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

ag+(aq) + e– ag(s) eo = +0.80 v
pb2+(aq) + 2e– pb(s) eo = –0.13 v

which value represents the overall electric cell potential?
use ecell = ered, cathode – ered, anode.

–0.13 v

–0.93 v

+0.67 v

+0.93 v

Reduction reactions are those reactions that reduce the oxidation number of a substance. Hence, the product side of the reaction must contain excess electrons. The opposite is true for oxidation reactions. When you want to determine the potential difference expressed in volts between the cathode and anode, the equation would be: E,reduction - E,oxidation. 

To cancel out the electrons, the e- in the reactions must be in opposite sides. To do this, you reverse the equation with the negative E0, then replacing it with the opposite sign. 

Pb(s) --> Pb2+ +2e-      E0 = +0.13 V
Ag+ + e-  ---> Ag           E0 = +0.80 V

Adding up the E0's would yield an overall electric cell potential of +0.93 V.

About 8 gallons of fuel will be used. (8.08)