Concentration polarisation (electrochemistry)

Concentration polarisation is the change in electrode potential from its equilibrium value when a certain level of current flows, due to inefficiencies in electrode reactions caused by sluggish mass transfer between the electrode surface and the solution.

When a piece of metal is immersed in its ionic solution, an equilibrium electrode potential is established with no net current flowing. If the metal is connected to another half-cell, e.g. an SHE, and the metal is below hydrogen in the electrochemical series, a net cathodic current flows.

$M^{z+}+ze^- \begin{matrix} r_{red}\\\rightleftharpoons \\ r_{ox} \end{matrix}M$

Furthermore, if the metal electrode is consuming species faster than reactants reaching the electrode surface, a concentration gradient arises between the bulk solution and the electrode surface. Since diffusion is a slow process, the depletion of reactants at the electrode surface leads to a lower electrode potential (similarly, when the electrode is producing species faster than the products leaving the electrode surface, the electrode again becomes polarised). The difference between the modified potential and the equilibrium potential is called the concentration overpotential η_con.

To derive a mathematical relationship between current (or current density) and concentration overpotential, we have to assume that activation overpotential of the cell is negligible. This is achieved by using of non-polarisable electrodes.

At equilibrium, the potential of the cathode is given by the Nernst equation:

$E=E^{\, o}+\frac{RT}{zF}ln[M^{z+}]\; \; \; \; \; \; \; \; 62$

When a current flows and polarisation of the electrode occurs, [M^z+] becomes [M^z+’] where [M^z+’] < [M^z+]. The corresponding open-circuit electrode potential is

$E'=E^{\, o}+\frac{RT}{zF}ln[M^{z+'}]\; \; \; \; \; \; \; \; 63$

The concentration overpotential of the polarised electrode is

$\eta_{con}=E'-E=\frac{RT}{zF}ln\frac{\left [ M^{z+'} \right ]}{\left [ M^{z+} \right ]}\; \; \; \; 64$

According to Fick’s first law of diffusion, the flux of ions, r, towards the cathode is

$r=-D\frac{\left [ M^{z+'} \right ]-\left [ M^{z+} \right ]}{x}\; \; \; \; \; \; \; \; 65$

where D is the coefficient of diffusion with SI units m²s^-1 and x is the distance between the bulk solution and the electrode surface.

Question

How is Fick’s first law derived?

Answer

In diffusion, particles move from a region of higher concentration to one of lower concentration. The flux of particles (i.e. the movement of the number of particles N per unit area per unit time) across a distance x is therefore proportional to the difference of [N]_x=x and [N]_x=0 divided by the change in distance, i.e.

$r=D\frac{d[N]}{dx}$

Knowing that d[N] < 0, if we defined r as positive in the direction of x, the above equation becomes

$r=-D\frac{d[N]}{dx}$

Substituting eq65 in eq39 of the previous article, the cathodic current density is:

$j=zFD\frac{\left [ M^{z+} \right ]-\left [ M^{z+'} \right ]}{x}\; \; \; \; \; \; \; \; 66$

Note that current in a cell not only flows from one electrode to the another electrode along a wire, but also continues to flow back to the first electrode through the electrolyte, and eq66 represents this flow. Rearranging eq66,

$\left [ M^{z+'} \right ]=\left [ M^{z+} \right ]-\frac{jx}{zFD}\; \; \; \; \; \; \; \; 67$

Substituting eq67 in eq64 and rearranging,

$j=\frac{zFD\left [ M^{z+} \right ]}{x}\left ( 1-e^{zf\eta_{\, con}} \right )\; \; \; \; \; \; \; \; 68$

where f = F/RT.

Concentration overpotential, like activation overpotential, leads to a lower potential than the equilibrium potential of a galvanic cell when a current flows.

$E_{galvanic}=E_{cell,open}-\Pi$

where $\Pi=\left | \eta_{con,R} \right |+\eta_{con,L}$ . In the case of an electrolytic cell, a higher applied potential is needed to maintain the desired current. If we include IR drop, the formula is:

$E_{applied}=E_{cell,open}-Ir-\Pi$

Concentration polarisation (electrochemistry)

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