Question

The current flowing in a copper voltammeter is 3.2 A. The number of copper ions \[C{u^{2 + }}\]deposited at the cathode per minute is.
A. \[0.5 \times {10^{20}}\]
B. \[1.5 \times {10^{20}}\]
C. \[3 \times {10^{20}}\]
D. \[6 \times {10^{20}}\]

Answer 1

Hint: The electric current is the charge flow rate. When there is a flow of electrons then the rate of flow of electrons constitutes the electric current. If we have given electric current then we can calculate the amount of charge flown per unit of time.

Formula Used:\[i = \dfrac{q}{t}\], where i is the electric current measured when q amount of charge flown through a cross-section in time t.

Complete answer:The amount of the electric current flowing in the copper voltammeter is given as 3.2A.
Using the definition of the electric current, the amount of charge flown per unit time to constitute 3.2A of electric current will be calculated as,
\[i = \dfrac{q}{t}\]
\[q = it\]
\[q = 3.2 \times 1C\]
\[q = 3.2C\]
So, there is a 3.2C charge is transferred in one second when the electric current flowing is 3.2A.
The time throughout which the electric current is flowing through the copper voltammeter is given as 1 minute.
\[t = 1\min \]
\[t = 60s\]
As there is 3.2C of charge is flowing in 1 second,
So, the amount of charge flown in 60 seconds will be, \[q = 3.2 \times 60C = 192C\]
In one copper ion, \[C{u^{2 + }}\]there is transfer of two electrons.
The magnitude of charge on one electron is \[1.6 \times {10^{ - 19}}C\]
So, the total charge transferred when one copper ion is deposited will be \[3.2 \times {10^{ - 19}}C\]
So, total number of \[C{u^{2 + }}\]ion deposited is,
\[n = \dfrac{{192}}{{3.2 \times {{10}^{ - 19}}}}\]
\[n = 6 \times {10^{20}}\]

the correct option is (D).

Note: We should be careful about the time throughout which the electrolysis is carried out because the mass deposited or liberated during electrolysis is proportional to the time. And the time unit should be in seconds.