Question

Sketch the galvanic cells based on the following half-reactions. Show the direction of electron flow, show the direction of ion migration through the salt bridge, and identify the cathode and anode. Give the overall balanced equation, and determine ${\mathcal{E}}^{\circ }$ for the galvanic cells. Assume that all concentrations are $1.0M$ and that all partial pressures are $1.0\mathrm{atm}.$ $\begin{array}{ll}\text{ a. }{\mathrm{H}}_2{\mathrm{O}}_2+2{\mathrm{H}}^{+}+2{\mathrm{e}}^{-}\to 2{\mathrm{H}}_2\mathrm{O}& {\mathcal{6}}^{\circ }=1.78\mathrm{V}\\ {\mathrm{O}}_2+2{\mathrm{H}}^{+}+2{\mathrm{e}}^{-}\to {\mathrm{H}}_2{\mathrm{O}}_2& {\mathcal{6}}^{\circ }=0.68\mathrm{V}\end{array}$ b. ${\mathrm{Mn}}^{2+}+2{\mathrm{e}}^{-}\to \mathrm{Mn}$ ${\mathcal{6}}^{\circ }=-1.18\mathrm{V}$ ${\mathrm{Fe}}^{3+}+3{\mathrm{e}}^{-}\to \mathrm{Fe}{\mathcal{6}}^{\circ }=-0.036\mathrm{V}$

Short answer

In the first galvanic cell, the anode (oxidation) half-reaction is ${\mathrm{O}}_2+2{\mathrm{H}}^{+}\to {\mathrm{H}}_2{\mathrm{O}}_2$, and the cathode (reduction) half-reaction is ${\mathrm{H}}_2{\mathrm{O}}_2+2{\mathrm{H}}^{+}\to 2{\mathrm{H}}_2\mathrm{O}$. The overall balanced equation is ${\mathrm{O}}_2+2{\mathrm{H}}^{+}\to 2{\mathrm{H}}_2\mathrm{O}$, with ${\mathcal{E}}_{cell}^{\circ }=1.1\mathrm{V}$. In the second galvanic cell, the anode half-reaction is ${\mathrm{Mn}}^{2+}+2{\mathrm{e}}^{-}\to \mathrm{Mn}$, and the cathode half-reaction is ${\mathrm{Fe}}^{3+}+3{\mathrm{e}}^{-}\to \mathrm{Fe}$. The overall balanced equation is $3{\mathrm{Mn}}^{2+}+2{\mathrm{Fe}}^{3+}\to 3\mathrm{Mn}+2\mathrm{Fe}$, with ${\mathcal{E}}_{cell}^{\circ }=1.144\mathrm{V}$. In both cells, electrons flow from the anode to the cathode, and ions migrate through the salt bridge towards their respective electrode.

Step-by-step solution

Identifying the half-reactions

Compare the standard cell potentials of the given reactions; the reaction with the higher standard cell potential will be the reduction half-reaction (cathode), while the other reaction will be the oxidation half-reaction (anode). Reduction half-reaction: $\mathrm{H}_{2} \mathrm{O}_{2}+2 \mathrm{H}^{+}+2 \mathrm{e}^{-} \rightarrow 2 \mathrm{H}_{2} \mathrm{O}$($\mathscr{E}^{\circ}=1.78 \mathrm{~V}$) Oxidation half-reaction: $\mathrm{O}_{2}+2 \mathrm{H}^{+}+2 \mathrm{e}^{-} \rightarrow \mathrm{H}_{2} \mathrm{O}_{2}$($\mathscr{E}^{\circ}=0.68 \mathrm{~V}$)

Electron flow and ion migration direction

Electrons flow from the anode to the cathode. Anions from the salt bridge move towards the anode and cations move towards the cathode.

Overall balanced equation

To get the overall balanced equation, we add the oxidation and reduction half-reactions: ${\mathrm{O}}_2+2{\mathrm{H}}^{+}+2{\mathrm{e}}^{-}\to {\mathrm{H}}_2{\mathrm{O}}_2+{\mathrm{H}}_2{\mathrm{O}}_2+2{\mathrm{H}}^{+}+2{\mathrm{e}}^{-}\to 2{\mathrm{H}}_2\mathrm{O}$ Simplified: ${\mathrm{O}}_2+2{\mathrm{H}}^{+}\to 2{\mathrm{H}}_2\mathrm{O}$

Standard cell potential

To find the standard cell potential, subtract the standard cell potential of the anode (oxidation) from the standard cell potential of the cathode (reduction): ${\mathcal{E}}_{cell}^{\circ }={\mathcal{E}}_{cathode}^{\circ }-{\mathcal{E}}_{anode}^{\circ }=1.78\mathrm{V}-0.68\mathrm{V}=1.1\mathrm{V}$ ##Sketching the cell for part b##

Identifying the half-reactions

Compare the standard cell potentials of the given reactions; the reaction with the higher standard cell potential will be the reduction half-reaction (cathode), while the other reaction will be the oxidation half-reaction (anode). Reduction half-reaction: ${\mathrm{Fe}}^{3+}+3{\mathrm{e}}^{-}\to \mathrm{Fe}$ (${\mathcal{E}}^{\circ }=-0.036\mathrm{V}$) Oxidation half-reaction: ${\mathrm{Mn}}^{2+}+2{\mathrm{e}}^{-}\to \mathrm{Mn}$ (${\mathcal{E}}^{\circ }=-1.18\mathrm{V}$)

Electron flow and ion migration direction

Electrons flow from the anode to the cathode. Anions from the salt bridge move towards the anode and cations move towards the cathode.

