Question 11.15: In electrochemistry, the observed current is generally deter...

In electrochemistry, the observed current is generally determined by ion diffusion in the electrolyte. It was found that diffusion-limited currents can be avoided by using very small electrodes, known as ‘ultramicroelectrodes’. In order to capture how conductive current densities (also called diffusion current densities) vary with the size of the electrode, consider a spherical electrode and a conductive matter current density with spherical symmetry, j_A = j_{Ar} \hat{r} ≡ j_r \hat{r}. Show that when the system reaches a stationary state, the conductive matter current density is non-zero. The analysis of the transient behaviour would show that the stationary state is reached faster when the electrode is smaller. In spherical coordinates (r, θ, φ), taking into account the spherical symmetry of the conductive matter current density, i.e. ∂/∂θ = 0 and ∂/∂φ = 0, the matter diffusion equation (11.54) for a solute of concentration c (r, t) reads,

∂_t n_A = D ∇^2 n_A                           (11.54)

\frac{∂c (r, t)}{∂t} = D \Bigl(\frac{∂^2c (r, t)}{∂r^2} + \frac{2}{r} \frac{∂c (r, t)}{∂r} \Bigr)

The boundary conditions are,

c (r > r_0, t = 0) = c^∗           and      \lim _{r\rightarrow \infty }c(r, t) = c^∗

where c^∗ is the concentration very far away from the electrode and r_0 is the radius of the electrode. According to relation (11.51), the conductive matter current density scalar j_r that characterises this electrode is,

j_A = −D(s, μ_A) ∇n_A              (11.51)

j_r(r_0,t) = -D \frac{\partial c (r,t)}{\partial r } \mid _{r=r_0}

Establish the following results :

a) The diffusion equation recast in terms of the function w(r, t) = r c (r, t) has the structure of a diffusion equation where the spherical coordinate r plays an analogous role to a Cartesian coordinate.]

b) The diffusion equation,

\frac{∂w(r, t)}{∂t} = D \frac{∂^2w(r, t)}{∂r^2}

admits the solution,

w (r, t) = B \int_{\nu _0}^{\nu }{\exp (-\acute{\nu }^2 )d \acute{\nu } }         where        \nu = \frac{r}{2\sqrt{Dt} }

First, write w(r, t) = f (η) where the variable η is a dimensionless function of r and t given by,

η (r, t) = \frac{r^2}{Dt}

c) In the limit where the radius of the electrode is negligible, i.e. r = 0, the scalar conductive matter current density is given by,

  j_r (0, t) = \frac{B}{8\sqrt{D} t^{3/2}}

d) After a transient behaviour, the scalar conductive matter current density reaches a stationary value,

  j_r (r_0,∞) = − \frac{Dc^∗}{r_0}
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a) To show that the function w(r, t) = r c (r, t) satisfies a diffusion equation for the spatial variable r, we compute the partial derivatives taking into account the fact that the variables r and t are independent. Using the diffusion equation for the matter concentration c (r, t), the partial time derivative of w(r, t) can be written as,

\frac{∂w(r, t)}{∂t} = \frac{r ∂c (r, t)}{∂t} = D \Bigl(r\frac{∂^2c (r, t)}{∂r^2} +2 \frac{∂c (r, t)}{∂r}\Bigr) .

The second-order partial spatial derivative is given by,

\frac{∂^2w (r, t)}{∂r^2} = \frac{∂}{∂r} \Bigl(\frac{∂}{∂r} (r c (r, t))\Bigr) = \frac{∂c (r, t)}{∂r} + \frac{∂}{∂r} \Bigl(r\frac{∂c (r, t)}{∂r}\Bigr) = r \frac{∂^2c (r, t)}{∂r^2} +2 \frac{∂c (r, t)}{∂r}

Thus,

\frac{∂w(r, t)}{∂t} = D \frac{∂^2 w(r, t)}{∂r^2}

b) The partial derivatives of the function w(r, t) have to be recast in terms of the partial derivatives of the function f (η) where η (r, t) = r^2/Dt.^32 The partial derivatives of the function η (r, t) = r^2/Dt are,

