   Chapter 18, Problem 99E

Chapter
Section
Textbook Problem

# What volume of F2 gas, at 25°C and 1.00 atm, is produced when molten KF is electrolyzed by a current of 10.0 A for 2.00 h? What mass of potassium metal is produced? At which electrode does each reaction occur?

Interpretation Introduction

Interpretation:

The electrolysis of molten KF is given. The volume of fluorine, the mass of potassium metal that is produced and the reaction at each electrode is to be stated.

Concept introduction:

The non-spontaneous reaction takes place in an electrolytic cell in which there occurs conversion of electrical energy into chemical energy and this is used for the electrolysis of a metal.

The charge generated in the cell is calculated as,

Q=It

When electricity is passed through an electrolytic cell, at that time the amount of the substance that is liberated at an electrode is given by,

W=ZQ=ZIt

The value of Z is given as,

Z=Atomicmassn×96,485

The mass of metal is calculated by multiplying the moles of metal with its atomic molar mass. This gives the mass of the metal to be plated out in the electrolytic cell.

To determine: The effect of change in concentration of Zn2+ and Ag+ over E and E° of the cell.

Explanation

Given,

The temperature is given as 25°C .

The pressure is given as 1atm .

The value of Z is given as,

Z=Atomicmassn×96,485=E96,485

Where,

• n is the number of electrons exchanged.
• E is the equivalent weight.

Substitute the values of atomic mass and n to obtain the value of the equivalent of F2 .

Z=38gmole2×96,485CmoleZ=19gmole96,485Cmole=1.9×104g/C

When electricity is passed through an electrolytic cell, at that time the amount of the substance that is liberated at an electrode is given by,

W=ZQ=ZIt

Where,

• Q is the charge carried in the cell.
• I is the current in amperes
• t is the time for which current is passed through the cell.
• Z is the electrochemical equivalent.

Substitute the value of Z , I and t in the above equation to obtain the amount of substance liberated at the electrode,

W=1.9×104gC×10Cs×7200s=13.68g

The moles of fluorine are calculated as,

Moles=GivenmassMolarmassMoles=13.68g38.0gmol=0.36mol

The ideal gas equation is,

PV=nRT

Where,

• P is the pressure of the gas
• V is the volume of the gas.
• n is the number of moles of the gas.
• R is the universal gas constant.
• T is the temperature in Kelvin.

Rearrange the equation and substitute the values for n , R , T and P to obtain the volume of fluorine,

V=nRTP=0.36mol×0.0821L.atmmolK×298K1.0atm=8.8L_

Volume of fluorine is calculated as 8

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