Overall balanced equation

To get the overall balanced equation, first balance the electrons by multiplying the manganese half-reaction by 3 and the iron half-reaction by 2. Then, add the oxidation and reduction half-reactions: $3({\mathrm{Mn}}^{2+}+2{\mathrm{e}}^{-}\to \mathrm{Mn})+2({\mathrm{Fe}}^{3+}+3{\mathrm{e}}^{-}\to \mathrm{Fe})$ $3{\mathrm{Mn}}^{2+}+6{\mathrm{e}}^{-}+2{\mathrm{Fe}}^{3+}+6{\mathrm{e}}^{-}\to 3\mathrm{Mn}+2\mathrm{Fe}$ Simplified: $3{\mathrm{Mn}}^{2+}+2{\mathrm{Fe}}^{3+}\to 3\mathrm{Mn}+2\mathrm{Fe}$

Standard cell potential

To find the standard cell potential, subtract the standard cell potential of the anode (oxidation) from the standard cell potential of the cathode (reduction): ${\mathcal{E}}_{cell}^{\circ }={\mathcal{E}}_{cathode}^{\circ }-{\mathcal{E}}_{anode}^{\circ }=-0.036\mathrm{V}-(-1.18\mathrm{V})=1.144\mathrm{V}$

Key concepts

Electron Flow

In a galvanic cell, the movement of electrons is crucial for generating electrical energy. Electrons always flow from the anode to the cathode. This is because the anode is the site of oxidation, where electrons are released, while the cathode is where reduction occurs, accepting electrons.

In practical terms, if you imagine a wire connecting two different metals immersed in their respective ion solutions, electrons will naturally move through this wire.

• From the anode, where oxidation happens and electrons are generated.
• Towards the cathode, where reduction occurs as the electrons get used up.

This flow of electrons from one electrode to the other forms an electrical current that can be harnessed to do work, like lighting a bulb or powering a device.

Ion Migration

Ions maintain charge neutrality in the galvanic cell. While electrons move through the external circuit, ions move internally, through the ion conductor, often a salt bridge. The migration of ions is essential to balance the flow of electrons and sustain the cell’s reaction.

Let's take a closer look:

• **Anions** (negative ions) from the salt bridge migrate towards the **anode**. This is because the anode, losing electrons, becomes positively charged and needs negative ions to balance it.
• **Cations** (positive ions) migrate towards the **cathode**. Here, the cathode gains electrons, becoming negatively charged; thus, positive ions are needed to neutralize this charge.

Without ion migration through the salt bridge, the cell would quickly reach equilibrium and stop functioning due to charge imbalance.

Cathode and Anode Identification

Identifying the cathode and anode in a galvanic cell is key to understanding its operation. This identification depends on the cell potentials of the half-reactions involved.

To determine which is which:

• The half-reaction with the higher standard reduction potential takes place at the cathode, undergoing reduction.
• The half-reaction with the lower standard reduction potential occurs at the anode, undergoing oxidation.

For example, if you have two reactions where one has a potential of 1.78 V and the other 0.68 V, the reaction with 1.78 V would be the cathode (reduced), and the 0.68 V the anode (oxidized). Remember, **reduction** at the **cathode** (Red Cat) and **oxidation** at the **anode** (An Ox).

Standard Cell Potential

The standard cell potential, denoted ${\mathcal{E}}_{cell}^{\circ }$, is the measure of the cell's ability to drive an electric current under standard conditions. It is calculated by taking the difference between the reduction potentials of the cathode and anode.

The formula is:
$${\mathcal{E}}_{cell}^{\circ }={\mathcal{E}}_{cathode}^{\circ }-{\mathcal{E}}_{anode}^{\circ }$$
To obtain the standard cell potential, ensure that both half-reactions are balanced and consider their respective standard reduction potentials.

For instance, if the standard potentials are 1.78 V at the cathode and 0.68 V at the anode, the calculation will be:
$$1.78\mathrm{V}-0.68\mathrm{V}=1.10\mathrm{V}$$This value indicates the driving force of the cell reaction.

Half-Reactions Balancing

Balancing half-reactions is an important step in determining the overall reaction in a galvanic cell. Each half-reaction must be adjusted so that the number of electrons lost in the oxidation process equals the number gained in the reduction process.

Here's how to balance them:

• Write the oxidation half-reaction and the reduction half-reaction separately.
• If the number of electrons in both reactions is not equal, multiply each reaction by a factor that will equalize the electron transfer.

In practical terms, if one reaction loses 2 electrons and another gains 3, you'll want to balance them by finding a common multiple (such as 6), adjusting coefficients accordingly.

For example, given:

• Oxidation: ${\mathrm{Mn}}^{2+}+2e^{-}\to \mathrm{Mn}$
• Reduction: ${\mathrm{Fe}}^{3+}+3e^{-}\to \mathrm{Fe}$

Multiply the manganese reaction by 3 and the iron reaction by 2 to equal 6 electrons exchanged.
This ensures that all elements and charges are balanced in the complete cell equation.