Question

If 1 A of current is passed through $CuS{O}_4$ solution for 10 s, the number of copper atoms deposited at the cathode will be

• A. $8\times {10}^{19}$
• B. $3.1\times {10}^{19}$
• C. $6.2\times {10}^{19}$
• D. $1.6\times {10}^{20}$

Answer 1

The correct option is B $3.1\times {10}^{19}$

Given, $I=1A,t=It,q=10C$
Charge of $C{u}^{2+}=2e$
$=2\times 1.6\times {10}^{-19}C$
The number of copper atoms deposited at the cathode
$=\frac{10}{2\times 1.6\times {10}^{-19}}$
$=3.1\times {10}^{19}$