Question

A voltaic cell similar to that shown in Figure \(20.5\) is constructed. One electrode compartment consists of a silver strip placed in a solution of \(\mathrm{AgNO}_{3}\), and the other has an iron strip placed in a solution of \(\mathrm{FeCl}_{2}\). The overall cell reaction is $$ \mathrm{Fe}(s)+2 \mathrm{Ag}^{+}(a q) \longrightarrow \mathrm{Fe}^{2+}(a q)+2 \mathrm{Ag}(s) $$ (a) What is being oxidized, and what is being reduced? (b) Write the half-reactions that occur in the two electrode compartments. (c) Which electrode is the anode, and which is the cathode? (d) Indicate the signs of the electrodes. (e) Do electrons flow from the silver electrode to the iron electrode, or from the iron to the silver? (f) In which directions do the cations and anions migrate through the solution?

Short answer

(a) Iron (Fe) is being oxidized and silver ion (Ag+) is being reduced. (b) Oxidation half-reaction: \(\mathrm{Fe}(s) \longrightarrow \mathrm{Fe}^{2+}(a q)+2 e^{-}\); Reduction half-reaction: \(\mathrm{Ag}^{+}(a q)+ e^{-} \longrightarrow \mathrm{Ag}(s)\). (c) Iron electrode is the anode, and silver electrode is the cathode. (d) Iron electrode (anode) is negative, and silver electrode (cathode) is positive. (e) Electrons flow from the iron electrode to the silver electrode. (f) Fe2+ and Ag+ migrate towards the silver electrode (cathode), while NO3- and Cl- migrate towards the iron electrode (anode).

Step-by-step solution

Identify the Oxidation and Reduction Processes

Recall that oxidation is the process of losing electrons, while reduction is the process of gaining electrons. In the given overall cell reaction: $$ \mathrm{Fe}(s)+2 \mathrm{Ag}^{+}(a q) \longrightarrow \mathrm{Fe}^{2+}(a q)+2 \mathrm{Ag}(s) $$ We can notice that Fe is losing 2 electrons as it becomes Fe2+ and Ag+ is gaining 1 electron as it becomes Ag. Therefore, Fe is being oxidized and Ag+ is being reduced.

Write the Half-Reactions

The half-reactions occurring in the two electrode compartments can be written as: Oxidation half-reaction: $$ \mathrm{Fe}(s) \longrightarrow \mathrm{Fe}^{2+}(a q)+2 e^{-} $$ Reduction half-reaction: $$ \mathrm{Ag}^{+}(a q)+ e^{-} \longrightarrow \mathrm{Ag}(s) $$

Identify the Anode and Cathode

The anode is the electrode where oxidation takes place. Since Fe is being oxidized, the iron electrode is the anode. The cathode is the electrode where reduction takes place. Since Ag+ is being reduced, the silver electrode is the cathode.

Indicate the Signs of the Electrodes

In a voltaic cell, the anode is the negative electrode, and the cathode is the positive electrode. Therefore, the iron electrode (anode) is negative, and the silver electrode (cathode) is positive.

Determine the Direction of Electron Flow

Electrons flow from the anode to the cathode in a voltaic cell. Since the iron electrode is the anode and the silver electrode is the cathode, electrons flow from the iron electrode to the silver electrode.

Determine the Directions of Cation and Anion Migration

In a voltaic cell, cations (positively charged ions) migrate towards the cathode, and anions (negatively charged ions) migrate towards the anode. In this cell, Fe2+ and Ag+ are cations, while NO3- and Cl- are anions. Therefore, Fe2+ and Ag+ will migrate towards the silver electrode (cathode), while NO3- and Cl- will migrate towards the iron electrode (anode).