Question

Derive relation between ionic conductance and ionic mobility

Answer 1

let's attempt to relate the ionic conductivity to the ionic mobility (it clearly makes sense to do so). So I am assuming we have a dilute solution of completely disassociated strong electrolyte with molar concentration c. Each formula unit of this electrolyte gives rise to ν+ cations of charge z+e and ν− anions of charge z−e.

The concentration of each type of ion is therefore, νc (ν could either mean ν+, or ν− to avoid unnecessary notational complexity).

the number density of each type of ions is:

No. density=νcNA

Now picture a small window of area A in your solution. The number of ions that can pass through said window in a time interval Δt is equal to the number of ions within the distance sΔt (where s is the drift speed of the ions)

sΔtA, happens to define a volume element and the number of ions in this interval is given by: V×no. density, i.e

sΔtAνcNA

Thus, the flux of ions is

J(ions)=sΔtAνcNAΔtA=sνcNA

remember flux always has dimensions of "something" per area per time

Flux of charge is

J(charge)=zesνcNA=zsνcF

again, I am omitting the +/- signs for the sake of clarity. Also, note F:=e×NA

Now, drift speed is s=μE, so the flux of charge (henceforth, simply J) is

J=zμνcFE

The current is charge flux times area, so

I=(zμνcFE)A

Additionally, E=Δϕl (i.e the potential gradient)

I=(zμνcFA)Δϕl(I)

Using Ohm's Law (i.e Δϕ=IR}, one obtains

I=ΔϕR=GΔϕ=κAΔϕl(II)

Comparing the (I) and (II) one is lead to

κ=zμνcF

This is the ionic conductivity. We need to divide by concentration of ions to the get the molar quantity. The concentration of ions however is not c, but rather νc and our final result is

λ=zμF

Answer 2

Ionic mobility or Ionic conductance is the conductivity of a solution containing 1 g ion, at infinite dilution, when two sufficiently large electrodes are placed 1 cm apart. Absolute ionic mobilityis the mobility with which the ion moves under unit potential gradient. It's unit is cm sec–1.