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Question

108g silver (molar mass 108gmol-1) is deposited at cathode from AgNO3(aq)solution by a certain quantity of electricity. The volume (in L) of oxygen gas produced at 273K and 1bar pressure from water by the same quantity of electricity is:


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Solution

Step 1: Given data and assumptions:

Weight of silver Ag deposited at cathode is WAg=108g

Molecular weight of Silver is 108gmol-1

Temperature is T=273K

Pressure is P=1bar

Let the quantity of electricity passed through the system be Q

Volume of oxygen gas produced from water H2O is VO2=V

Step 2: Write the Chemical reactions involved during the process:

  1. Ag++e-Ag
  2. 2H2OO2+4H++4e-

The number of electrons involved during the formation of Ag is nAg=1

The number of electrons involved during the formation of O2 is nO2=4

Step 3: Find the number of moles of O2:

According to Faraday's first law of electrolysis;

1) WAg=MAgnAg×F×Q...(I) where, M(Ag) is the molecular weight of AgNO3

Where, nAg=1 is the number of electrons involved in the formation of Ag

F=1Faraday

Q=Amount of electricity passed through a system

2) No.ofmolesofO2=1nO2×F×Q...(II)

Divide eq.(I) and eq.(II);

WAgNo.ofmolesofO2=M(Ag)×Q/(n(Ag)×F)1×Q/(n(O2)×F)108gNo.ofmolesofO2=108gmol-1×Q/(n×F)1×Q/(4×F)NumberofmolesofO2=14

Step 4: Find the volume of O2:

From the Ideal gas equation:

PV=nRTV=nO2×R×TPV=14mol×0.0822LatmK-1mol-1×273K1bar1atm=1.01325barV=14×0.0822×1.01325×273V=5.684L

Hence, the volume of oxygen gas produced at 273K and 1bar pressure from the water is VO2=5.684L.


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