Question

Calculate the voltages of the following cells at \(25^{\circ} \mathrm{C}\) and under the following conditions: (a) \(\mathrm{Cu}\left|\mathrm{Cu}^{+}(0.80 \mathrm{M}) \| \mathrm{Hg}_{2}^{2+}(0.10 \mathrm{M})\right| \mathrm{Hg} \mid \mathrm{Pt}\) (b) \(\mathrm{Cr}\left|\mathrm{Cr}^{3+}(0.615 \mathrm{M}) \| \mathrm{Ni}^{2+}(0.228 \mathrm{M})\right| \mathrm{Ni}\)

Short answer

Question: Determine the cell voltages of two electrochemical cells. The first cell consists of Cu being oxidized to Cu+ and Hg2+ being reduced to Hg. The concentrations are [Cu+] = 0.8 M and [Hg2+] = 0.1 M. The second cell has Cr being oxidized to Cr3+ and Ni2+ being reduced to Ni, with [Cr3+] = 0.615 M and [Ni2+] = 0.228 M. Both cases are at 25°C. Solution: The cell voltages were calculated using the Nernst equation. For the first cell, the voltage is approximately 0.203 V and for the second cell, it is approximately 0.449 V.

Step-by-step solution

Identify the half-reactions

For the given cell, the half-reactions are: Oxidation: \(\mathrm{Cu} \to \mathrm{Cu}^+ + e^-\) Reduction: \(\mathrm{Hg}_2^{2+} + 2e^- \to 2 \mathrm{Hg}\)

Determine the standard cell voltage, \(E_0\)

Look up the standard reduction potentials from a table: \(E^0(\mathrm{Cu^+} \to \mathrm{Cu}) = +0.522\: \mathrm{V}\) \(E^0(\mathrm{Hg} \to \mathrm{Hg}_2^{2+}) = +0.851\: \mathrm{V}\) The standard cell voltage is found by taking the difference between the reduction potentials: \(E_{0} = E^0(\mathrm{Hg} \to \mathrm{Hg}_2^{2+}) - E^0(\mathrm{Cu^+} \to \mathrm{Cu}) = 0.851 - 0.522 = 0.329\: \mathrm{V}\)

Find the number of moles of electrons transferred, \(n\)

In this cell, the half-reaction for reduction has 2 moles of electrons transferred: $$ n = 2 $$

Calculate the reaction quotient, \(Q\)

From the balanced half-reactions, the reaction quotient is given by: $$Q = \frac{[\mathrm{Cu}^{+}]}{[\mathrm{Hg}_2^{2+}]}$$ Substitute the given concentrations: $$Q = \frac{0.80}{0.10} = 8$$

Plug the values into the Nernst equation and calculate the cell voltage, \(E_{cell}\)

Substitute the values into the Nernst equation: $$E_{cell} = E_{0} - \frac{RT}{nF} \ln{Q}$$ $$E_{cell} = 0.329 - \frac{8.314 \times 298}{2 \times 96485} \ln{8}$$ $$E_{cell} \approx 0.203\: \mathrm{V}$$ The cell voltage for part (a) is approximately 0.203 V. ##Part (b)##

Identify the half-reactions

For the given cell, the half-reactions are: Oxidation: \(\mathrm{Cr} \to \mathrm{Cr}^{3+} + 3e^-\) Reduction: \(\mathrm{Ni}^{2+} + 2e^- \to \mathrm{Ni}\)

Determine the standard cell voltage, \(E_0\)

Look up the standard reduction potentials from a table: \(E^0(\mathrm{Cr^{3+}} \to \mathrm{Cr}) = -0.744\: \mathrm{V}\) \(E^0(\mathrm{Ni^{2+}} \to \mathrm{Ni}) = -0.257\: \mathrm{V}\) The standard cell voltage is found by taking the difference between the reduction potentials: \(E_{0} = E^0(\mathrm{Ni^{2+}} \to \mathrm{Ni}) - E^0(\mathrm{Cr^{3+}} \to \mathrm{Cr}) = -0.257 - (-0.744) = 0.487\: \mathrm{V}\)

Find the number of moles of electrons transferred, \(n\)

In this cell, the half-reactions have 2 and 3 moles of electrons transferred. Taking the least common multiple, we get: $$ n = 6 $$

