This question has been asked several times and there appaer to be two different answer and no clear answer on who is right. (One response says the answer is wrong)
Please clarifiy,
Use the following voltaic cell reaction;
PdCl4(aq)+Cd(s)--> Pd(s) + 4 Cl^-(aq) + Cd^2-(aq) E cell= +1.03V
a) half reactions
b) Calcutlate E red involving Pd
c) ID the anode and cathode