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

Step-by-step 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 =
`(E - V_(t))/(r_(int))`

Putting the given values into the above formula as follows.

I =
`(E - V_(t))/(r_(int))`

=
`(12 - 10)/(2)`

= 1 A

Atomic weight of copper is 63.54 g/mol. Therefore, equivalent weight of copper is
`(M)/(2)`.

That is,
`(M)/(2)`

=
`(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 * 10^(-4)` g/C

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

It = Q

=
`1 * (1)/(2) * 60 times 60`

= 1800 C

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

M = ZQ

=
`3.29 * 10^(-4) g/C * 1800 C`

= 0.5927 g

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