Question

Give the balanced cell equation and determine \(\mathscr{E}^{\circ}\) for the galvanic cells based on the following half-reactions. Standard reduction potentials are found in Table 18.1. a. \(\mathrm{Cr}_{2} \mathrm{O}_{7}^{2-}+14 \mathrm{H}^{+}+6 \mathrm{e}^{-} \rightarrow 2 \mathrm{Cr}^{3+}+7 \mathrm{H}_{2} \mathrm{O}\) \(\mathrm{H}_{2} \mathrm{O}_{2}+2 \mathrm{H}^{+}+2 \mathrm{e}^{-} \rightarrow 2 \mathrm{H}_{2} \mathrm{O}\) b. \(2 \mathrm{H}^{+}+2 \mathrm{e}^{-} \rightarrow \mathrm{H}_{2}\) \(\mathrm{Al}^{3+}+3 \mathrm{e}^{-} \rightarrow \mathrm{Al}\)

Short answer

a. The balanced cell equation for cell a is: \(\mathrm{Cr}_{2} \mathrm{O}_{7}^{2-} + 6\mathrm{H}_{2} \mathrm{O}_{2} + 20\mathrm{H}^{+} \rightarrow 2\mathrm{Cr}^{3+} + 13\mathrm{H}_{2} \mathrm{O}\) with a standard cell potential of \(\mathscr{E}^{\circ}= 3.10 \,\text{V}\). b. The balanced cell equation for cell b is: \(\mathrm{H}_{2} + \mathrm{Al}^{3+} \rightarrow 2 \mathrm{H}^{+} + \mathrm{Al}\) with a standard cell potential of \(\mathscr{E}^{\circ}= 1.66 \,\text{V}\).

Step-by-step solution

Balancing Half-Reactions

First, let's figure out the balanced reaction for cell a. We have: Oxidation: \(\mathrm{H}_{2} \mathrm{O}_{2}+2 \mathrm{H}^{+}+2 \mathrm{e}^{-} \rightarrow 2 \mathrm{H}_{2} \mathrm{O}\) Reduction: \(\mathrm{Cr}_{2} \mathrm{O}_{7}^{2-}+14 \mathrm{H}^{+}+6 \mathrm{e}^{-} \rightarrow 2 \mathrm{Cr}^{3+}+7 \mathrm{H}_{2} \mathrm{O}\) We need to balance the number of electrons in both half-reactions so that the total number of electrons in oxidation and reduction are equal. To do this, we can multiply the oxidation half-reaction by 3.

Balancing the Cell Reaction

Now our balanced half-reactions are: Oxidation: \(3\mathrm{(H}_{2} \mathrm{O}_{2}+2 \mathrm{H}^{+}+2 \mathrm{e}^{-} \rightarrow 2 \mathrm{H}_{2} \mathrm{O})\) Reduction: \(\mathrm{Cr}_{2} \mathrm{O}_{7}^{2-}+14 \mathrm{H}^{+}+6 \mathrm{e}^{-} \rightarrow 2 \mathrm{Cr}^{3+}+7 \mathrm{H}_{2} \mathrm{O}\) Now we can add the half-reactions together to get the balanced cell reaction: \(3(\mathrm{H}_{2} \mathrm{O}_{2}+2 \mathrm{H}^{+}+2 \mathrm{e}^{-}) + \mathrm{Cr}_{2} \mathrm{O}_{7}^{2-}+14 \mathrm{H}^{+}+6 \mathrm{e}^{-} \rightarrow 3(2 \mathrm{H}_{2} \mathrm{O}) + 2 \mathrm{Cr}^{3+}+7 \mathrm{H}_{2} \mathrm{O}\) Simplify the equation: \(\mathrm{Cr}_{2} \mathrm{O}_{7}^{2-} + 6\mathrm{H}_{2} \mathrm{O}_{2} + 20\mathrm{H}^{+} \rightarrow 2\mathrm{Cr}^{3+} + 13\mathrm{H}_{2} \mathrm{O}\)

Calculating \(\mathscr{E}^{\circ}\) for Cell a

Using Table 18.1, we can find the standard reduction potentials for the half-reactions: Oxidation (reverse of the original reaction): \(-\mathscr{E}^{\circ}(\mathrm{H}_{2} \mathrm{O}_{2}) = -1.77 \,\text{V}\) Reduction: \(\mathscr{E}^{\circ}(\mathrm{Cr}_{2} \mathrm{O}_{7}^{2-}) = 1.33 \,\text{V}\) Now we can calculate the total cell potential: \(\mathscr{E}^{\circ}_{total}=\mathscr{E}^{\circ}_{reduction} + (-\mathscr{E}^{\circ}_{oxidation})\) \(\mathscr{E}^{\circ}_{total}= 1.33 \,\text{V} - (-1.77 \,\text{V})\) \(\mathscr{E}^{\circ}_{total}= 3.10 \,\text{V}\) _Cell b_

Balancing Half-Reactions

Similarly, let's balance the half-reactions for cell b: Oxidation: \(\mathrm{H}_{2} \longrightarrow 2 \mathrm{H}^{+}+2 \mathrm{e}^{-}\) Reduction: \(\mathrm{Al}^{3+}+3 \mathrm{e}^{-} \longrightarrow \mathrm{Al}\) Both half-reactions are already balanced in terms of electrons.

