Question

Given the following reaction: 2 Ag(s) + Fe₂⁺(aq) → 2 Ag(aq) + Fe(s), the standard reduction potentials are: Ag(aq) + e- → Ag(s) E0 = 0.80 V and Fe₂⁺(aq) + 2 e- → Fe(s) E0 = 0.77 V. What is the standard reduction potential (E0) for the reaction?

1) 0.03 V
2) 0.77 V
3) 0.80 V
4) 1.57 V

Answer 1

The standard reduction potential for the reaction 2 Ag(s) + Fe2+(aq) → 2 Ag+(aq) + Fe(s) is 1.57 V, which is the sum of the standard reduction potentials of the silver half-reaction and the reversed standard reduction potential of the iron half-reaction.

Step-by-step explanation:

The standard reduction potential (E0) for the given reaction 2 Ag(s) + Fe2+(aq) → 2 Ag+(aq) + Fe(s) can be calculated using the provided standard reduction potentials for the half-reactions. The half-reaction at the cathode is the reduction of Ag+ to Ag which has an E0 of 0.80 V. The half-reaction at the anode is the oxidation of Fe to Fe2+, and by convention, its E0 is the opposite of the E0 for its reduction, which is given as 0.77 V. Since oxidation occurs at the anode, we need to take the negative of this value. Therefore, the E0 of the anode reaction is -0.77 V. To find the standard cell potential (Ecell0), we subtract the anode's potential from the cathode's potential:

Ecell0 = Ecathode0 - Eanode0 = 0.80 V - (-0.77 V) = 0.80 V + 0.77 V = 1.57 V.

Therefore, the standard reduction potential for the reaction is 1.57 V, which corresponds to answer option 4.