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Question: Passage of the current for \( 548 \) seconds through a silver coulometer results in the deposition o...

Passage of the current for 548548 seconds through a silver coulometer results in the deposition of 0.746g0.746g of silver. What is the current in AA ?
(A) 1.221.22
(B) 1.161.16
(C) 1.071.07
(D) 1.001.00

Explanation

Solution

Hint : Here, the current is set to pass through a silver coulometer and this results in the deposition of silver there. So, at first we have to find the comparison between the farads and grams so that we can calculate the required current. As we know that the charge flowing in the electrochemical cell is equivalent to n times farad.
Use: 1F=96500C(mole)11F = 96500C{\left( {mole} \right)^{ - 1}}
Q=nFQ = nF ; nn no. of electrons, QQ is the charge and FF is the faraday constant and it is the charge carried by 1mole1mole of electrons.
Q=ItQ = It ; II is current, tt is the time for which the current is flowing.
Molecular weight of the silver: 108g108g .

Complete Step By Step Answer:
Let us consider the reaction when the current is passed through the silver coulometer as:
Ag++eAgA{g^ + } + {e^ - } \to Ag
Now, let us consider the comparison between the molecular weight of silver AgAg and the charge as below:
1Fmolecular weight of Ag1F \to {\text{molecular weight of }}Ag
Therefore, charge carried by the total weight of the silver is
1F108g1F \to 108g
But 1F=96500C(mole)11F = 96500C{\left( {mole} \right)^{ - 1}}
Thus, we have
96500C108g96500C \to 108g
Charge carried by one gram of silver is:
1g96500108C1g \to \dfrac{{96500}}{{108}}C
Now, according to the given condition the deposition of silver is 0.746g0.746g and we are asked to find the current through this weight of silver. So, we will multiply both the sides by 0.746g0.746g and we get:
0.746g0.746×96500108C0.746g \to 0.746 \times \dfrac{{96500}}{{108}}C
This is called the charge flowing through the deposition.
On calculation we get
Q=0.746×96500108CQ = 0.746 \times \dfrac{{96500}}{{108}}C …. (by Q=nFQ = nF )
t=548sect = 548\sec
We know the current flowing through the deposition of silver is given by:
Q=ItQ = It
I=Qt\Rightarrow I = \dfrac{Q}{t}
I=0.746×96500108C548sec\Rightarrow I = \dfrac{{0.746 \times \dfrac{{96500}}{{108}}C}}{{548\sec }}
On calculating the above equation for current we have
I=1.22AI = 1.22A
Thus, the current flowing in the deposition of silver is 1.22A1.22A
The correct answer is option A.

Note :
Remember the concept of the electrochemistry that the charge flowing in the given material is equal to the product of no. of electrons flowing in that material and the charge per mole of electron i.e. faraday constant. We must remember what one farad equivalent is to. Be careful in the calculation part.