Question

Calculate voltages of the following cells at \(25^{\circ} \mathrm{C}\) and under the following conditions. (a) \(\mathrm{Zn}\left|\mathrm{Zn}^{2+}(0.50 \mathrm{M}) \| \mathrm{Cd}^{2+}(0.020 \mathrm{M})\right| \mathrm{Cd}\) (b) \(\mathrm{Cu}\left|\mathrm{Cu}^{2+}(0.0010 \mathrm{M}) \| \mathrm{H}^{+}(0.010 \mathrm{M})\right| \mathrm{H}_{2}(1.00 \mathrm{~atm}) \mid \mathrm{Pt}\)

Short answer

Question: Calculate the cell voltages for the two given electrochemical cells at \(25^{\circ} \mathrm{C}\): (a) Cell: \(\mathrm{Zn}\left|\mathrm{Zn}^{2+}(0.50 \mathrm{M})\|\mathrm{Cd}^{2+}(0.020 \mathrm{M})\right|\mathrm{Cd}\) (b) Cell: \(\mathrm{Cu}\left|\mathrm{Cu}^{2+}(0.0010 \mathrm{M})\|\mathrm{H}^{+}(0.010 \mathrm{M})\right|\mathrm{H}_{2}(1.00 \mathrm{~atm})\mid\mathrm{Pt}\) Answer: (a) Cell voltage: \(0.340\:V\) (b) Cell voltage: \(0.786\:V\)

Step-by-step solution

Identify half-reactions and standard reduction potential

Zn: Zn -> Zn\(^{2+}\) + 2e\(^-\) : \(E^{\circ}_1 = -0.76\:V\) (oxidation) Cd: Cd\(^{2+}\) + 2e\(^-\) -> Cd \(\:E^{\circ}_2 = -0.40\:V\) (reduction)

Calculate the standard cell voltage

\(E^{\circ}_{cell} = E^{\circ}_2 - E^{\circ}_1 = -0.40 - (-0.76) = 0.36\:V\)

Determine the number of moles of electrons transferred

In both reactions, two electrons are transferred; therefore, \(n = 2\).

Calculate the reaction quotient Q

For the reaction, the quotient is \(Q = \frac{[\mathrm{Cd}^{2+}]}{[\mathrm{Zn}^{2+}]} = \frac{0.020}{0.50}\).

Use Nernst equation to calculate the cell voltage

\(E_{cell} = E^{\circ}_{cell} - \frac{RT}{nF} \ln(Q) = 0.36 - \frac{(8.314)(298)}{(2)(96485)} \ln\left(\frac{0.020}{0.50}\right) = \boxed{0.340\:V}\) (b) Cell: \(\mathrm{Cu}\left|\mathrm{Cu}^{2+}(0.0010 \mathrm{M})\|\mathrm{H}^{+}(0.010 \mathrm{M})\right|\mathrm{H}_{2}(1.00 \mathrm{~atm})\mid\mathrm{Pt}\)

Identify half-reactions and standard reduction potential

Cu: Cu -> Cu\(^{2+}\) + 2e\(^-\) : \(E^{\circ}_1 = 0.34\:V\) (oxidation) H: 2H\(^+\) + 2e\(^-\) -> H\(_2\) : \(E^{\circ}_2 = -0.00\:V\) (reduction)

Calculate the standard cell voltage

\(E^{\circ}_{cell} = E^{\circ}_2 - E^{\circ}_1 = 0.79\:V\)

Determine the number of moles of electrons transferred

In each reaction, two electrons are transferred; therefore, \(n = 2\).

Calculate the reaction quotient Q

For the reaction, the quotient is \(Q = \frac{[\mathrm{H}^{+}]^2}{[\mathrm{Cu}^{2+}][\mathrm{H}_2]/P^{\circ} } = \frac{(0.010)^2}{(0.0010)(1.00)}\).

