The strength of an oxidizing agent is related to its standard reduction potential. A higher (more positive) standard reduction potential indicates a stronger oxidizing agent because the species is more likely to gain electrons (be reduced).

Given the options:

Option A: $$E^o_{\text{Sn}^{4+}/\text{Sn}^{2+}} = +1.15 \text{ V}$$

Option B: $$E^o_{\text{Al}^{3+}/\text{Al}} = -1.66 \text{ V}$$

Option C: $$E^o_{\text{Pb}^{4+}/\text{Pb}^{2+}} = +1.67 \text{ V}$$

Option D: $$E^o_{\text{Tl}^{3+}/\text{Tl}} = +1.26 \text{ V}$$

The species with the highest standard reduction potential is the strongest oxidizing agent. In this case, Option C, $$E^o_{\text{Pb}^{4+}/\text{Pb}^{2+}} = +1.67 \text{ V}$$, is the highest value among the given potentials.

Therefore, lead(IV) ion ($\text{Pb}^{4+}$) with a reduction potential of $$+1.67 \text{ V}$$ is the strongest oxidizing agent among the given options.