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

Question 49

Multiple Choice

Given the following two half-reactions,write the overall reaction in the direction in which it is spontaneous and calculate the standard cell potential.
Pb²⁺(aq) + 2 e^- $\to$ Pb(s) $~~~~~~~~$ $~~~~~~~~$ E^$\circ$ = -0.126 V
Fe³⁺(aq) + e^- $\to$ Fe²⁺(s) $~~~~~~~~$ $~~~~~~~~$ E^$\circ$ = +0.771 V

A) Pb²⁺(aq) + 2 Fe²⁺(s) $\to$ Pb(s) + 2 Fe³⁺(aq) $Given the following two half-reactions,write the overall reaction in the direction in which it is spontaneous and calculate the standard cell potential. Pb2+(aq) + 2 e- \to Pb(s) ~~~~~~~~~~~~~~~~ E \circ = -0.126 V Fe3+(aq) + e- \to Fe2+(s) ~~~~~~~~~~~~~~~~ E \circ = +0.771 V A) Pb2+(aq) + 2 Fe2+(s) \to Pb(s) + 2 Fe3+(aq) = +0.897 V B) Pb2+(aq) + Fe2+(s) \to Pb(s) + Fe3+(aq) = +0.645 V C) Pb(s) + 2 Fe3+(aq) \to Pb2+(aq) + 2 Fe2+(s) = +1.416 V D) Pb(s) + 2 Fe3+(aq) \to Pb2+(aq) + 2 Fe2+(s) = +0.897 V E) Pb(s) + Fe3+(aq) \to Pb2+(aq) + Fe2+(s) = +0.645 V$ = +0.897 V
B) Pb²⁺(aq) + Fe²⁺(s) $\to$ Pb(s) + Fe³⁺(aq) $Given the following two half-reactions,write the overall reaction in the direction in which it is spontaneous and calculate the standard cell potential. Pb2+(aq) + 2 e- \to Pb(s) ~~~~~~~~~~~~~~~~ E \circ = -0.126 V Fe3+(aq) + e- \to Fe2+(s) ~~~~~~~~~~~~~~~~ E \circ = +0.771 V A) Pb2+(aq) + 2 Fe2+(s) \to Pb(s) + 2 Fe3+(aq) = +0.897 V B) Pb2+(aq) + Fe2+(s) \to Pb(s) + Fe3+(aq) = +0.645 V C) Pb(s) + 2 Fe3+(aq) \to Pb2+(aq) + 2 Fe2+(s) = +1.416 V D) Pb(s) + 2 Fe3+(aq) \to Pb2+(aq) + 2 Fe2+(s) = +0.897 V E) Pb(s) + Fe3+(aq) \to Pb2+(aq) + Fe2+(s) = +0.645 V$ = +0.645 V
C) Pb(s) + 2 Fe³⁺(aq) $\to$ Pb²⁺(aq) + 2 Fe²⁺(s) $Given the following two half-reactions,write the overall reaction in the direction in which it is spontaneous and calculate the standard cell potential. Pb2+(aq) + 2 e- \to Pb(s) ~~~~~~~~~~~~~~~~ E \circ = -0.126 V Fe3+(aq) + e- \to Fe2+(s) ~~~~~~~~~~~~~~~~ E \circ = +0.771 V A) Pb2+(aq) + 2 Fe2+(s) \to Pb(s) + 2 Fe3+(aq) = +0.897 V B) Pb2+(aq) + Fe2+(s) \to Pb(s) + Fe3+(aq) = +0.645 V C) Pb(s) + 2 Fe3+(aq) \to Pb2+(aq) + 2 Fe2+(s) = +1.416 V D) Pb(s) + 2 Fe3+(aq) \to Pb2+(aq) + 2 Fe2+(s) = +0.897 V E) Pb(s) + Fe3+(aq) \to Pb2+(aq) + Fe2+(s) = +0.645 V$ = +1.416 V
D) Pb(s) + 2 Fe³⁺(aq) $\to$ Pb²⁺(aq) + 2 Fe²⁺(s) $Given the following two half-reactions,write the overall reaction in the direction in which it is spontaneous and calculate the standard cell potential. Pb2+(aq) + 2 e- \to Pb(s) ~~~~~~~~~~~~~~~~ E \circ = -0.126 V Fe3+(aq) + e- \to Fe2+(s) ~~~~~~~~~~~~~~~~ E \circ = +0.771 V A) Pb2+(aq) + 2 Fe2+(s) \to Pb(s) + 2 Fe3+(aq) = +0.897 V B) Pb2+(aq) + Fe2+(s) \to Pb(s) + Fe3+(aq) = +0.645 V C) Pb(s) + 2 Fe3+(aq) \to Pb2+(aq) + 2 Fe2+(s) = +1.416 V D) Pb(s) + 2 Fe3+(aq) \to Pb2+(aq) + 2 Fe2+(s) = +0.897 V E) Pb(s) + Fe3+(aq) \to Pb2+(aq) + Fe2+(s) = +0.645 V$ = +0.897 V
E) Pb(s) + Fe³⁺(aq) $\to$ Pb²⁺(aq) + Fe²⁺(s) $Given the following two half-reactions,write the overall reaction in the direction in which it is spontaneous and calculate the standard cell potential. Pb2+(aq) + 2 e- \to Pb(s) ~~~~~~~~~~~~~~~~ E \circ = -0.126 V Fe3+(aq) + e- \to Fe2+(s) ~~~~~~~~~~~~~~~~ E \circ = +0.771 V A) Pb2+(aq) + 2 Fe2+(s) \to Pb(s) + 2 Fe3+(aq) = +0.897 V B) Pb2+(aq) + Fe2+(s) \to Pb(s) + Fe3+(aq) = +0.645 V C) Pb(s) + 2 Fe3+(aq) \to Pb2+(aq) + 2 Fe2+(s) = +1.416 V D) Pb(s) + 2 Fe3+(aq) \to Pb2+(aq) + 2 Fe2+(s) = +0.897 V E) Pb(s) + Fe3+(aq) \to Pb2+(aq) + Fe2+(s) = +0.645 V$ = +0.645 V

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