Why Sn(II) is a reducing agent while Pb(II)is not?

The reducing character of element in 4^th group is given by
M+2 ------->M+4 +2e-
As stability of the +4 oxidation state decreases down the group,therefore Sn(IV)is more stable as compared to Pb(IV).
For the elements like Sn and Pb to be reducing in (II) state ; they must have stable (IV) state as;

In reducing agent M+2 -> M+4 + 2e-

Now, we know that the inert pair effect increases down the group , so the stability of +4 oxidation state decreases down the group and hence the +4 oxidation state of Pb is less stable than that of Sn .

So, it becomes little difficult for Pb to get converted into its +4 state from its +2 state, so it does not behave as reducing agent.

Because the stability of +4 oxidation state decreases as we go from top to bottom of a group, therefore Pb(||) is not a reducing agent while Sn(||) isCSiGeSnPb