It requires 40 ml of 0.5(M) Ce4+ to titrate 10 ml of 1(M) Sn2+ whereby Sn2+ is oxidized Sn4+. What is the oxidation state of cerium in the reduction product ?

Moles of Ce4+ reacted = 20 × 10-3 mol2 mol of Ce4+ reacts with (4 - x) mol of Sn2+∴ 20×10−3 mol of Ce4+ reacts with (4−x)×10−2mol of Sn2+∴ (4−x)×10−2=10×10−3 ∴ 4 - x = 1   or  x = + 3.