Given the following two half-reactions,write the overall reaction in the direction in which it is product-favored,and calculate the standard cell potential. Pb²⁺(aq)+ 2 e^- → Pb(s) E° = -0.126 V Fe³⁺(aq)+ e^- → Fe²⁺(s) E° = +0.771 V A) Pb²⁺(aq)+ 2 Fe²⁺(s)→ Pb(s)+ 2 Fe³⁺(aq); = +0.897 V B) Pb²⁺(aq)+ Fe²⁺(s)→ Pb(s)+ Fe³⁺(aq); = +0.645 V C) Pb(s)+ 2 Fe³⁺(aq)→ Pb²⁺(aq)+ 2 Fe²⁺(s); = +1.416 V D) Pb(s)+ 2 Fe³⁺(aq)→ Pb²⁺(aq)+ 2 Fe²⁺(s); = +0.897 V E) Pb(s)+ Fe³⁺(aq)→ Pb²⁺(aq)+ Fe²⁺(s); = +0.645 V

Question

Quizplus · Accepted Answer

The overall reaction involves the reduction of PbS to Pb and the oxidation of FeS to FeS2. The half-reactions must be flipped and multiplied by factors to balance the number of electrons transferred.

PbS + 2e- -> Pb (E° = -0.126 V)
FeS -> FeS2 + 2e- (E° = -0.771 V)

Adding the two reactions gives:

PbS + FeS -> Pb + FeS2

The standard cell potential can be found using the formula:

E°cell = E°reduction + E°oxidation
E°cell = (-0.126 V) + (-0.771 V)
E°cell = -0.897 V

Choice D gives the correct overall reaction and standard cell potential.

Given the Following Two Half-Reactions,write the Overall Reaction in the Direction