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. Balanced cell equation: \(\mathrm{Cr}_{2} \mathrm{O}_{7}^{2-}+3\mathrm{H}_{2}\mathrm{O}_{2}+20\mathrm{H}^{+} \rightarrow 2 \mathrm{Cr}^{3+}+13\mathrm{H}_{2}\mathrm{O}\), Standard cell potential: \(\mathscr{E}^{\circ}=-0.44\,\text{V}\) b. Balanced cell equation: \(6\mathrm{H}^{+}+\mathrm{Al}^{3+}+9\mathrm{e}^{-} \rightarrow 3\mathrm{H}_{2}+\mathrm{Al}\), Standard cell potential: \(\mathscr{E}^{\circ}=1.66\,\text{V}\)

Step-by-step solution

Identify the oxidation and reduction half-reactions.

The two half-reactions are: 1. \(\mathrm{Cr}_{2} \mathrm{O}_{7}^{2-}+14 \mathrm{H}^{+}+6 \mathrm{e}^{-} \rightarrow 2 \mathrm{Cr}^{3+}+7 \mathrm{H}_{2} \mathrm{O}\) (Reduction half-reaction) 2. \(\mathrm{H}_{2} \mathrm{O}_{2}+2 \mathrm{H}^{+}+2 \mathrm{e}^{-} \rightarrow 2 \mathrm{H}_{2} \mathrm{O}\) (Oxidation half-reaction)

Balance the electrons in the half-reactions.

We need to multiply each half-reaction to make the number of electrons equal in both. Multiply the second reaction with 3, so we'll have 6 electrons in each reaction: 1. \(\mathrm{Cr}_{2} \mathrm{O}_{7}^{2-}+14 \mathrm{H}^{+}+6 \mathrm{e}^{-} \rightarrow 2 \mathrm{Cr}^{3+}+7 \mathrm{H}_{2} \mathrm{O}\) 2. \(3(\mathrm{H}_{2} \mathrm{O}_{2}+2 \mathrm{H}^{+}+2 \mathrm{e}^{-} \rightarrow 2 \mathrm{H}_{2} \mathrm{O}) \Rightarrow 3\mathrm{H}_{2}\mathrm{O}_{2}+6\mathrm{H}^{+}+6\mathrm{e}^{-}\rightarrow 6\mathrm{H}_{2} \mathrm{O}\)

Add half-reactions to get the overall cell reaction.

Add the two balanced half-reactions to get the overall cell reaction: \(\mathrm{Cr}_{2} \mathrm{O}_{7}^{2-}+14 \mathrm{H}^{+}+6 \mathrm{e}^{-}+3\mathrm{H}_{2}\mathrm{O}_{2}+6\mathrm{H}^{+}+6\mathrm{e}^{-} \rightarrow 2 \mathrm{Cr}^{3+}+7 \mathrm{H}_{2} \mathrm{O}+6\mathrm{H}_{2} \mathrm{O}\) Simplify the equation: \(\mathrm{Cr}_{2} \mathrm{O}_{7}^{2-}+3\mathrm{H}_{2}\mathrm{O}_{2}+20\mathrm{H}^{+} \rightarrow 2 \mathrm{Cr}^{3+}+13\mathrm{H}_{2}\mathrm{O}\)

Calculate the standard cell potential \(\mathscr{E}^{\circ}\) using the given standard reduction potentials:

The standard cell potential \(\mathscr{E}^{\circ}\) is the difference between the standard reduction potentials of the two half-reactions. Table 18.1 values: \(\mathscr{E}^{\circ}_\text{Cr}=1.33\,\text{V}\) (for the first half-reaction) \(\mathscr{E}^{\circ}_\text{H2O2}=1.77\,\text{V}\) (for the second half-reaction) Since \(\mathrm{H}_{2} \mathrm{O}_{2}\) is the oxidizing agent (it's being reduced), we'll have: \(\mathscr{E}^{\circ}=\mathscr{E}^{\circ}_\text{Cr}-\mathscr{E}^{\circ}_\text{H2O2}=1.33\,\text{V}-1.77\,\text{V}=-0.44\,\text{V}\) For the second galvanic cell:

Identify the oxidation and reduction half-reactions.

The two half-reactions are: 1. \(2 \mathrm{H}^{+}+2 \mathrm{e}^{-} \rightarrow \mathrm{H}_{2}\) (Reduction half-reaction) 2. \(\mathrm{Al}^{3+}+3 \mathrm{e}^{-} \rightarrow \mathrm{Al}\) (Oxidation half-reaction)

Balance the electrons in the half-reactions.

Multiply the first reaction by 3 to equalize the number of electrons in each half-reaction: 1. \(3(2 \mathrm{H}^{+}+2 \mathrm{e}^{-} \rightarrow \mathrm{H}_{2})\Rightarrow 6\mathrm{H}^{+}+6 \mathrm{e}^{-} \rightarrow 3\mathrm{H}_{2}\) 2. \(\mathrm{Al}^{3+}+3 \mathrm{e}^{-} \rightarrow \mathrm{Al}\)

Add half-reactions to get the overall cell reaction.

