Question

An electrochemical cell is powered by the half reactions shown below.

Ag+(aq) + e– mc017-1.jpg Ag(s) Eo = +0.80 V
Pb2+(aq) + 2e– mc017-2.jpg Pb(s) Eo = –0.13 V

Which value represents the overall electric cell potential?
Use Emc017-3.jpgcell = Emc017-4.jpgred, cathode – Emc017-5.jpgred, anode.
–0.13 V
–0.93 V
+0.67 V
+0.93 V

Answer 1

0.93

Step-by-step explanation:

Answer 2

Ag+(aq) + e– → Ag(s) Eo = +0.80 V
Pb2+(aq) + 2e– → Pb(s) Eo = –0.13 V

=>

2Ag+(aq) + 2e– → 2Ag(s) Eo = +0.80 V

Pb(s) → Pb2+(aq) + 2e– - Eo = – (-0.13 V) = +0.13 V
----------------------------------

2Ag(+) (aq) + Pb(s) -------> 2Ag(s) + Pb(2+)


E cell = E0 cathode – Eo anode = 0.80V - (- 0.13V) = 0.80V + 0.13V = 0.93V.

Answer: + 0.93V