A strip of iron is placed in a 1 M solution of iron(II) sulfate,and a strip of copper is placed in a 1 M solution of copper(II) chloride.The two solutions are connected with a salt bridge,and the two metals are connected by a wire.
Reduction Half-Reaction
E° (V)
Fe²⁺(aq) + 2e^- Fe(s)
-0) 41Cu²⁺(aq) + 2e^- Cu(s)
0) 34
Which of the following takes place?

Question

A strip of iron is placed in a 1 M solution of iron(II) sulfate,and a strip of copper is placed in a 1 M solution of copper(II) chloride.The two solutions are connected with a salt bridge,and the two metals are connected by a wire.
Reduction Half-Reaction
E° (V)
Fe²⁺(aq) + 2e^-

Fe(s)
-0) 41Cu²⁺(aq) + 2e^-

Cu(s)
0) 34
Which of the following takes place?

Quizplus · Accepted Answer

According to the given reduction half-reactions and their standard reduction potentials, copper has a higher tendency to be reduced compared to iron. Hence, copper(II) ions present in the copper(II) chloride solution will be reduced at the cathode (copper electrode) and form copper metal. At the same time, iron(II) ions in the iron(II) sulfate solution will be oxidized at the anode (iron electrode) and release electrons to the external circuit. Therefore, the Fe(II) concentration of the iron half-cell decreases due to the formation of Fe(III) ions. However, sulfur does not participate directly in the redox reactions and chlorine is not produced at either electrode.

A Strip of Iron Is Placed in a 1 M