Question

Peak current in a cyclic voltammogram is 3.3 μA. The electrode area is 0.45 cm², the diffusion coefficient for the redox couple is 9 x 10^-6 cm²/sec, and the scan rate is 50 mV/sec. Assuming that you began with only the oxidized species and scanned negatively and that n=1, calculate the concentration of the oxidized species in the bulk solution in mol/L.

NOTE:

Though they are not explicitely stated, the units on the number 2.69 x 10⁵ are C/V^1/2mol.

Answer 1

Peak current = 3.3 µA * 10^-3= 0.0033 mA

Electrode area = 0.45 cm2 = 45 mm2

Diffusion coefficient = D = 9*10^-6 cm2/sec = 9*10^-4 mm2/sec

Scan rate = 50 mV/sec

n=1 (number of electrons transferred)

2.69*10^5 C/V^(1/2)*mol

We will be calculating C in mol/L

Step 1:

Formula for calculating peak current is

where ip is the peak current, n is the number of electrons transferred, A is the electrode

area, D is the diffusion coefficient of the species, v is the scan rate and C* is the bulk

concentration of the species.

Solving for C i.e. bulk concentration of species.

C = 0.0033 / (2.69 * 10⁵) * 1^3/2*45*(9*10^-4)^1/2*50^1/2

    = 0.0033 / (2.69 * 10⁵) * 1 * 45 * 0.03 * 7.07

    = 0.0033/ 2567470

    = 1.285 * 10^-8 mol/L

Concentration of bulk species is 1.285 * 10^-8 mol/L.