Respuesta :
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 = [tex]\frac{E - V_{t}}{r_{int}}[/tex]
Putting the given values into the above formula as follows.
I = [tex]\frac{E - V_{t}}{r_{int}}[/tex]
= [tex]\frac{12 - 10}{2}[/tex]
= 1 A
Atomic weight of copper is 63.54 g/mol. Therefore, equivalent weight of copper is [tex]\frac{M}{2}[/tex].
That is, [tex]\frac{M}{2}[/tex]
= [tex]\frac{63.54 g/mol}{2}[/tex]
Hence, electrochemical equivalent of copper is as follows.
Z = (\frac{E}{96500}) g/C
= (\frac{63.54 g/mol}{2 \times 96500}) g/C
= [tex]3.29 \times 10^{-4}[/tex] g/C
Therefore, charge delivered from the battery in half-hour is calculated as follows.
It = Q
= [tex]1 \times \frac{1}{2} \times 60 times 60[/tex]
= 1800 C
So, copper deposited at the cathode in half-an-hour is as follows.
M = ZQ
= [tex]3.29 \times 10^{-4} g/C \times 1800 C[/tex]
= 0.5927 g
Thus, we can conclude that 0.5927 g of copper is deposited at the cathode in half an hour.
