Question

Find the mass of silver deposited when a current of 1.5 ampere is passed through silver nitrate solution for 5 min (atomic mass of Ag is 108 g)

Answer 1

To solve this question, we must first calculate the total charge used in the experiment. Then we must derive a relation the number of electrons required in the experiment and then correlate it with the total amount of charge required. Then finally, a relation between the mass deposited and the charge used must be derived to get the final answer.

Complete Step-by-Step Answer:
Before we move forward with the solution of this question, let us first understand some important basic concepts.
Charge can be understood as the electrical polarity on any substance that causes an imbalance in the electron arrangement of the substance. This charge is a derived property and depends on the amount of work done to produce the charge. Charge can be measured in different units like coulombs and faraday. The relation between these two units can be identified as:
1 faraday = 96500 C
Charge can also be represented as a product of current and time. Hence, the mathematical representation of the same can be given as:
Charge = (current) (time)
Q = (I) (t)
Hence, the total charge used in the given experiment is:
Q = (1.5 A) (5 minutes) = (1.5 A) (300 seconds) = 450 A.s = 450 C
Since the given compound silver nitrate dissociates as follows: $AgN{O_3} \to A{g^ + } + N{O_3}^ -$ , the reactions at the cathode and anode can be given as:
Anode: $N{O_3}^ - \to 1{e^ - } + N{O_2} + \dfrac{1}{2}{O_2}$
Cathode: $A{g^ + } + 1{e^ - } \to Ag$
Hence, we can see that 1 electron is required to deposit 1 mole of Ag. This means that charge on one electron is sufficient to deposit one mole of Ag. This relation can be given as:
Charge of one electron $\to$ 1 mole of Ag
Charge of one electron $\to$ 108 g of Ag
1F $\to$ 108 g of Ag
96500C $\to$ 108 g of Ag
Hence,
450 C$\to$ $\dfrac{{450 \times 108}}{{96500}}$ g of Ag
450 C $\to$ 0.503 g of Ag
Hence, 0.503 g of Ag is deposited in this experiment.

Note: It is much less common than the coulomb, but sometimes used in electrochemistry. One faraday of charge is the magnitude of the charge of one mole of electrons, i.e. 96485. 33212... C. Expressed in faradays, the Faraday constant F equals "1 faraday of charge per mole".