Question

The potential difference across the terminals of a battery of e.m.f. 12 V and internal resistance 2 ohm drops to 10 V when it is connected to a Copper voltameter. Calculate the Copper deposited at the cathode in half an hour. Atomic weight of Copper is 63.546 g mol-1

Answer 1

Explanation:

The given data is as follows.

      E.m.f = 12 V,        Voltage = 10 V,        Resistance = 2 ohm

Hence, calculate the current as follows.

          I = $\frac{E - V_{t}}{r_{int}}$

Putting the given values into the above formula as follows.

         I = $\frac{E - V_{t}}{r_{int}}$

           = $\frac{12 - 10}{2}$

           = 1 A

Atomic weight of copper is 63.54 g/mol. Therefore, equivalent weight of copper is $\frac{M}{2}$.

That is,          $\frac{M}{2}$

                = $\frac{63.54 g/mol}{2}$

Hence, electrochemical equivalent of copper is as follows.

                   Z = (\frac{E}{96500}) g/C

                       = (\frac{63.54 g/mol}{2 \times 96500}) g/C

                       = $3.29 \times 10^{-4}$ g/C

Therefore, charge delivered from the battery in half-hour is calculated as follows.

                     It = Q

                       = $1 \times \frac{1}{2} \times 60 times 60$  

                       = 1800 C

So, copper deposited at the cathode in half-an-hour is as follows.

                     M = ZQ

                = $3.29 \times 10^{-4} g/C \times 1800 C$

                = 0.5927 g

Thus, we can conclude that 0.5927 g of copper is deposited at the cathode in half an hour.