Question

In an electrolyte $3.2\times 10^{18}$ bivalent positive ions drift to the right per second while $3.6\times 10^{18}$ monovalent negative ions drift to the left per second. Then the current is

• A. 1.6 A to the left
• B. 1.6 A to the right
• C. 0.45 A to the right
• D. 0.45 A to the left

Answer 1

Answer: (B)

Net current $i_{\text{net }}=i_{(+)}+i_{(-)}$

$=\frac{n_{(+)}{q}_{(+)}}{t}+\frac{n_{(-)}{q}_{(-)}}{t}$
$=\frac{n_{(+)}}{t}\times 2e+\frac{n_{(-)}}{t}\times e$
$=3.2\times 10^{18}\times 2\times 1.6\times 10^{-19}+3.6\times 10^{18}\times 1.6\times 10^{-19}$
$=1.6A$ (towards right)