The balanced half cell reaction at cathode of copper voltameter is,
$C{u^{2 + }} + 2{e^ - }\xrightarrow{{}}Cu$
The number of moles of copper formed at cathode is calculated by using the formula as,
$Number\,of\,moles = \dfrac{1}{{Number\,of\,electrons}}$
At the cathode of copper voltameter, $2$ electrons are involved so the number of electrons is $2$. Substituting the value we get,
$Number\,of\,moles = \dfrac{1}{2}$
2) The balanced half cell reaction at cathode of silver voltameter is,
$A{g^ + } + {e^ - }\xrightarrow{{}}Ag$
The number of moles of silver formed at cathode is calculated by using the formula as,
$Number\,of\,moles = \dfrac{1}{{Number\,of\,electrons}}$
At a cathode of silver voltameter, only one electron is involved so the number of electrons is $1$. Substituting the value we get,
$Number\,of\,moles = 1$
3) The balanced half cell reaction at cathode of water voltameter is,
$2{H^ + } + 2{e^ - }\xrightarrow{{}}{H_2}$
The number of moles of silver formed at cathode is calculated by using the formula as,
$Number\,of\,moles = \dfrac{1}{{Number\,of\,electrons}}$
At a cathode of water voltameter, $2$ electrons are involved so the number of electrons is $2$. Substituting the value in formula we get,
$Number\,of\,moles = \dfrac{1}{2}$
The ratio of the number of moles of copper, silver and hydrogen formed at the cathode is $1:2:1$

So, the correct answer is Option C .

Note:
We know that the amount of moles in a given amount of any substance is equal to the grams of the substance divided by its molecular weight.
The mathematically expressed as,
$Mole = \dfrac{{weight\,of\,the\,subs\tan ce}}{{\,Molecular\,weight}}$.