Pulse radiolytic investigation of the reduction of cadmium(II) ions

Abstract

In γ-irradiated aqueous CdSO₄ solutions containing alcohols or formate ions, the hydrated electron reacts with Cd²⁺ to form Cd⁺ and the OH and H radicals react with the organic solute to form organic free radicals. The 100 eV yield of metallic cadmium was 2.5 in the presence of formate and close to zero in the presence of t-butanol and in the absence of organic additives. In the presence of other alcohols, the highest yield observed was with methanol [G(Cd)= 1.3] and the lowest with propan-2-ol (G= 0.4).

Pulse radiolytic studies with optical absorption, conductimetric and light scattering techniques for detection of intermediates yielded the following results: Cd⁺ is oxidized by the radical CH₂(CH₃)₂-COH. α-Alcohol radicals react with Cd⁺ ions to form products which hydrolyse in the millisecond time scale to regenerate Cd²⁺. The reaction 2 Cd⁺→ Cd²⁺₂(2k= 3.0 × 10⁹ dm³ mol^–1 s^–1) is responsible for the formation of metallic cadmium. The rate constants for the reactions of Cd⁺ with OH and various organic radicals and the absorption spectra of Cd⁺ and Cd²⁺₂ were also determined. During the first few seconds of the disappearance of Cd²⁺₂, small agglomerates of cadmium that scatter u.v.-radiation are formed. Further agglomeration to produce colloidal particles that scatter 514.5 nm laser light occurs within a few minutes after the high energy electron pulse.

Scattering measurements were also carried out with irradiated Ni²⁺ solutions. The mechanism for colloidal nickel formation was found to be different from that for colloidal cadmium formation.