\frac{∂η}{∂t} = – \frac{r^2}{Dt^2} = – \frac{η}{t}       and    \frac{∂η}{∂r} = \frac{2r}{Dt} = \frac{2η}{r}

Since f (η) = w(r, t), the first-order partial derivatives of the function w(r, t) are recast
in terms of the first-order derivative of the function f (η) as,

\frac{\partial w}{\partial t} =\frac{df}{d\eta } \frac{\partial \eta }{\partial t} = – \frac{r^2}{Dt^2} \frac{df}{d\eta } = – \frac{\eta }{t} \frac{df}{d\eta } .

\frac{\partial w}{\partial r} =\frac{df}{d\eta } \frac{\partial \eta }{\partial r} = \frac{2 r}{Dt} \frac{df}{d\eta } =  \frac{2\eta }{r} \frac{df}{d\eta }

The second-order partial derivative of the function w(r, t) is recast in terms of the second-order derivative of the function f (η) as,

\frac{\partial^2 w}{\partial r^2} = \frac{\partial}{\partial r}\biggl(\frac{2r}{Dt}\frac{df}{d\eta } \biggr) =\frac{2}{Dt} \frac{df}{d\eta }+ \frac{2r}{Dt} \frac{d^2f}{d\eta ^2} \frac{\partial \eta }{\partial r} = \frac{2\eta }{r^2} \frac{df}{d\eta } + \frac{4\eta ^2}{r^2}\frac{d^2f}{d\eta ^2}

Thus, the diffusion equation becomes,

-\frac{\eta }{t} \frac{df}{d\eta } = \frac{2D\eta }{r^2} \frac{df}{d\eta } + \frac{4D\eta ^2}{r^2} \frac{d^2f}{d\eta ^2}

Using the definition of the dimensionless function η, this differential equation is recast as,

4η\frac{d^ 2 f}{dη^2}+ (η + 2)\frac{df}{dη} = 0

Here, we introduce a function g (η) as the derivative of f (η) with respect to η,

g (η) = \frac{df}{dη} 

Thus, the differential equation becomes,

4 η \frac{dg (η)}{dη} + (η + 2) g (η) = 0 

which is recast as,

\frac{dg (η)}{g (η)} = – \biggl(\frac{1}{4} + \frac{1}{2\eta } \biggr) dη 

When integrating from η_0 to η we find,

\int_{g_0}^{g(\eta )}{\frac{d\acute{g}(\acute{\eta } ) }{\acute{g}(\acute{\eta } ) } } = \int_{\eta _0}^{\eta }{\biggl(\frac{1}{4}+\frac{1}{2\acute{\eta } } \biggr) } d\acute{\eta }

where g_0 = g (η_0) . The solution is,

\ln \biggl(\frac{g(\eta )}{g_0}\biggr) = -\frac{1}{2} \ln \Bigl(\frac{\eta }{\eta _0}\Bigr) -\frac{1}{4} (\eta -\eta _0)

which is recast as,

\ln \Biggl(\frac{g(\eta )\eta ^{1/2}}{g_0 \eta ^{1/2}_0}\Biggr) = -\frac{1}{4} (\eta -\eta _0)

and implies that,

g (η) = \frac{A}{η^{1/2}} \exp \bigl(-\frac{\eta }{4}\bigr)

where the constant A= g_0 η^{1/2}_0 \exp (η_0/4) .When integrating from η_0 to η we find,

f(\eta ) = \int_{\eta _0}^{\eta }{\acute{g}(\acute{\eta } )d\acute{\eta } } = A \int_{\eta _0}^{\eta }{\frac{1}{\acute{\eta }^{1/2} } }\exp \Bigl(-\frac{\acute{\eta } }{4}\Bigr) d\acute{\eta }

