Q3(b). Stokes radius and diffusion coefficient of Zn²⁺

Given: E=5V cm⁻¹=500V m⁻¹; v=2.74×10⁻⁵m s⁻¹; η=0.890 mPa s=8.90×10⁻⁴kg m⁻¹ s⁻¹; z=2; T=298.15 K.

(i) Radius of the hydrated ion

1. The field pulls the ion with force zeE. The water holds it back with the Stokes drag 6πη rv, which grows with speed. Within a very short time the two are equal and the ion moves at a constant drift speed: zeE=6πη rv
2. Mobility: u=v/E=2.74×10⁻⁵500=5.48×10⁻⁸m² V⁻¹ s⁻¹
3. r=(zeE)/(6πη v)=2×1.602×10⁻¹⁹×5006×3.142×8.90×10⁻⁴×2.74×10⁻⁵=3.49×10⁻¹⁰ m=349 pm

Reading the answer: the crystal radius of Zn²⁺ is only 75 pm. The small, doubly charged ion has a strong field and drags a hydration shell of water with it, so its hydrodynamic radius is about 4.6 times larger. Stokes' law on a molecular scale gives the right order of magnitude, not an exact size.

(ii) Diffusion coefficient

1. Diffusion and drift are both resisted by the same friction, which gives the Einstein relation: D=(u kT)/(ze)=(uRT)/(zF)
2. D=5.48×10⁻⁸×1.38×10⁻²³×298.152×1.602×10⁻¹⁹=7.04×10⁻¹⁰m² s⁻¹
3. Check with the Stokes–Einstein equation, D=kT/6πη r, using r from (i): the same 7.04×10⁻¹⁰m² s⁻¹ (7.04×10⁻⁶cm² s⁻¹), as it must be.

Reading the answer: the tabulated mobility of Zn²⁺ at 298 K is 5.47×10⁻⁸m² V⁻¹ s⁻¹, within 0.2 % of the value here. D is of the usual order for ions, about 10⁻⁹m² s⁻¹, and smaller than for Na⁺ (1.33×10⁻⁹) because the hydrated Zn²⁺ is bigger. The ionic conductivity is λ=zuF=105.7 S cm² mol⁻¹.

Application. the diffusion current of a polarographic wave (Ilkovič equation) depends on D^(1/2), and D links the conductivity, the mobility and the size of an ion through the Nernst–Einstein relation.