Calculate the reaction quotient, \(Q\)

From the balanced half-reactions, the reaction quotient is given by: $$Q = \frac{[\mathrm{Cr}^{3+}]^2}{[\mathrm{Ni}^{2+}]^3}$$ Substitute the given concentrations: $$Q = \frac{(0.615)^2}{(0.228)^3} \approx 29.034$$

Plug the values into the Nernst equation and calculate the cell voltage, \(E_{cell}\)

Substitute the values into the Nernst equation: $$E_{cell} = E_{0} - \frac{RT}{nF} \ln{Q}$$ $$E_{cell} = 0.487 - \frac{8.314 \times 298}{6 \times 96485} \ln{29.034}$$ $$E_{cell} \approx 0.449\: \mathrm{V}$$ The cell voltage for part (b) is approximately 0.449 V.

Key concepts

Nernst Equation

The Nernst Equation is a fundamental formula used to determine the potential of an electrochemical cell under non-standard conditions. It correlates the measurable cell potential with the temperature, the number of moles of electrons transferred during the electrochemical reaction (), the concentrations of the reactants and products (expressed through the reaction quotient, Q), and the standard cell potential (₀). The general form of the Nernst equation at temperature T is:\[\begin{equation}E_{cell} = E^0_{cell} - \frac{RT}{nF} \ln Q\end{equation}\]where:

• \(E_{cell}\) is the cell potential under non-standard conditions,
• \(E^0_{cell}\) is the standard cell potential,
• \(R\) is the universal gas constant (8.314 J/mol·K),
• \(T\) is the temperature in kelvins,
• \(n\) is the number of moles of electrons transferred per mole of reaction,
• \(F\) is the Faraday's constant (96485 C/mol), and
• \(Q\) is the reaction quotient.

When applying the Nernst equation, you can calculate the exact voltage of an electrochemical cell under given conditions, just like in the textbook example.

Standard Reduction Potential

Standard reduction potentials (\(E^0\)) play a pivotal role in determining the feasibility and direction of redox reactions. To find the standard cell potential (\(E^0_{cell}\)), we use the values for each electrode under standard conditions, which are usually provided in tables. Standard conditions are defined as 1 molar concentration for solutions, a pressure of 1 atm for gases, and a temperature of 25°C (or 298 K). The standard cell potential is the difference between the standard reduction potentials of the cathode (reduction half-reaction) and the anode (oxidation half-reaction).
\[\begin{equation}E^0_{cell} = E^0_{cathode} - E^0_{anode}\end{equation}\]Remember, a positive \(E^0_{cell}\) indicates that the redox reaction is spontaneous under standard conditions. The higher the reduction potential, the greater the tendency of the species to be reduced. In the example provided, the difference in standard reduction potentials between the cathode and anode determined the direction of electron flow and the cell's overall voltage.

Reaction Quotient

The reaction quotient (Q) is a dimensionless value that indicates the relative amounts of products and reactants present during a reaction at a given moment. It is similar to the equilibrium constant (_eq), but while _eq pertains to the state of equilibrium, Q can be calculated under any conditions. For a general reaction of the form ₂ → ₁, the reaction quotient is given by:\[\begin{equation}Q = \frac{[A]^a[B]^b}{[C]^c[D]^d}\end{equation}\]where [], [], [], and [] are the molar concentrations of the chemical species, and , , , and are their respective stoichiometric coefficients. When plugged into the Nernst equation, Q helps us determine how far the reaction has proceeded from the standard state and the effect this has on the cell potential.

Electrochemical Reaction Stoichiometry

Understanding electrochemical reaction stoichiometry is essential for accurate calculations in electrochemistry, especially when applying the Nernst equation. It involves balancing the redox reaction by ensuring that the number of electrons lost in oxidation equals the number caught by reduction. Finding the number of moles of electrons transferred () is crucial because it impacts several factors in the Nernst equation, such as the reaction quotient Q and the calculation of voltage.In an electrochemical cell, the stoichiometry is reflected through the half-reaction equations, where electrons should be balanced. For example, in the textbook solution, the stoichiometry of the half-reactions determines the value of used in the calculation. Stoichiometry also influences the reaction quotient as the stoichiometric coefficients of the reactants and products are the exponents in the formula for Q.