Using the change of variable,

\nu =\frac{\eta ^{1/2}}{2} = \frac{r}{2\sqrt{Dt} }     then      d\nu = \frac{d\eta }{4\eta ^{1/2}}

and taking into account the definition h (ν) = f (η) = w(r, t), we can recast the solution as,

w (r, t) = h (ν) = B \int_{\nu _0}^{\nu }{\exp (-\acute{\nu }^2 )d\acute{\nu } }

where the constant B = 4 A. For ν_0 = 0, the solution is the error function h (ν) = erf (ν) multiplied by a constant.

c) The matter concentration c (r, t) is given by,

c (r, t) = \frac{w(r, t)}{r} = \frac{B}{r} \int_{\nu _0}^{\nu }{\exp (-\acute{\nu }^2 )d\acute{\nu } }

where ν (r, t) and we choose ν_0 (r, t) = ν (r_0, t) . Thus, the conductive matter current density scalar at the electrode of radius r_0 is written as,

j_r(r_0,t) = -D \frac{∂c (r, t)}{∂r}\mid _{r=r_0} = \frac{BD}{r^2_0} \int_{\nu _0}^{\nu }{\exp (-\acute{\nu }^2 )} d\acute{\nu } \mid _{r=r_0} – \frac{BD}{r^2_0} \exp (-\nu ^2)\frac{\partial \nu }{\partial r} \mid _{r=r_0}

where we used the fact that the upper integration bound ν (r, t) is a function of r. In this relation, the integral vanishes since ν evaluated at r_0 is ν_0 , which means that the upper and lower integration bounds are equal. Taking into account that,

\exp (-\nu ^2) \frac{\partial \nu }{\partial r} \mid _{r=r_0} = \exp\Bigl(-\frac{r^2}{4Dt}\Bigr)\frac{\partial}{\partial r} \Bigl(\frac{r}{2\sqrt{Dt} }\Bigr) \mid _{r=r_0} = \frac{1}{2\sqrt{Dt} } \exp \Bigl(-\frac{r^2_0}{Dt}\Bigr)

we find,

j_r(r_0,t) = -\frac{B}{2r^2_0} \sqrt{\frac{D}{t} } \exp\biggl(-\frac{r^2_0}{4Dt}\biggr)

In the limit where the radius of the electrode is negligible, i.e. r_0 = 0 , the scalar conductive matter current density is given by,

j_r(0,t) \lim _{r_0\rightarrow 0} j_r(r_0,t) = \lim _{r_0\rightarrow 0}\Biggl(-\frac{B}{2r^2_0}\sqrt{\frac{D}{t}}\biggl(1-\frac{r^2_0}{4Dt}\biggr)  \Biggr) = \frac{B}{8 \sqrt{D} t^{3/2}}

d) The stationary state is reached in the limit where t→∞. In the stationary limit,

\lim _{t\rightarrow \infty }\nu (r_0,t) = \lim _{t\rightarrow \infty }\frac{r_0}{2\sqrt{Dt} } =0

and initially,

\nu _0 = \lim _{t\rightarrow 0}\nu (r_0,t) = \lim _{t\rightarrow \infty }\frac{r_0}{2\sqrt{Dt} } =\infty

Thus, in the stationary state, the general expression for the scalar conductive matter current density obtained in c) reduces to,

j_r(r_0,\infty ) = -D \frac{∂c (r,∞)}{∂r} \mid _{r=r_0} = -\frac{BD}{r^2_0} \int_{0}^{\infty }{\exp (-\nu ^2)}D\nu

The error function erf (x) is defined as,

 erf (x) =\frac{2}{\sqrt{\pi } } \int_{0}^{x}{\exp (-\nu ^2)} d\nu

and erf (∞) = 1. Thus, the scalar conductive matter current density is recast as,

j_r(r_0,\infty )=-\frac{\sqrt{\pi }BD }{2r^2-0} erf (∞)==-\frac{\sqrt{\pi }BD }{2r^2-0}= -\frac{Dc^*}{r_0}

where

B = \frac{2r_0 c^*}{\sqrt{\pi } } .

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