Question

Calculate the charge in coulombs required for oxidation of 2 moles of water to oxygen

Answer 1

) 2H2O→4H+O2+4e− Q=4F =4×96500Cmol−1 =38600cmol−1 (b) Zn is oxidized and Ag2O is reduced to Ag E0cell=E0Ag+/Ag−E0zn2+/zn =[0.344−(−0.76)V]V =1.104V AG0=−nFE0cell =−2×96500cmol−1×1.10V =−213×105Jmol−1