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}}](https://tex.z-dn.net/?f=%5Cfrac%7BE%20-%20V_%7Bt%7D%7D%7Br_%7Bint%7D%7D)
Putting the given values into the above formula as follows.
I = ![\frac{E - V_{t}}{r_{int}}](https://tex.z-dn.net/?f=%5Cfrac%7BE%20-%20V_%7Bt%7D%7D%7Br_%7Bint%7D%7D)
= ![\frac{12 - 10}{2}](https://tex.z-dn.net/?f=%5Cfrac%7B12%20-%2010%7D%7B2%7D)
= 1 A
Atomic weight of copper is 63.54 g/mol. Therefore, equivalent weight of copper is
.
That is, ![\frac{M}{2}](https://tex.z-dn.net/?f=%5Cfrac%7BM%7D%7B2%7D)
= ![\frac{63.54 g/mol}{2}](https://tex.z-dn.net/?f=%5Cfrac%7B63.54%20g%2Fmol%7D%7B2%7D)
Hence, electrochemical equivalent of copper is as follows.
Z = (\frac{E}{96500}) g/C
= (\frac{63.54 g/mol}{2 \times 96500}) g/C
=
g/C
Therefore, charge delivered from the battery in half-hour is calculated as follows.
It = Q
=
= 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](https://tex.z-dn.net/?f=3.29%20%5Ctimes%2010%5E%7B-4%7D%20g%2FC%20%5Ctimes%201800%20C)
= 0.5927 g
Thus, we can conclude that 0.5927 g of copper is deposited at the cathode in half an hour.