Balancing the Cell Reaction

Now we can add the half-reactions together to get the balanced cell reaction: \(\mathrm{H}_{2} + \mathrm{Al}^{3+} \longrightarrow 2 \mathrm{H}^{+}+2 \mathrm{e}^{-} + \mathrm{Al}\) Simplify the equation: \(\mathrm{H}_{2} + \mathrm{Al}^{3+} \longrightarrow 2 \mathrm{H}^{+} + \mathrm{Al}\)

Calculating \(\mathscr{E}^{\circ}\) for Cell b

Using Table 18.1, we can find the standard reduction potentials for the half-reactions: Oxidation (reverse of the original reaction): \(-\mathscr{E}^{\circ}(\mathrm{H}_{2}) = 0 \,\text{V}\) Reduction: \(\mathscr{E}^{\circ}(\mathrm{Al}^{3+}) = -1.66 \,\text{V}\) Now we can calculate the total cell potential: \(\mathscr{E}^{\circ}_{total}=-\mathscr{E}^{\circ}_{oxidation} + \mathscr{E}^{\circ}_{reduction}\) \(\mathscr{E}^{\circ}_{total}= 0 \,\text{V} - (-1.66 \,\text{V})\) \(\mathscr{E}^{\circ}_{total}= 1.66 \,\text{V}\) So, the balanced cell equations and their corresponding standard cell potentials are: a. \(\mathrm{Cr}_{2} \mathrm{O}_{7}^{2-} + 6\mathrm{H}_{2} \mathrm{O}_{2} + 20\mathrm{H}^{+} \rightarrow 2\mathrm{Cr}^{3+} + 13\mathrm{H}_{2} \mathrm{O}\) with \(\mathscr{E}^{\circ}= 3.10 \,\text{V}\) b. \(\mathrm{H}_{2} + \mathrm{Al}^{3+} \rightarrow 2 \mathrm{H}^{+} + \mathrm{Al}\) with \(\mathscr{E}^{\circ}= 1.66 \,\text{V}\)

Key concepts

Standard Reduction Potentials

In the world of galvanic cells, standard reduction potentials are key to understanding how different substances react in an electrochemical cell. The standard reduction potential, typically represented as \( \mathscr{E}^{\circ} \), indicates the tendency of a chemical species to gain electrons and be reduced. This is a critical factor that influences the direction and feasibility of electrochemical reactions.

To find these potentials, we turn to tables like the one mentioned in the original exercise, Table 18.1, which lists various half-reaction potentials under standard conditions (1M concentration, 1 atm pressure, and 25°C). Values for standard reduction potentials are measured in volts (V) and help predict the behavior of substances in electrochemical reactions.

• Positive values indicate a greater likelihood to be reduced (gain electrons).
• Negative values suggest a lesser tendency to gain electrons; thus, the substance might prefer to lose electrons.

Knowing these potentials allows us to determine which half-reaction serves as the oxidation reaction and which one results in reduction.
Standard reduction potentials also guide us in calculating the overall cell potential, helping design systems with desirable electrical outputs.

Cell Potential Calculation

Cell potential, or electromotive force (EMF), is crucial in assessing the energy output of a galvanic cell. The cell potential is calculated using the standard reduction potentials of the involved half-reactions. It essentially measures the potential energy difference between the cathode and anode.

Here's a simplified process to calculate it:

• Identify and Reverse: For the overall cell reaction, identify the oxidation and reduction reactions using the reduction potentials. The oxidation process requires reversing a standard reduction reaction, changing its sign.
• Add Potentials: Add the standard potentials of the cathode's reduction and the reversed anode's oxidation half-reaction to get the total cell potential.

As in the original exercise:

• For cell a, where hydrogen peroxide oxidizes and dichromate reduces, calculate \(\mathscr{E}^{\circ}_{total} = 1.33 \,\text{V} + 1.77 \,\text{V} = 3.10 \,\text{V}\).
• For cell b, with hydrogen oxidizing and aluminum reducing, it's \(\mathscr{E}^{\circ}_{total} = 0 \,\text{V} + 1.66 \,\text{V} = 1.66 \,\text{V}\).

Thus, understanding and calculating cell potential enables us to determine the efficiency and feasibility of electrochemical reactions in galvanic cells.

Half-Reaction Balancing

Balancing half-reactions is a foundational step in analyzing galvanic cell reactions. Each half-reaction represents either the oxidation or the reduction process, and balancing them ensures that matter and charge are conserved in the overall cell reaction.

Start by writing the unbalanced half-reactions from the problem. Check that:

• Atoms are balanced. Ensure each element's total count matches on both sides of the half-reaction.
• Charges are balanced. This involves ensuring the same net charge on both sides.

When electrons are involved, it's crucial to balance the number of electrons lost in oxidation with those gained in reduction so that they cancel out when reactions are combined.

For example, in cell a from the exercise, the oxidation reaction initially provides 2 electrons. The reduction half-reaction consumes 6 electrons. To balance, multiply the oxidation reaction's coefficients by 3 so both reactions involve 6 electrons. Once electroneutrality and atomic balance are ensured, combine the reactions to form the balanced equation. Half-reaction balancing helps in evaluating the consistency of a proposed chemical process and in calculating the cell's electromotive force.