Use Nernst equation to calculate the cell voltage

\(E_{cell} = E^{\circ}_{cell} - \frac{RT}{nF} \ln(Q) = 0.79 - \frac{(8.314)(298)}{(2)(96485)} \ln\left(\frac{(0.010)^2}{(0.0010)(1.00)}\right) = \boxed{0.786\:V}\)

Key concepts

Standard Reduction Potential

The standard reduction potential, represented by the symbol \(E^\circ\), is a measure of the tendency of a chemical species to gain electrons and thereby be reduced. It is an essential concept in electrochemistry as it helps predict the direction of electron flow in an electrochemical cell. Each half-cell in an electrochemistry setup has its standard reduction potential, which is based on a scale where the standard hydrogen electrode (SHE) has been assigned a potential of 0 volts. The standard reduction potentials for zinc and cadmium as part of the exercise are \(-0.76\text{ V}\) and \(-0.40\text{ V}\), respectively. These values are crucial for calculating the standard cell voltage and the overall cell potential under standard conditions.

When referring to standard conditions, it means that the concentrations of all the aqueous species are at 1M concentration, the gases are at a pressure of 1 atm, and the temperature is usually 25 degrees Celsius (298 K). With these in place, the potential of the electrochemical cell can be calculated under standard conditions using the standard reduction potentials of the involved half-reactions.

Standard Cell Voltage

The standard cell voltage, \(E^\circ_{cell}\), is the difference in potential between two half-cells in an electrochemical cell when all species are at standard conditions (usually 1M concentration, 1 atm pressure, 298 K). It represents the maximum potential difference the cell can achieve without any non-standard conditions such as altered concentrations or pressures. To calculate the standard cell voltage, subtract the standard reduction potential of the anode (oxidation half-reaction) from the cathode (reduction half-reaction). For example, in the zinc-cadmium cell, the standard cell voltage is calculated as \(E^\circ_{cell} = E^\circ_{\text{Cd}} - E^\circ_{\text{Zn}} = -0.40\text{ V} - (-0.76 \text{ V}) = 0.36\text{ V}\). This gives a picture of the driving force behind the electron transfer under standard conditions.

Reaction Quotient

The reaction quotient, \(Q\), is a function that denotes the instant ratio of concentrations of products to reactants, each raised to the power of their respective coefficients in the balanced chemical equation for the reaction taking place in the electrochemical cell. In essence, it reflects the current state of a reaction versus its equilibrium state. It plays a central role in the Nernst equation, allowing for the calculation of the cell's voltage under non-standard conditions. For the exercise involving zinc and cadmium, the reaction quotient is calculated using the given concentrations: \(Q = \frac{[\text{Cd}^{2+}]}{[\text{Zn}^{2+}]} = \frac{0.020}{0.50}\). The Nernst equation uses \(Q\) to determine how far the cell potential deviates from the standard cell potential due to changes in concentration.

Electrochemical Cells

Electrochemical cells are the basic units of a battery, where chemical energy is converted into electrical energy through redox reactions. They consist of anode and cathode compartments, which are half-cells containing electrodes submerged in an electrolyte. The flow of electrons is driven from the anode to the cathode through an external circuit, while ions move through the electrolyte to balance the charge. Electrochemical cells are categorized as galvanic (or voltaic) when they produce electrical energy spontaneously, or electrolytic when electrical energy is required to drive a non-spontaneous reaction. The zinc-cadmium and copper-hydrogen combinations from the exercise represent two different galvanic cells where spontaneous electron flow can be predicted and measured by the concepts of standard reduction potentials and standard cell voltages.

Electron Transfer

Electron transfer is the core of electrochemical reactions, where electrons move from one species to another. This transfer is the basis of the oxidation-reduction (redox) reactions happening in electrochemical cells. The species that loses electrons is oxidized (anode), while the species that gains electrons is reduced (cathode). The ability to transfer electrons between chemical species is quantified by their reduction potentials. In the given exercise scenarios, zinc is oxidized to zinc ions, releasing two electrons, and cadmium ions gain two electrons to form cadmium metal. Similarly, copper is oxidized, and protons are reduced to hydrogen gas. Understanding electron transfer is critical for analyzing and predicting the behavior of electrochemical cells, as well as for harnessing electrical energy from chemical reactions.