Question

When a copper voltmeter is connected with a battery of e.m.f. 12 volts then in 30 minutes 2 gm of copper is deposited. If the same voltmeter is connected across a 6 volt battery, then the mass of copper deposited in 45 minutes would be
A 1 gm
B 1.5 gm
C 2 gm
D 2.5 gm

Answer 1

Hint: In this question, we need to compare two situations in the first case when we applied 12 volts of voltage (can calculate current with help of resistance) in an electrolytic solution than the copper (cation) deposited at the cathode in 30 min and the amount of deposition is 2 gm. In another condition, we change the voltage supply and that is 6. In this case, we need to find the amount of mass deposited on the cathode of copper in 45 minutes. To find the mass of copper deposited in the second case can be derived with the help of the faraday law of electrolysis such as
\[M\text{ }\propto \text{ }Q\]
\[M\text{ }=\text{ }ZIt\]
where Z is electrochemical equivalent, I is current and t is time.

Complete Step by Step Answer:
The copper voltmeter is an instrument used to determine the current and electrochemical equivalent weight. The copper voltmeter consists of two electrodes of copper, one electrode is the anode (attached to the +ve terminal of the battery) and the other is the cathode (attached to the -ve terminal of the cell). Both electrodes are placed in a copper sulfate solution.

In the first case when the current started to flow the applied voltage is 12 volts the solution started to dissociate and the cation moved towards the cathode and the anion moved towards the anode. The copper is cation so it is stated to be deposited on the cathode electrode and the deposited copper amount is 2 gm. Now as per faraday law of electrolysis, the amount of deposition on the electrode (m) depends on the amount of charge passed (Q) to the electrolytic solution such as
\[m\text{ }\propto \text{ }Q\]
\[m=\text{ }Z\text{ }\left( It \right)\]as \[I\text{ }=\text{ }Q/t\]
Where Z is electrochemical equivalent
\[m\text{ }=\text{ }Z\text{ }\left( V/R \right)t\] as \[I\text{ }=\text{ }V/R\]
As we need to find the relationship between m, V, and t we can write it as
\[m\text{ }\propto \text{ }Vt\]
Now in the second case, \[{{m}_{1}}\]is not given but V changes from 12 volts to 6 volts (\[{{V}_{1}}\]) and t changes from 30 min to 45min (\[{{t}_{1}}\]) such as
\[{{m}_{1}}\text{ }\propto \text{ }{{V}_{1}}\text{ }{{t}_{1}}\]
Now by finding the ratio of m to \[{{m}_{1}}\], we will get the weight of copper deposited on the cathode in the second case such as
\[m/{{m}_{1}}\text{ }=\text{ }Vt/{{V}_{1}}{{t}_{1}}\]
\[2/{{m}_{1}}\text{ }=\text{ }\left( 12\text{ }\times \text{ }30 \right)/\left( 6\text{ }\times \text{ }45 \right)\]
\[{{m}_{1}}\text{ }=\text{ }1.5\text{ }gm\]
Thus, the correct option is B.

Note: A copper voltmeter is an electrolytic cell in which both electrodes are placed in a beaker in the same electrolytic solution (conducting solution) and also the electrode which are connected to the positive terminal of the battery are known as anode and with the negative terminal is the cathode electrode which is opposite to galvanic cell. In a galvanic cell, two electrodes are placed in different beakers, and the cathode and anode connection to the battery is opposite to the electrolytic cell.