Question

E.C.E. of Cu and Ag are \[7 \times {10^{ - 6}}\]and \[1.2 \times {10^{ - 6}}\]. A certain current deposits 14 gm of Cu. The amount of Ag deposited is
A. 1.2 gm
B. 1.6 gm
C. 2.4 gm
D. 1.8 gm

Answer 1

Hint: The copper and Silver will get deposited on the cathode by liberating 2 and 1 electron respectively from its valence shell. The amount of the ion deposited is determined using Faraday’s 1st law of electrolysis.

Formula Used:\[m = Zit\], where m is the mass of the ion deposited, i is the current through the electrolyte and t is the time taken.

Complete answer:It is given that 14 gm of copper is deposited on the cathode,
\[m = 0.99g\]
The electrochemical equivalent of copper is given as \[7 \times {10^{ - 6}}\]
\[{Z_{Cu}} = 7 \times {10^{ - 6}}\]
The electrochemical equivalent of Silver is given as \[1.2 \times {10^{ - 6}}\]
\[{Z_{Ag}} = 1.2 \times {10^{ - 6}}\]
The current through both electrolytes is the same and for the same period of time.
Using Faraday’s law of electrolysis, the amount of Cu and Ag deposited on the cathode is,
\[{m_{Cu}} = {Z_{Cu}}it \ldots \ldots \left( i \right)\]
\[{m_{Ag}} = {Z_{Ag}}it \ldots \ldots \left( {ii} \right)\]
On dividing the first equation by the second, we get
\[\dfrac{{{m_{Cu}}}}{{{m_{Ag}}}} = \dfrac{{{Z_{Cu}}it}}{{{Z_{Ag}}it}}\]
\[\dfrac{{{m_{Cu}}}}{{{m_{Ag}}}} = \dfrac{{{Z_{Cu}}}}{{{Z_{Ag}}}}\]
Putting the mass of the copper deposited in the equation and the electric equivalence of Copper and Silver, we get the amount of the Ag to be deposited on the cathode,
\[\dfrac{{14gm}}{{{m_{Ag}}}} = \dfrac{{7 \times {{10}^{ - 6}}}}{{1.2 \times {{10}^{ - 6}}}}\]
\[{m_{Ag}} = \left( {\dfrac{{1.2 \times {{10}^{ - 6}}}}{{7 \times {{10}^{ - 6}}}}} \right) \times 14gm\]
\[{m_{Ag}} = 2.40gm\]
So, 2.4 gm of Ag is deposited on the cathode on the same amount of current is passed through the electrolyte of silver.
Therefore,

the correct option is (C).

Note: We should be careful about the quantities used in Faraday’s law of electrolysis. As it is given that the current in both the electrolytes are same but the time period is not given, so we should assume the time period to be the same as we need to find the ratio of the masses deposited.