Question

Sketch the galvanic cells based on the following overall reactions. Show the direction of electron flow, the direction of ion migration through the salt bridge, and identify the cathode and anode. Give the overall balanced equation. Assume that all concentrations are 1.0 \(M\) and that all partial pressures are 1.0 atm. a. \(\mathrm{IO}_{3}^{-}(a q)+\mathrm{Fe}^{2+}(a q) \Longrightarrow \mathrm{Fe}^{3+}(a q)+\mathrm{I}_{2}(a q)\) b. \(\mathrm{Zn}(s)+\mathrm{Ag}^{+}(a q) \rightleftharpoons \mathrm{Zn}^{2+}(a q)+\mathrm{Ag}(s)\)

Short answer

a. For reaction a, the cathode is the IO3- electrode and the anode is the Fe2+ electrode. Electrons flow from the Fe2+ electrode to the IO3- electrode through the wire. The salt bridge allows positive ions to migrate towards the cathode and negative ions towards the anode. The overall balanced equation is \(6\mathrm{Fe}^{2+}(aq) + 2\mathrm{IO}_{3}^{-}(aq) \rightarrow 6\mathrm{Fe}^{3+}(aq) + I_{2}(aq)\). b. For reaction b, the cathode is the Ag+ electrode and the anode is the Zn electrode. Electrons flow from the Zn electrode to the Ag+ electrode through the wire. The salt bridge allows positive ions to migrate towards the cathode and negative ions towards the anode. The overall balanced equation is \(\mathrm{Zn}(s) + 2\mathrm{Ag}^{+}(aq) \rightleftharpoons \mathrm{Zn}^{2+}(aq) + 2\mathrm{Ag}(s)\).

Step-by-step solution

Identify reduction and oxidation half-reactions

First, we need to identify the reduction and oxidation half-reactions for each overall reaction. a. For the reaction \(\mathrm{IO}_{3}^{-}(a q)+\mathrm{Fe}^{2+}(a q) \Longrightarrow \mathrm{Fe}^{3+}(a q)+\mathrm{I}_{2}(a q)\) Oxidation half-reaction: \(\mathrm{Fe}^{2+}(aq) \rightarrow \mathrm{Fe}^{3+}(aq) + e^-\) Reduction half-reaction: \(\mathrm{IO}_{3}^{-}(aq) + 6 e^{-} \rightarrow \mathrm{I}_{2}(aq)\) b. For the reaction \(\mathrm{Zn}(s)+\mathrm{Ag}^{+}(a q) \rightleftharpoons \mathrm{Zn}^{2+}(a q)+\mathrm{Ag}(s)\) Oxidation half-reaction: \(\mathrm{Zn}(s) \rightarrow \mathrm{Zn}^{2+}(aq) + 2 e^-\) Reduction half-reaction: \(\mathrm{Ag}^{+}(aq) + e^- \rightarrow \mathrm{Ag}(s)\)

Determine electron flow, ion migration direction, cathode, and anode

Electrons flow from the anode (oxidation) to the cathode (reduction). Ions in the salt bridge migrate to maintain electrical neutrality. a. For reaction a, Anode: \(\mathrm{Fe}^{2+}(aq) \rightarrow \mathrm{Fe}^{3+}(aq) + e^-\) Cathode: \(\mathrm{IO}_{3}^{-}(aq) + 6 e^{-} \rightarrow \mathrm{I}_{2}(aq)\) Electron flow: from Fe to IO3- Ion migration: positive ions towards the cathode and negative ions towards the anode. b. For reaction b, Anode: \(\mathrm{Zn}(s) \rightarrow \mathrm{Zn}^{2+}(aq) + 2 e^-\) Cathode: \(\mathrm{Ag}^{+}(aq) + e^- \rightarrow \mathrm{Ag}(s)\) Electron flow: from Zn to Ag+ Ion migration: positive ions towards the cathode and negative ions towards the anode.

Sketch the galvanic cell

Now we will sketch the galvanic cell based on the defined electron flow, ion migration, cathode and anode. a. For reaction a: - Anode: Fe2+ electrode - Cathode: IO3- electrode - Salt bridge connecting the two half-cells with positive ions going towards the cathode and negative ions going towards the anode. - Electron flow: from Fe2+ electrode to IO3- electrode through the wire. b. For reaction b: - Anode: Zn electrode - Cathode: Ag+ electrode - Salt bridge connecting the two half-cells with positive ions going towards the cathode and negative ions going towards the anode. - Electron flow: from Zn electrode to Ag+ electrode through the wire.

