Answer:
E° = 1.24 V
Explanation:
Let's consider the following galvanic cell: Fe(s) | Fe²⁺(aq) || Ag⁺(aq) | Ag(s)
According to this notation, Fe is in the anode (where oxidation occurs) and Ag is in the cathode (where reduction occurs). The corresponding half-reactions are:
Anode: Fe(s) ⇒ Fe²⁺(aq) + 2 e⁻
Cathode: Ag⁺(aq) + 1 e⁻ ⇒ Ag(s)
The standard cell potential (E°) is the difference between the standard reduction potential of the cathode and the standard reduction potential of the anode.
E° = E°red, cat - E°red, an
E° = 0.80 V - (-0.44 V) = 1.24 V
