Question

A copper voltammeter is connected in series with a heater coil of resistance \[0.1\Omega \]. A steady current flows in the circuit for twenty minutes and a mass of 0.99 g of copper is deposited at the cathode. If the electrochemical equivalent of copper is 0.00033 gm/C, then the heat generated in the coil is
A. 750J
B. 650J
C 350J
D. 250J

Answer 1

Hint: The copper will get deposited on the cathode by liberating electrons from its valence shell. When there will be a flow of electrons then there will be electricity and when electricity will flow through a resistor then using Joule’s law of heating there will be heat energy generated.

Formula Used:\[m = Zit\], where m is the mass of the ion deposited, Z is the electrochemical equivalent, i is the electric current and t is the time.
\[P = {i^2}R\], where P is the rate of heat generation when i current flows in a resistor of resistance R.

Complete answer:It is given that 0.99 g of copper is deposited on the cathode,
\[m = 0.99g\]
The electrochemical equivalent of copper is given as \[0.00033gm/C\]
\[Z = 0.00033gm/C\]
The time is given as 20 minutes,
\[t = 20\min \]
\[t = 20 \times 60s\]
\[t = 1200s\]
Then the electric current is,
\[i = \dfrac{m}{{Zt}}\]
Putting the values, we get
\[i = \dfrac{{0.99}}{{0.00033 \times 1200}}\]
\[i = 2.5A\]
The electric current is 2.5A.
The value of the resistance of the resistor connected through the copper voltammeter is given as \[0.1\Omega \]
So, the rate of heat energy generated using Joule’s law formula is,
\[P = {i^2}R\]
\[P = {2.5^2} \times 0.1W\]
\[0.625W\]
As we know that power is the rate of heat energy generated,
\[P = \dfrac{E}{t}\]
So, the heat energy generated during the given time interval will be calculated as,
\[E = pt\]
\[E = 0.625 \times 1200J\]
\[E = 750J\]
So, the heat energy generated during the electrolysis is 750J
Therefore,

the correct option is (A).

Note: The power is the energy per second. We need to find the total energy, so we have to find the energy throughout the process was running. We should be careful about the units of the physical quantities given in the question.