Question

if 1.00×10^18 singly - charged positive sodium ions flow from the anode of an electrochemical cell each second and an equal number of singly - charged negative chlorine ions flow from the cathode of the cell, how much current flows through the cell?

Explanation:

Step1: Recall the formula for current

Current $I=\frac{Q}{t}$, where $Q$ is the total charge and $t$ is the time. The charge of a single - ion is equal to the elementary charge $e = 1.6\times10^{- 19}\ C$.

Step2: Calculate the total charge

The number of positive sodium ions $n = 1.00\times10^{18}$ and the number of negative chlorine ions is also $n = 1.00\times10^{18}$. The total charge $Q$ is the sum of the charge of positive and negative ions. Since each ion has a charge of magnitude $e$, the total charge $Q=(n\times e + n\times e)$. Substituting $n = 1.00\times10^{18}$ and $e = 1.6\times10^{-19}\ C$, we get $Q=(1.00\times10^{18}\times1.6\times10^{-19}+1.00\times10^{18}\times1.6\times10^{-19})\ C$.
$Q=(1.00\times10^{18}\times1.6\times10^{-19})\times2\ C=0.32\ C$.

Step3: Calculate the current

The time $t = 1\ s$. Using the formula $I=\frac{Q}{t}$, substituting $Q = 0.32\ C$ and $t = 1\ s$, we get $I=\frac{0.32\ C}{1\ s}=0.32\ A$.

Answer:

$0.32\ A$