Write the overall balanced equation

Finally, we write the overall balanced equation for both reaction a and b: a. For reaction a, multiply the oxidation half-reaction by 6 and add the two half-reactions: \(6(\mathrm{Fe}^{2+}(aq) \rightarrow \mathrm{Fe}^{3+}(aq) + e^-) + (\mathrm{IO}_{3}^{-}(aq) + 6 e^{-} \rightarrow \mathrm{I}_{2}(aq))\) The overall balanced equation is: \(6\mathrm{Fe}^{2+}(aq) + 2\mathrm{IO}_{3}^{-}(aq) \rightarrow 6\mathrm{Fe}^{3+}(aq) + I_{2}(aq)\) b. For reaction b, simply add the oxidation and reduction half-reactions: \(\mathrm{Zn}(s) \rightarrow \mathrm{Zn}^{2+}(aq) + 2 e^- + 2(\mathrm{Ag}^{+}(aq) + e^- \rightarrow \mathrm{Ag}(s))\) The overall balanced equation is: \(\mathrm{Zn}(s) + 2\mathrm{Ag}^{+}(aq) \rightleftharpoons \mathrm{Zn}^{2+}(aq) + 2\mathrm{Ag}(s)\)

Key concepts

Galvanic cells

Galvanic cells, also known as voltaic cells, are electrochemical cells that convert chemical energy into electrical energy through spontaneous redox reactions. These reactions occur in two separate compartments, known as half-cells, that are connected by a wire and a salt bridge. In a galvanic cell, one electrode acts as the anode where oxidation occurs, and the other acts as the cathode where reduction takes place.
Electrons produced during the oxidation process at the anode travel through the external circuit to the cathode, generating an electric current that can power devices. Meanwhile, the salt bridge allows ions to migrate between the two half-cells to maintain electrical neutrality. This flow of ions is critical for the continuity of the electron flow and thus, for keeping the cell operational.
If you imagine a simple zinc-copper galvanic cell, where

• the zinc electrode serves as the anode, undergoing oxidation,
• and the copper electrode functions as the cathode, undergoing reduction,

this setup effectively demonstrates how a galvanic cell works. The spontaneous chemical reaction drives the flow of electrons from the zinc to the copper, creating electricity in the process.

Oxidation-reduction reactions

Oxidation-reduction reactions, or redox reactions, are chemical processes in which the oxidation state of atoms changes through the transfer of electrons. These reactions are essential for the functioning of galvanic cells and are broadly split into two interconnected half-reactions: oxidation and reduction.
In the oxidation half-reaction, a substance loses electrons and increases its oxidation state. Take for example zinc in a zinc-copper galvanic cell, where zinc metal

• loses electrons and transforms into zinc ions ( \( \text{Zn}(s) \rightarrow \text{Zn}^{2+}(aq) + 2e^- \) ).

Conversely, in the reduction half-reaction, a substance gains electrons and decreases its oxidation state. Continuing with the zinc-copper cell, copper ions

• in solution gain electrons to form solid copper ( \( \text{Cu}^{2+}(aq) + 2e^- \rightarrow \text{Cu}(s) \) ).

The redox reactions in galvanic cells release energy, manifesting as electricity, as one reactant is subjected to oxidation and the other to reduction. Understanding these reactions is pivotal in deciphering how energy conversion occurs within electrochemical setups.

Half-reaction balancing

Half-reaction balancing is a critical skill when dealing with redox reactions, as it ensures that the chemical process obeys the conservation of mass and charge. It involves separately balancing oxidation and reduction reactions and then combining them to yield a balanced chemical equation.
To balance half-reactions, follow these steps:

• First, identify the elements undergoing redox changes in the reaction.
• Balance all elements except for hydrogen and oxygen.
• Balance oxygen atoms by adding water molecules (\( \text{H}_2\text{O} \)).
• Balance hydrogen by adding hydrogen ions (\( \text{H}^+ \)).
• Finally, balance the charge by adding electrons.

After balancing the individual half-reactions, they are added together such that the electrons cancel out, yielding the overall balanced chemical equation.
This method is convenient, especially in galvanic cell calculations, enabling chemists and students alike to systematically solve complex electrochemical equations and predict the feasibility of reactions.