Question

The weight of Ag deposited by passing 241.25 coulombs of current through ${\text{AgN}}{{\text{O}}_{\text{3}}}$ solution is:
A. 0.27 g
B. 2.7 g
C. 27 g
D. 0.54 g

Answer 1

Hint: This question is based on the passage of current through the silver nitrate solution. We know that on passing the current silver nitrate will break into silver ions, and nitrate ions. The formula used to calculate the moles of electrons passed from the solution is $\dfrac{{{\text{Q(C)}}}}{{{\text{96500C}}}}$. Further we can calculate the mass of silver.

Complete step by step answer:
-Now, we will calculate the weight of silver deposited on passing the current.
-First, we will write the chemical reaction of silver nitrate, which occurs on passing the current; i.e.
${\text{AgN}}{{\text{O}}_{\text{3}}} \to {\text{A}}{{\text{g}}^{\text{ + }}}{\text{ + N}}{{\text{O}}_{\text{3}}}^{\text{ - }}$
-As mentioned the current passed through the solution is 241.25 coulombs. So, we can calculate the moles of electrons passed, i.e.
$\dfrac{{{\text{Q(C)}}}}{{{\text{96500C}}}}$ = $\dfrac{{{\text{241}}{\text{.25}}}}{{{\text{96500}}}}{\text{mol}}{\text{.}}{{\text{e}}^{\text{ - }}}$
-We know that 1 mole of Ag is equal to the 108 g of Ag. Here, 108 g represents the molar mass of Ag per mole.
-Thus, the mass of silver produced will be equal to the molar mass of silver into the moles of electrons passed through the solution. It can be written as:
-Mass of silver produced = (molar mass of Ag) (moles of electrons passed)
-Mass of silver produced = (108 g/ mol) ( $\dfrac{{{\text{241}}{\text{.25}}}}{{{\text{96500}}}}{\text{mol}}{\text{.}}{{\text{e}}^{\text{ - }}}$) (1 mol Ag/ mol electron)
By solving this we get
-Mass of silver produced = 0.27 g
-In the last, we conclude that the amount of Ag deposited on passing the 241.25 coulombs of current through the solution is 0.27 g.
Hence, the correct option is A.

Note:
Sometimes, there could be a confusion while calculating the amount of silver using this formula mentioned above. So, the alternative method to calculate the amount of silver deposited is:
-We know that, 1 Faraday of current is equal to 1 mole of Ag. So, 96500 C of current is also equal to 1 mole of Ag.
-Thus, 241.25 C of current is equal to the 241.25/96500 moles of Ag.
-We already mentioned that 1 mole of Ag is equal to the 108 g. Then,
241.25/96500 moles of Ag will correspond to $\dfrac{{24.25}}{{96500}} \times 108$.
-So, the amount of Ag deposited is 0.27 g
Therefore, this method can also be used to calculate the amount of silver.