Respuesta :
Explanation:
Given that,
Number of sodium ions at the negative electrode, [tex]Na^+=2.68\times 10^{16}[/tex]
Number of chloride ions at the positive electrode, [tex]Cl^-=3.92\times 10^{16}[/tex]
(a) The current flowing in the circuit is due to the positive as well as negative charges such that total charge becomes:
[tex]Q=(Na^++Cl^-)e[/tex]
[tex]Q=(2.68\times 10^{16}+3.92\times 10^{16})(1.6\times 10^{-19})[/tex]
Q = 0.01056 C
The current is given by :
[tex]I=\dfrac{Q}{t}[/tex]
[tex]I=\dfrac{0.01056}{1}=10.56\ mA[/tex]
So, the current passing between the electrodes is 10.56 mA.
(b) The direction of electric current is towards negative electrodes.
Explanation:
(a) Â First, we will calculate the charge of sodium ions as follows.
       q = ne
         = [tex]2.68 \times 10^{16} \times 1.6 \times 10^{-19} C[/tex]
         = [tex]4.288 \times 10^{-3} C[/tex]
Now, charge of chlorine ions is calculated as follows.
      q' = ne
        = [tex]3.92 \times 10^{16} \times 1.6 \times 10^{-19} C[/tex]
        = [tex]6.272 \times 10^{-3} C[/tex]
Hence, the current will be calculated as follows.
       i = [tex]\frac{q}{t} + \frac{q'}{t}[/tex]
        = [tex]\frac{4.288 \times 10^{-3} C}{1.00} + \frac{6.272 \times 10^{-3} C}{1.00}[/tex]
        = [tex]10.56 \times 10^{-3} A[/tex]
        = 10.56 mA
Therefore, current passing between the electrodes is 10.56 mA.
(b) Â Since, positive ions are moving towards the negative electrode. And, current is the flow of ions or electrons therefore, the direction of current is towards the negative electrode.
