Question

In an ionic solution, a current consists of ${\mathbf{Ca}}^{\mathbf{2}\mathbf{+}}$ions (of charge +2e )and ${\mathbf{CI}}^{\mathbf{-}}$ions (of charge $\mathbf{-}\mathbf{e}$) traveling in opposite directions. If $\mathbf{5}\mathbf{.}\mathbf{11}\mathbf{\times }{\mathbf{10}}^{\mathbf{18}}\mathbf{}{\mathbf{CI}}^{-}$ ions go from Ato Bevery 0.50 min, while $\mathbf{3}\mathbf{.}\mathbf{4}\mathbf{\times }{\mathbf{10}}^{\mathbf{18}}\mathbf{}{\mathbf{Ca}}^{\mathbf{2}\mathbf{+}}$ ions move from Bto A, what is the current (in mA) through this solution, and in which direction (from A to B or from B to A ) is it going?

Short answer

The current through the solution is 64.80 A and the direction is going from B to A.

Step-by-step solution

Define the current I .

The charge Q per unit time t is known as the current I.

$I=\frac{Q}{t}$

Where ,I is current in ampere $\left(A\right)$, Q is charge in Columbus$\left(C\right)$ and $t$is time in sec $\left(s\right)$.

Determine current and direction of current.

Given that,

$$\begin{gathered} t=0.5\min \\ q=-1.6\times {10}^{-19}C \\ {n}_{CI}=5.11\times {10}^8\mathrm{electron}/{\mathrm{m}}^3 \\ {n}_{Ca}=3.24\times {10}^8\mathrm{electron}/{\mathrm{m}}^3 \end{gathered}$$

The current between the electrodes is:

$$\begin{gathered} I=\frac{Q}{t} \\ =\frac{\left(n_{CI}+2n_{Ca}\right)\times \left|q\right|}{t} \\ =\frac{\left(5.11\times {10}^{18}+2\times \left(3.24\times {10}^{18}\right)\right)\times \left|-1.6\times {10}^{-19}\right|}{0.5\times 60} \\ =64.80\mathrm{mA} \end{gathered}$$

The current moves from positive side to negative side, so the direction of current is from B to A in the direction of calcium ions.

Hence, current through the solution is 64.80 A and the direction is going from B to A .