Question

Two voltammeters, one of copper and another of silver, are joined in parallel. When a total charge q flows through the voltammeters, an equal amount of metals is deposited. If the electrochemical equivalents of copper and silver are \[{Z_1}\]and \[{Z_2}\] respectively the charge which flows through the silver voltammeter is
A. \[q\left( {\dfrac{{{Z_1}}}{{{Z_2}}}} \right)\]
B. \[q\left( {\dfrac{{{Z_2}}}{{{Z_1}}}} \right)\]
C. \[\dfrac{q}{{1 + \left( {\dfrac{{{Z_1}}}{{{Z_2}}}} \right)}}\]
D. \[\dfrac{q}{{1 + \left( {\dfrac{{{Z_2}}}{{{Z_1}}}} \right)}}\]

Answer 1

Hint: When two electrical components are connected in parallel then the electricity gets divided among the branches based on the resistances of the branches. As the electricity is the flow of charge, so the charges get divided amount the copper branch and the silver branch. We use faraday’s first law of electrolysis to determine the amount of charge flown.

Formula Used:\[m = Zit\], where m is the mass of 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:It is given that the total amount of charge flown to the combination of the copper and silver branch is q.
Let the charge flown to the copper is \[q1\]and the charge flown to the silver branch is \[{q_2}\]
Using conservation of charge,
\[{q_1} + {q_2} = q\]
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,
\[m = Zit\]
Using the definition of electric current, the electric current is the flow rate of charge,
\[i = \dfrac{q}{t}\]
\[q = it\]
So, Faraday’s law becomes,
\[m = Zq\]
For the copper and silver branches respectively,
\[{m_{Cu}} = {Z_1}{q_1}\]
\[{q_1} = \dfrac{{{m_{Cu}}}}{{{Z_1}}}\]
And,
\[{m_{Ag}} = {Z_1}{q_2}\]
\[{q_2} = \dfrac{{{m_{Ag}}}}{{{Z_2}}}\]
Putting in the charge equation,
It is given that the mass of the copper and silver deposited is equal, i.e. \[{m_{Cu}} = {m_{Ag}}\]
On dividing the charges, we get
\[\dfrac{{{q_1}}}{{{q_2}}} = \dfrac{{{Z_2}}}{{{Z_1}}}\]
\[{q_1} = \left( {\dfrac{{{Z_2}}}{{{Z_1}}}} \right){q_2}\]
Putting in the charge equation, we get
\[\left( {\dfrac{{{Z_2}}}{{{Z_1}}}} \right){q_2} + {q_2} = q\]
\[\left( {\left( {\dfrac{{{Z_2}}}{{{Z_1}}}} \right) + 1} \right){q_2} = q\]
\[{q_2} = \dfrac{q}{{\left( {\dfrac{{{Z_2}}}{{{Z_1}}}} \right) + 1}}\]
\[{q_2}\]is the charge through the silver.
Therefore,

the correct option is (D).

Note: We should be careful about the combination in which different voltammeter is connected. If they are connected in series then the charge flown through both will be the same and the mass deposited will be different.