= Solution
All three marked points are at strongly oxidizing potentials, where X is depleted at the electrode: $c_X(0)\simeq0$. Assuming only X initially in bulk, $c_{X^+}(\infty)=0$ and $c_X(\infty)=c_b$. Oxidation produces X$^+$ at the interface.
* \b[A:] the largest steady limiting current corresponds to the fastest rotation and thinnest <diffusion> boundary layer. X rises steeply from zero to $c_b$; X$^+$ falls steeply from its surface value to zero.
* \b[B:] slower rotation gives a thicker layer and shallower gradients, hence a smaller limiting current.
* \b[C:] the nonrotating electrode gives the peaked, subsequently decaying current. Its <diffusion> layer continues to broaden with time, approximately as $\sqrt{Dt}$, giving the broadest and shallowest profiles at the indicated late-scan point.
For equal <diffusion coefficients>, the two concentration profiles are complementary, $c_X+c_{X^+}=c_b$. More generally equality of their interfacial molar fluxes requires $D_Xc_X'(0)=-D_{X^+}c_{X^+}'(0)$, so the product surface concentration need not equal $c_b$. A and B are approximately steady hydrodynamic profiles; C is a time-dependent <diffusion> profile.
\Image[/past-exam-of-the-chemistry-course-of-the-year-3-of-the-natural-sciences-course-of-the-university-of-cambridge/2022/paper-4-transport.png]
{title=Illustrative reactant and product diffusion profiles at the three transport conditions}
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