Question

Based on the standard EMF series and your knowledge of half-reactions, determine the cell potential and spontanei ty of a cell that consists of a pure cobalt electrode in a solution of Co^2+ ions; the other half is a lead electrode immersed in a Pb^2+ solution.

Pb +2e- Pb Sn +2e Sn Ni 2e Ni Co 2e -0.126 -0.136 -0.250 -0.277 Co
a. +0.403, spontaneous
b. -0.403, nonspontaneous
c. +0.151, spontaneous
d. -0.151, nonspontaneous

Answer 1

+0.151, spontaneous

Step-by-step explanation:

Given that;

Co^2+(aq) + 2e ---->Co(s) -0.28 V

Pb^2+(aq) + 2e ---->Pb(s). -0.13 V

Hence Co is the anode and Pb is the cathode

E°cell = E°cathode - E°anode

So;

E°cell = -0.13 V - (-0.28 V)

E°cell = 0.15 V

The cell reaction is spontaneous since E°cell is positive.