Question

The current inside the copper voltmeter
A. Is half the outside value
B. Is the same as the outside value
C. Is twice the outside value
D. Depends on the concentration of CuSO₄

Answer 1

Hint: The copper voltmeter is an instrument (containing two copper electrodes in the copper sulfate solution) used to determine the weight equivalent, mass and also current passed through the solution of the electrodes. To detect the current inside and outside the copper voltmeter we need to determine the mass of the electrode before and after the current passed because as per faraday law of electrolysis, increased or deposited mass or decreased mass is directly proportional to charge passed through the solution such as
\[M\text{ }\propto \text{ }Q\] or \[M\text{ }=\text{ }Zit\]
Where Z is electrochemical equivalent, i is current and t is time.

Complete Step by Step Answer:
The copper voltmeter as discussed in hind is an instrument used to measure the equivalent mass of copper metal electrodes or amount of electricity passed through the process of electrolysis. In this instrument, two electrodes of copper are placed in the copper sulfate, \[CuS{{O}_{4}}\] conducting solution (in one beaker). One electrode (Anode) of copper is attached to the positive terminal and the other (cathode) to the negative terminal.

To check the current inside the copper voltmeter we need to weigh both electrodes before and after the electricity is passed to the solution.
Let whatever the mass before passing electricity but as soon as we supply the electricity to apparatus solution of copper sulfate solution stated to dissociates into ions \[C{{u}^{2+}}\]and \[S{{O}_{4}}^{2-}\]. The negatively charged \[S{{O}_{4}}^{2-}\]started to move towards the anode (+ve) to get oxidize and reduce \[C{{u}^{2+}}\] (present at electrode) and the resultant compound \[CuS{{O}_{4}}\] goes into the solution. On the cathode (-ve), \[C{{u}^{2+}}\]started to deposit.

And as we determine the mass after electrolysis, again it has been noticed that mass before and after electricity passed is the same. In this way we can say there is no change in weight. And according to the Faraday law of electrolysis mass of electrodes change with a change in electricity or charge supply (\[M\text{ }\propto \text{ }Q\]). So we can say if the weight doesn’t change then the current will be the same inside as that outside.
Thus, the correct option B.

Note: In this apparatus, anode reduction occurs (or can say anode acts like an oxidizing agent) and cathode oxidation occurs or can say it acts like a reducing agent. As discussed above, the mass that is released from the copper plate (cathode) is the same as the mass that is deposited on the copper plate at the anode side thus this process follows the law of mass and as well as conservation of charge flow.