Question

The atomic weight of silver and copper are $108$ and $64$ . A silver voltameter and a copper voltameter are connected in series and when current is passed $10.8$gm of silver is deposited. The mass of copper deposited will be
A. $6.4$ gm
B. $12.8$ gm
C. $3.2$ gm
D. $10.8$ gm

Answer 1

Hint: According to Faraday’s second law of electrolysis, When the same amount of electricity is passed through different electrolytes, the mass of the deposited substances is directly proportional to their respective chemical equivalent. We can use the second law of Faraday in electrolysis for the calculation of the mass of deposited copper.

Formula used: Mathematical expression of Faraday’s second law is:
$\dfrac{{{\text{m}}_{\text{1}}}}{{{\text{m}}_{\text{2}}}}\text{=}\dfrac{{{\text{E}}_{\text{1}}}}{{{\text{E}}_{\text{2}}}}$
Here ${{m}_{1}}\And {{m}_{2}}$are the mass of deposited substances $1\And 2$
${{E}_{1}}\And {{E}_{2}}$ are chemical equivalents of deposited substances $1\And 2$

Complete Step by Step Answer:
In the given question the values of the atomic weight of silver and copper are given. When current is passed silver is deposited, here we have to calculate the mass of deposited copper.
Chemical equivalent,$E=\dfrac{M}{{{n}_{f}}}$

Here $M=$ Atomic weight of deposited chemicals
${{n}_{f}}=$ n-factor
$A{{g}^{+}}(aq.)+{{e}^{-}}\to Ag(s)$
${{n}_{f}}$of $Ag$$=1$

Chemical equivalents of silver,${{E}_{Ag}}=\dfrac{{{M}_{Ag}}}{{{n}_{f}}}=\dfrac{108}{1}=108$
$C{{u}^{2+}}(aq.)+2{{e}^{-}}\to Cu(s)$
${{n}_{f}}$ of $Cu$$=2$
Chemical equivalents of copper, ${{E}_{Cu}}=\dfrac{{{M}_{Cu}}}{{{n}_{f}}}=\dfrac{64}{2}=32$

According to the second law of Faraday in electrolysis,
$\dfrac{{{m}_{Cu}}}{{{m}_{Ag}}}=\dfrac{{{E}_{Cu}}}{{{E}_{Ag}}}$
Or, ${{m}_{Cu}}={{m}_{Ag}}\times \dfrac{{{E}_{Cu}}}{{{E}_{Ag}}}$
Or, ${{m}_{Cu}}=10.8g\times \dfrac{32}{108}=3.2g$
Therefore, the mass of deposited copper is $3.2g$.
Thus, option (C) is correct.
Additional information: According to Faraday’s first law of electrolysis is that the chemical deposited caused by current flow through electrolysis is directly proportional to the quantity of electricity passing through the electrolyte. Therefore by using the first law we can also determine the amount of charge passing through the electrolyte and the number of liberated chemicals at any electrode.

Note: To approach this type of problem we should remember both laws of Faraday in electrolysis. We should know how to calculate chemical equivalents. Thus to find out the Chemical equivalent, the atomic weight of the substances is divided by their n-factor.