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and<br />

c 2 = 0 A2.34<br />

c1 = c / exp ! a2<br />

3 z3<br />

" %<br />

$ '<br />

# &<br />

dz<br />

(<br />

) A2.35<br />

0<br />

Equation (1.120) becomes<br />

F(!) = c<br />

(<br />

(<br />

!<br />

0<br />

)<br />

0<br />

exp ! a2 "<br />

$<br />

#<br />

exp ! a2 "<br />

$<br />

#<br />

3 z3<br />

3 z3<br />

%<br />

'<br />

&<br />

dz<br />

%<br />

'<br />

&<br />

dz<br />

The normalising integral is finite and equal to<br />

exp ! a2<br />

3 z3<br />

" %<br />

$ '<br />

# &<br />

dz<br />

(<br />

) =<br />

0<br />

2 3<br />

9<br />

5/6<br />

a 2/3 * 2 " %<br />

#<br />

$<br />

3&<br />

'<br />

The concentration profiles are therefore given by<br />

! y $<br />

c(x, y) = F<br />

"<br />

#<br />

! (x) %<br />

&<br />

= F(") =<br />

To calculate the thickness of the diffusion layer, we can write<br />

and<br />

" !c%<br />

#<br />

$<br />

!y&<br />

'<br />

y=0<br />

! dF $<br />

"<br />

#<br />

d! %<br />

& !=0<br />

= !! " dF %<br />

!y #<br />

$<br />

d! &<br />

'<br />

!=0<br />

=<br />

3 1/6 a 2/3 ' 2 ! $<br />

"<br />

#<br />

3%<br />

& c b<br />

2<br />

= 1 " dF %<br />

" #<br />

$<br />

d! &<br />

'<br />

!=0<br />

This equation allows us to calculate a= 0.87116.<br />

By integration of eq.(1.118), we have<br />

! =<br />

∫<br />

l<br />

3<br />

0<br />

3a 2 Dx<br />

dx<br />

!<br />

0<br />

l<br />

∫ dx<br />

=<br />

A2.36<br />

A2.37<br />

3 1/6 a 2/3 c' 2 ! $<br />

"<br />

#<br />

3%<br />

&<br />

exp (<br />

2<br />

a2<br />

3 z3<br />

! $<br />

# &<br />

" %<br />

dz<br />

"<br />

) A2.38<br />

0<br />

= (c<br />

"<br />

= cb<br />

"<br />

A2.39<br />

= 0.8131 a 2/3 c b = c b A2.40<br />

∫<br />

l<br />

3<br />

0<br />

0.759Dhx<br />

dx<br />

< v ><br />

l<br />

∫ dx<br />

0<br />

=<br />

3<br />

4<br />

0.759Dh<br />

< v ><br />

l<br />

3 l 4/3<br />

The limiting current on the band electrode is then for the case illustrated in figure 13<br />

I = nFDLc b l dx<br />

0 ! (x)<br />

! = nFDLc b l<br />

!<br />

0<br />

dx<br />

3<br />

3a 2 Dx<br />

"<br />

= 3 –1/3 a "2/3 nFc b LD 2/3 " 1/3 l 2/3<br />

A2.41<br />

32

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