Question

Current passes through a solution of sodium chloride. In 1.00 s, 2.68×1016 Na+ ions arrive at the negative electrode and 3.92×1016 Cl−ions arrive at the positive electrode.

What is the current passing between the electrodes? (Answer in mA)

What is the direction of the current? (Outward the negative electrode or toward the negative electrode)

Answer 1

Concepts and reason

The main concept required to solve this problem is electric current.

Firstly, calculate the current due to movement of sodium ions and chlorine ions separately. Then, add them to find the total current produced.

Finally, find the direction of the current by analyzing the movement of the positive and negative charges.

Fundamentals

The expression for the charge is as follows:

$q = ne$

Here, n is the number of charges and e is the charge on each electron.

The expression for the current produced due to the movement of charges is as follows:

$i = \frac{{\Delta q}}{{\Delta t}}$

Here, $\frac{{\Delta q}}{{\Delta t}}$ is the number of charges moved in a time interval.

a)

For sodium ions:

The expression for the charge is as follows:

${q_1} = ne$

Substitute $2.68 \times {10^{16}}$ for n and $1.6 \times {10^{ - 19}}{\rm{ C}}$ for e in the above expression.

$\begin{array}{c}\\{q_1} = \left( {2.68 \times {{10}^{16}}} \right)\left( {1.6 \times {{10}^{ - 19}}{\rm{ C}}} \right)\\\\ = 4.288 \times {10^{ - 3}}{\rm{ C}}\\\end{array}$

For chlorine ions:

The expression for the charge is as follows:

${q_2} = ne$

Substitute $3.92 \times {10^{16}}$ for n and $1.6 \times {10^{ - 19}}{\rm{ C}}$ for e in the above expression.

$\begin{array}{c}\\{q_2} = \left( {3.92 \times {{10}^{16}}} \right)\left( {1.6 \times {{10}^{ - 19}}{\rm{ C}}} \right)\\\\ = 6.272 \times {10^{ - 3}}{\rm{ C}}\\\end{array}$

The total charge flown in the solution is as follows:

$q = {q_1} + {q_2}$

Substitute $4.288 \times {10^{ - 3}}{\rm{ C}}$ for q₁ and $6.272 \times {10^{ - 3}}{\rm{ C}}$ for q₂ in the above expression.

$\begin{array}{c}\\q = 4.288 \times {10^{ - 3}}{\rm{ C}} + 6.272 \times {10^{ - 3}}{\rm{ C}}\\\\ = 10.56 \times {10^{ - 3}}{\rm{ C}}\\\end{array}$

The expression for the current is as follows:

$i = \frac{{\Delta q}}{{\Delta t}}$

The change in the number of charges from the start of the time is equal to the number of charges arrived at the electrodes.

Substitute $10.56 \times {10^{ - 3}}{\rm{ C}}$ for q and 1.00 s for t in the above expression.

$\begin{array}{c}\\i = \left( {\frac{{10.56 \times {{10}^{ - 3}}{\rm{ C}}}}{{1.00{\rm{ s}}}}} \right)\left( {\frac{{1{\rm{ mC}}}}{{{{10}^{ - 3}}{\rm{ C}}}}} \right)\\\\ = 10.56{\rm{ mA}}\\\end{array}$

b)

The conventional direction of the current is the direction in which the positive charge move.

The positive ions in the solution move towards the negative electrode so that the flow of the positive charge is from positive to negative electrode.

Thus, the direction of the current through the solution is towards the negative electrode.

Ans: Part a

The current passing between the electrodes is $10.6{\rm{ mA}}$ .

Part b

The direction of the current is towards the negative electrode.

Answer 2

Part a

The current passing between the electrodes is $10.6{\rm{ mA}}$ .

Part b

The direction of the current is towards the negative electrode.

Answer 3

Concepts and reason

The concepts used to solve this problem are charge, and current.

First find the charge of the sodium and chloride ions.

Find the current passing between the electrodes using the charge.

Fundamentals

The charge of a collection of particles can be written as a whole number multiple of the charge of proton. The fundamental charges belong to protons and electrons. They both have same magnitude of charge.

Expression for charge of a substance is,

$q = ne$

Here, $q$ is the total charge, $n$ is the number of particles, and $e$ is the charge of protons.

Current is defined as the flow of electric charge. If a charge $q$ passes through a surface for a time period $t$, the current is expressed as,

$i = \frac{q}{t}$

Here, $i$ is the current, $q$ is the charge, and $t$ is the time.

The net charge that is arriving at the negative electrode is,

$q = ne$

Substitute $2.68 \times {10^{16}}$ for $n$ and $1.6 \times {10^{ - 19}}\,{\rm{C}}$ for $e$.

$\begin{array}{c}\\q = \left( {2.68 \times {{10}^{16}}} \right)\left( {1.6 \times {{10}^{ - 19}}\,{\rm{C}}} \right)\\\\ = 4.288 \times {10^{ - 3}}\,{\rm{C}}\\\end{array}$

Therefore the net charge of ${\rm{N}}{{\rm{a}}^ + }$ ions is $4.288 \times {10^{ - 3}}\,{\rm{C}}$.

The net charge that is arriving at the positive electrode is,

$q' = ne$

Substitute $3.92 \times {10^{16}}$for $n$, and $1.6 \times {10^{ - 19}}\,{\rm{C}}$ for $e$

$\begin{array}{c}\\q' = \left( {3.92 \times {{10}^{16}}} \right)\left( {1.6 \times {{10}^{ - 19}}\,{\rm{C}}} \right)\\\\ = 6.272 \times {10^{ - 3}}\,{\rm{C}}\\\end{array}$

Therefore the net charge of ${\rm{C}}{{\rm{l}}^ - }$ ions is $6.272 \times {10^{ - 3}}\,{\rm{C}}$.

Therefore, the current is expressed as

$i = \frac{q}{t} + \frac{{q'}}{t}$

Substitute $4.288 \times {10^{ - 3}}\,{\rm{C}}$ for $q$, $6.272 \times {10^{ - 3}}\,{\rm{C}}$ for $q'$, and $1.00\,{\rm{s}}$ for $t$.

$\begin{array}{c}\\i = \frac{{4.288 \times {{10}^{ - 3}}\,{\rm{C}}}}{{1.00\,{\rm{s}}}} + \frac{{6.272 \times {{10}^{ - 3}}\,{\rm{C}}}}{{1.00\,{\rm{s}}}}\\\\ = 10.56 \times {10^{ - 3}}\,{\rm{A}}\\\\{\rm{ = 10}}{\rm{.56}}\,{\rm{mA}}\\\end{array}$

Ans:

The current that is flowing in the sodium chloride solution is ${\rm{10}}{\rm{.6}}\,{\rm{mA}}$

Answer 4

Answer 5

From the definition of current

I = Q / t, where I is the current, Q is the charge, and t is the amount of time

charge (Q) = N*e

Q =(2.68*10^16 + 3.92*10^16)*1.60^-19 C

=0.01056 C

current I =(0.01056 C)/(1 sec) =10.56 mA

I =10.6 mA

B)

the direction of a current is the direction in which the protons flow. This is in opposite to the direction of an electric field (away from the negative charge). So the direction of the current is towards the negative electrode.

answer is toward the negative electrode