Question

How many electrons flow through a metallic wire if a current of ${\rm{0}}{\rm{.5 \,A}}$ is passed for ${\rm{2 \,hours}}$? (Given: $1\;{\rm{F}} = {\rm{96,500 \,C}} \cdot {\rm{mo}}{{\rm{l}}^{ - 1}}$)

Answer 1

Current is carried as charge on electrons so the relationship between the charge transferred with the given current and number of electrons that would carry that amount of charge can be used to determine the same.

Complete step by step solution:
Given:
->Current passed through wire is: $I = {\rm{0}}{\rm{.5 \,A}}$
->Duration for which current flowed through the wire is: $t = {\rm{2 \,hr}}$
->$1\;{\rm{F}} = {\rm{96,500 \,C}} \cdot {\rm{mo}}{{\rm{l}}^{ - 1}}$

We have Faraday’s law that gives a quantifying relationship between the charge and current. Mathematically, it can be expressed as per the following expression:
$Q = It$
Here, $Q$ is the quantity of charge that has been produced by the current, $I$ during a time interval of $t$.
Before calculating the charge, we have to convert the units of time. We can write the conversion factor for time as follows:
$\dfrac{{3600\,{\rm{ s}}}}{{{\rm{1\, hr}}}}$
Now, we can convert the units of given time from hours to seconds by using the above conversion factor as follows:
$\left( {\dfrac{{3600\, {\rm{ s}}}}{{{\rm{1 \,hr}}}}} \right) \times {\rm{2 \,hr}} = 7200\, {\rm{ s}}$
Let’s calculate the charge produced as per our given value of current and the calculated value of time as follows:
$Q = {\rm{0}}{\rm{.5\, A}} \times {\rm{7200}}\;{\rm{s}}\\\ = 3600\,{\rm{ C}}$
As we have given the value of Faraday’s constant to be $1\;{\rm{F}} = {\rm{96,500 \,C}} \cdot {\rm{mo}}{{\rm{l}}^{ - 1}}$, we can define as the charge carried by $1\,{\rm{ mole}}$ of electrons. So, we can write this in the form of a unit formula as follows:
$\dfrac{{{\rm{1 \,mole \,}}{e^ - }}}{{{\rm{96,500 \,C}}}}$
Let’s calculate the number of moles of electrons that would carry the calculated quantity of charge by using the above unit formula as follows:
$\left( {\dfrac{{{\rm{1 \,mole \,}}{e^ - }}}{{{\rm{96,500 \,C}}}}} \right) \times {\rm{3,600 \,C}} = {\rm{0}}{\rm{.0373 \,mol \,}}{e^ - }$
Now, we know that ${\rm{1 \,mole}}$ is the amount of any substance that has $6.02 \times {10^{23}}$ particles. Mathematically, we can write this in the form of unit formula as follows:
$\dfrac{{6.02 \times {{10}^{23}}{\rm{ \,particles}}}}{{{\rm{1 \,mole}}}}$
Here, we have electrons as particles so we can write the unit formula as follows:
$\dfrac{{6.02 \times {{10}^{23}}{\rm{ }\,}{e^{ - 1}}}}{{{\rm{1\, mole\, }}{e^{ - 1}}}}$
Finally, we can find out the number of electrons in the calculated number of moles by using the above unit formula as follows:
$\left( {\dfrac{{6.02 \times {{10}^{23}\,}{\rm{ }}{e^{ - 1}}}}{{{\rm{1 \,mole\, }}{e^{ - 1}}}}} \right) \times {\rm{0}}{\rm{.0373 \,mol \,}}{e^ - } = {\rm{2}}{\rm{.24}} \times {10^{22}\,}{\rm{ }}{e^ - }$
Hence, the number of electrons that flew through the metallic wire is ${\rm{2}}{\rm{.24}} \times {10^{22}}$.

Note:
We could have calculated the number of electrons by simply dividing the total quantity of charge by unit charge carried by one electron.