Answer:
three electrons were transferred in the process
Step-by-step explanation:
The electrode potential of the cathode is
E°cathode= E°cell + E°anode
E°cathode = 2.35V + (-0.74V)
E°cathode= 1.61 V
Let us look at the reduction half equation; the oxidation half equation must be;
Oxidation half equation;
Cr(s) ----> Cr^3+(aq) + 3e
The reduction half equation must now be
Reduction half equation;
3Ce^4+(aq) + 3e ----> 3Ce^3+(aq)
This implies that three electrons were transferred in the process as shown by the balanced half cell reaction equations.