Problem 53 - Chemistry and Chemical Reactivity

Question

You want to set up a series of voltaic cells with specific cell potentials. The \(\mathrm{Ag}^{+}(\mathrm{aq}, 1.0 \mathrm{M}) | \mathrm{Ag}(\mathrm{s})\) half-cell is one of the compartments. Identify several half-cells that you could use so that the cell potential will be close to (a) \(1.7 \mathrm{V}\) and (b) \(0.50 \mathrm{V} .\) Consider cells in which the silver cell can be either the cathode or the anode.

Step-by-Step Solution

Answer

Use Al³⁺/Al or Ni²⁺/Ni for potentials near 1.7 V and Zn²⁺/Zn or NiO₂/Ni²⁺ for 0.50 V.

Step 1: Determine Standard Reduction Potential of Ag⁺/Ag

The standard reduction potential for the half-cell \( \mathrm{Ag}^{+} + e^{-} \rightarrow \mathrm{Ag} \) is \(E^{\circ} = +0.80\, \mathrm{V}.\) This value will be used to find the cell potential when paired with other half-cells.

Step 2: Calculate Cell Potentials for (a) 1.7 V

For a cell potential of 1.7 V, the equation is:\[ E^{\circ}_{\text{cell}} = E^{\circ}_{\text{cathode}} - E^{\circ}_{\text{anode}} = 1.7\, \mathrm{V}.\]When Ag⁺/Ag is the cathode, we set\[ +0.80\, \mathrm{V} - E^{\circ}_{\text{anode}} = 1.7\, \mathrm{V},\] yielding \(E^{\circ}_{\text{anode}} = -0.90\, \mathrm{V}.\)Alternatively, when Ag⁺/Ag is the anode,\[ E^{\circ}_{\text{cathode}} - (+0.80\, \mathrm{V}) = 1.7\, \mathrm{V},\]resulting in \(E^{\circ}_{\text{cathode}} = 2.50\, \mathrm{V}.\)

Step 3: Identify Compatible Half-Cells for (a) 1.7 V

When \(\mathrm{Ag}^{+}/ \mathrm{Ag}\) is the cathode (0.80 V), the anode should be a species with \(-0.90\, \mathrm{V}\), such as the \(\mathrm{Al}^{3+}/ \mathrm{Al} \) half-cell \((E^{\circ} = -1.66 \mathrm{V}).\) When \(\mathrm{Ag}^{+}/ \mathrm{Ag}\) is the anode (0.80 V), the cathode should have \(2.50\, \mathrm{V},\) which is not available in standard tables.

Step 4: Calculate Cell Potentials for (b) 0.50 V

For a cell potential of 0.50 V, use:\[ E^{\circ}_{\text{cell}} = E^{\circ}_{\text{cathode}} - E^{\circ}_{\text{anode}} = 0.50\, \mathrm{V}.\]When Ag⁺/Ag is the cathode,\[ +0.80\, \mathrm{V} - E^{\circ}_{\text{anode}} = 0.50\, \mathrm{V},\] giving \(E^{\circ}_{\text{anode}} = +0.30\, \mathrm{V}.\)As the anode,\[ E^{\circ}_{\text{cathode}} - (+0.80\, \mathrm{V}) = 0.50\, \mathrm{V},\]resulting in \(E^{\circ}_{\text{cathode}} = 1.30\, \mathrm{V}.\)

Step 5: Identify Compatible Half-Cells for (b) 0.50 V

With \(\mathrm{Ag}^{+}/ \mathrm{Ag}\) as the cathode, choose an anode like the \( \mathrm{Zn}^{2+}/ \mathrm{Zn} \) half-cell \((E^{\circ} = -0.76 \mathrm{V}). \)With \(\mathrm{Ag}^{+}/ \mathrm{Ag}\) as the anode, opt for a cathode such as \( \mathrm{NiO}_{2}/ \mathrm{Ni}^{2+} \) \((E^{\circ} = 1.37 \mathrm{V}).\)

Key Concepts

Standard Reduction Potential

The concept of standard reduction potential is integral in understanding how voltaic cells work. It represents the tendency of a chemical species to gain electrons and thereby be reduced. The standard reduction potentials are measured under standard conditions, which involve solute concentrations of 1 M, a pressure of 1 atm for any gases, and a temperature of 25°C (298 K).

These potentials are expressed in volts and are typically determined using a standard hydrogen electrode (SHE) as the reference. The SHE has a reduction potential defined as zero volts. Other half-cells are then measured relative to this reference. In the exercise above, the standard reduction potential for the silver half-cell (\( \mathrm{Ag}^{+} + e^{-} \rightarrow \mathrm{Ag} \)) is given as \( E^{\circ} = +0.80 \, \mathrm{V} \). This positive value indicates that silver ions have a relatively strong tendency to be reduced to metallic silver.

Understanding these values helps in predicting how different half-cells can interact in terms of electron flow and the overall cell potential they can produce when combined.

Half-Cells

Half-cells are the building blocks of voltaic cells. A half-cell consists of a metal electrode in contact with a solution containing its ions. In the context of electrochemistry, half-cells operate via oxidation and reduction reactions. One half-cell undergoes oxidation (loses electrons), while the other undergoes reduction (gains electrons).

For example, in the exercise, the silver/silver ion half-cell \( (\mathrm{Ag}^{+}/\mathrm{Ag}) \) can operate either as an anode or cathode depending on which other half-cell it is paired with. An anode half-cell will lose electrons (oxidation), and the cathode half-cell will gain electrons (reduction). These paired reactions are critical as they determine the direction of electron flow which produces the electrical energy seen in voltaic cells.

Half-cells are connected through a salt bridge or a porous partition that allows ions to flow between them, maintaining the electrical neutrality that is essential for continuous operation of the cell.

Cell Potential

Cell potential, commonly referred to as electromotive force (EMF), is an essential measure of the capability of a voltaic cell to drive an electric current. The cell potential indicates the voltage available from the voltaic cell when it is at equilibrium and no current is flowing.

Calculated based on the standard reduction potentials of the involved half-cells, the cell potential is computed as:

• \( E^{\circ}_{\text{cell}} = E^{\circ}_{\text{cathode}} - E^{\circ}_{\text{anode}} \)

This equation assists in predicting the feasibility and duration a voltaic cell can sustain a current. In the given exercise, two target cell potentials (1.7 V and 0.50 V) are used as a guideline to determine which half-cells to pair with a silver/silver ion half-cell.

When deciding on feasible half-cell pairings, this calculated potential needs to reflect favored electron flow, determining how the voltaic cell configurations should be constructed.

Electrochemical Series

The electrochemical series, or activity series, is a crucial tool in electrochemistry. It systematically orders elements based on their standard reduction potentials, highlighting their reactivity. In essence, the series allows chemists to predict which metal or substance will be oxidized or reduced when competing reactions occur.

Higher ranked substances in the series are more adept in gaining electrons, meaning they are more readily reduced. Conversely, substances ranked lower are more likely to lose electrons, making them oxidized more easily. This understanding aids in making informed predictions about half-cell reactions in voltaic or electrochemical cells.

In exercises like the one provided, referencing the electrochemical series helps identify suitable pairings to achieve desired cell potentials. The chosen half-cell pairs need to match the necessary potential difference required for specific volatge targets. This process underscores the practicality of the electrochemical series in real-world applications, ensuring effective design and management of voltaic cells for varied purposes.