Add the two balanced half-reactions to get the overall cell reaction: \(6\mathrm{H}^{+}+6 \mathrm{e}^{-}+\mathrm{Al}^{3+}+3 \mathrm{e}^{-} \rightarrow 3\mathrm{H}_{2}+\mathrm{Al}\)

Calculate the standard cell potential \(\mathscr{E}^{\circ}\) using the given standard reduction potentials:

The standard cell potential \(\mathscr{E}^{\circ}\) is the difference between the standard reduction potentials of the two half-reactions. Table 18.1 values: \(\mathscr{E}^\circ_\text{H}=0\,\text{V}\) (for the first half-reaction) \(\mathscr{E}^\circ_\text{Al}=-1.66\,\text{V}\) (for the second half-reaction) Since \(\mathrm{H}^{+}\) is the oxidizing agent (it's being reduced), we'll have: \(\mathscr{E}^{\circ}=\mathscr{E}^{\circ}_\text{H}-\mathscr{E}^{\circ}_\text{Al}=0\,\text{V}-(-1.66\,\text{V})=1.66\,\text{V}\) In summary for both galvanic cells, the balanced cell equations and standard cell potentials are: a. \(\mathrm{Cr}_{2} \mathrm{O}_{7}^{2-}+3\mathrm{H}_{2}\mathrm{O}_{2}+20\mathrm{H}^{+} \rightarrow 2 \mathrm{Cr}^{3+}+13\mathrm{H}_{2}\mathrm{O}\), \(\mathscr{E}^{\circ}=-0.44\,\text{V}\) b. \(6\mathrm{H}^{+}+\mathrm{Al}^{3+}+9\mathrm{e}^{-} \rightarrow 3\mathrm{H}_{2}+\mathrm{Al}\), \(\mathscr{E}^{\circ}=1.66\,\text{V}\)

Key concepts

Standard Reduction Potential

A key concept in understanding galvanic cells is the standard reduction potential, represented as \( \mathscr{E}^{\circ} \). This quantity measures the tendency of a species to be reduced, acting as an oxidizing agent. Measured in volts, it provides a benchmark to predict the direction of electron flow in electrochemical reactions.

These values are tabulated, with most tables setting the standard hydrogen electrode (SHE) at zero volts for reference. A positive \( \mathscr{E}^{\circ} \) means the substance is a strong oxidizing agent, favoring reduction, while a negative value suggests it prefers oxidation. For example, \( \mathrm{H}_{2}\mathrm{O}_{2} \) has a standard reduction potential of \( 1.77 \, \text{V} \), signifying it is a strong oxidizer.

This is important because in any galvanic cell, electrons flow from the half-cell with a lower (more negative) they will. \( \mathscr{E}^{\circ} \) to the half-cell with a higher (more positive) \( \mathscr{E}^{\circ} \). Calculating the cell potential \( \mathscr{E}^{\circ}_{\text{cell}} \) involves subtracting the \( \mathscr{E}^{\circ} \) of the oxidation reaction from that of the reduction reaction, guiding us on whether the reaction is spontaneous.

Half-Reaction Balancing

Balancing half-reactions is crucial when working with galvanic cells. These reactions occur at the anode and cathode, where one species donates electrons (oxidation) and another accepts them (reduction).

The aim of balancing is to ensure that the same number of electrons is transferred in both half-reactions, making them equal and allowing for their combination into a full cell reaction.

To balance, we modify the coefficients of the substances involved to match the number of electrons lost in the oxidation reaction to the number of electrons gained in the reduction reaction. For instance, in the problem's oxidation half-reaction \( \mathrm{H}_{2}\mathrm{O}_{2} + 2 \mathrm{H}^{+} + 2 \mathrm{e}^{-} \rightarrow 2 \mathrm{H}_{2}\mathrm{O} \), we multiplied by 3 to equalize it with the reduction half-reaction because it involves 6 electrons.

This step might also require balancing atoms like oxygen or hydrogen by adding \( \mathrm{H}^{+} \), \( \mathrm{OH}^{-} \), or water molecules, especially in acidic or basic solutions, to ensure mass and charge balance.

Cell Potential Calculation

After balancing the half-reactions, calculating the cell potential \( \mathscr{E}^{\circ}_{\text{cell}} \) helps determine the feasibility and direction of the overall redox process.

The cell potential is the voltage, or electromotive force (EMF), generated by the galvanic cell when in standard conditions, calculated by subtracting the standard reduction potential of the oxidizing agent (\( \mathscr{E}^{\circ}_{\text{red}} \)) from that of the reducing agent (\( \mathscr{E}^{\circ}_{\text{ox}} \)).

For example, with \( \mathrm{H}_{2}\mathrm{O}_{2} \) reducing, the calculation was \( \mathscr{E}^{\circ}_{\text{cell}} = \mathscr{E}^{\circ}_{\text{red}} (1.33 \, \text{V}) - \mathscr{E}^{\circ}_{\text{ox}} (1.77 \, \text{V}) = -0.44 \, \text{V} \), showing a non-spontaneous process under standard conditions for this specific cell setup.

In practical terms, a positive \( \mathscr{E}^{\circ}_{\text{cell}} \) indicates a spontaneous reaction and energy-producing galvanic cell, while a negative value suggests the need for energy input to drive the reaction, typical of electrolytic cells.