 Solution

ID: past-exam-of-the-chemistry-course-of-the-year-3-of-the-natural-sciences-course-of-the-university-of-cambridge/2022/paper-4/51/b/solution

All three marked points are at strongly oxidizing potentials, where X is depleted at the electrode:

c_{X} (0) ≃ 0

. Assuming only X initially in bulk,

c_{X^{+}} (\infty) = 0

and

c_{X} (\infty) = c_{b}

. Oxidation produces X

^{+}

at the interface.

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: slower rotation gives a thicker layer and shallower gradients, hence a smaller limiting current.
C: the nonrotating electrode gives the peaked, subsequently decaying current. Its diffusion layer continues to broaden with time, approximately as $D t$ , 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_{X} c_{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.

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