Posted by Steve on .
Calulate E naught cell for the reaction below
5 Cd(s) + 2 MnO4^-(aq) + 16H+(aq)+ 2 Mn2+(aq) + 8 HxsO(l)
The oxidation anode must be 5Cd(s) ==> 5Cd2+(aq) +10e^-
The reduction cathode must be 2MnO4^- + 16H2^=(aq) + 10e^- ==> 2Mn2+(aq) + 8H2O
E naught cell = E naught cathode - E naught anode = (+1.52V) - (-0.40V) = +1.92V
Inorganic chemistry -
I agree with 1.92 v.