Question

The chemical equivalent of copper and silver are 32 and 108 respectively. When copper and silver voltmeters are connected in series and an electric current is passed through for some time, 1.6 g of copper is deposited. Then, the mass of silver deposited will be
A. 3.5 g
B. 2.8 g
C. 5.4 g
D. None of these

Answer 1

Hint: When two electrical components are connected in series then the electricity in both components are same. As electricity is defined as the flow of charge, so, is the amount of the charge transferred to copper and silver. We use faraday’s first law of electrolysis to determine the amount of silver deposited.

Formula Used:\[m = Zit\], where m is the mass of the ion deposited, Z is the electrochemical equivalent, i is the electric current and t is the time and Z is the electric equivalence of the electrolyte.

Complete answer:As the copper and silver voltammeter are connected in series, so the charges to each will be the same. Let the equal electric current to copper and silver be I for the same interval of time.
As we know that the mass of the ion deposited is proportional to the current passing through the electrolyte and the time duration using Faraday’s first law of electrolysis,
\[m = Zit\]
We have given the chemical equivalence of copper and silver as 32 and 108.
\[{E_{Cu}} = 32\]
\[{E_{Ag}} = 108\]
Then the electrochemical equivalence can be determined as,
\[Z = \dfrac{E}{F}\]
For the copper and silver branches respectively,
\[{m_{Cu}} = {Z_{Cu}}it\]
And,
\[{m_{Ag}} = {Z_{Ag}}it\]
The amount of copper deposited is given as 1.6g.
On dividing the expression for the masses of the ion deposited, we get
\[\dfrac{{{m_{Cu}}}}{{{m_{Ag}}}} = \dfrac{{{Z_{Cu}}it}}{{{Z_{Ag}}it}}\]
\[\dfrac{{{m_{Cu}}}}{{{m_{Ag}}}} = \dfrac{{\left( {\dfrac{{{E_{Cu}}}}{F}} \right)}}{{\left( {\dfrac{{{E_{Ag}}}}{F}} \right)}}\]
\[\dfrac{{{m_{Cu}}}}{{{m_{Ag}}}} = \dfrac{{{E_{Cu}}}}{{{E_{Ag}}}}\]
\[{m_{Ag}} = \left( {\dfrac{{{E_{Ag}}}}{{{E_{Cu}}}}} \right){m_{Cu}}\]
\[{m_{Ag}} = \left( {\dfrac{{108}}{{32}}} \right) \times 1.6g\]
\[{m_{Ag}} = 5.4g\]
So, the mass of silver deposited is 5.4g.
Therefore,

the correct option is (C).

Note: We should be careful about how the copper and silver voltammeter are arranged in the circuit. If the circuit is series then the current in both will be the same and if the circuit is in parallel then the current in both will be different.