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3 years ago

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Current passes through a solution of sodium chloride. In 1.00 s, 2.68×1016Na+ ions arrive at the negative electrode and 3.92×101

6Cl− ions arrive at the positive electrode. (a) What is the current passing between the electrodes? (b) What is the direction of the current?

2 answers:

Tanzania [10]3 years ago

8 0

Explanation:

Given that,

Number of sodium ions at the negative electrode, $Na^+=2.68\times 10^{16}$

Number of chloride ions at the positive electrode, $Cl^-=3.92\times 10^{16}$

(a) The current flowing in the circuit is due to the positive as well as negative charges such that total charge becomes:

$Q=(Na^++Cl^-)e$

$Q=(2.68\times 10^{16}+3.92\times 10^{16})(1.6\times 10^{-19})$

Q = 0.01056 C

The current is given by :

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

$I=\dfrac{0.01056}{1}=10.56\ mA$

So, the current passing between the electrodes is 10.56 mA.

(b) The direction of electric current is towards negative electrodes.

Tresset [83]3 years ago

7 0

Explanation:

(a) First, we will calculate the charge of sodium ions as follows.

q = ne

= $2.68 \times 10^{16} \times 1.6 \times 10^{-19} C$

= $4.288 \times 10^{-3} C$

Now, charge of chlorine ions is calculated as follows.

q' = ne

= $3.92 \times 10^{16} \times 1.6 \times 10^{-19} C$

= $6.272 \times 10^{-3} C$

Hence, the current will be calculated as follows.

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

= $\frac{4.288 \times 10^{-3} C}{1.00} + \frac{6.272 \times 10^{-3} C}{1.00}$

= $10.56 \times 10^{-3} A$

= 10.56 mA

Therefore, current passing between the electrodes is 10.56 mA.

(b) 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.

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