Question

A voltaic cell utilizes the following reaction and operates at \(298 \mathrm{~K}\) : $$ 3 \mathrm{Ce}^{4+}(a q)+\mathrm{Cr}(s)-\rightarrow 3 \mathrm{Ce}^{3+}(a q)+\mathrm{Cr}^{3+}(a q) $$ (a) What is the emf of this cell under standard conditions? (b) What is the emf of this cell when \(\left[\mathrm{Ce}^{4+}\right]=3.0 M\), \(\left[\mathrm{Ce}^{3+}\right]=0.10 \mathrm{M}\), and \(\left[\mathrm{Cr}^{3+}\right]=0.010 \mathrm{M} ?\) (c) What is the emf of the cell when \(\left[\mathrm{Ce}^{4+}\right]=0.10 \mathrm{M},\left[\mathrm{Ce}^{3+}\right]=1.75 \mathrm{M}\), and \(\left[\mathrm{Cr}^{3+}\right]=2.5 \mathrm{M} ?\)

Short answer

The emf of the voltaic cell under different conditions are: (a) \(2.35 V\), (b) \(2.25 V\), and (c) \(-0.11 V\).

Step-by-step solution

Determine the standard reduction potentials

First, we need to identify the two half-reactions. For that, we can look up the standard reduction potentials for the involved reactions: \(Ce^{4+}(aq) + e^{-} \rightarrow Ce^{3+}(aq)\) with standard reduction potential, \(E_{Ce^{4+}/Ce^{3+}}^0\) \(Cr^{3+}(aq) + 3e^{-} \rightarrow Cr(s)\) with standard reduction potential, \(E_{Cr^{3+}/Cr(s)}^0\) Note that, in the provided reaction, \(\mathrm{Ce}^{4+}\) is reduced and \(\mathrm{Cr}(s)\) is oxidized. Hence, the latter reaction must be reversed: \(Cr(s) \rightarrow Cr^{3+}(aq) + 3e^{-}\) Now, we will have to look for the standard reduction potentials in a reference table. From the table, we get: \(E_{Ce^{4+}/Ce^{3+}}^0 = 1.61~V\) \(E_{Cr^{3+}/Cr(s)}^0 = -0.74~V\) Next, we need to determine the balanced equation for the overall reaction.

Balance the overall reaction

The overall reaction is given by: \[3\mathrm{Ce}^{4+}(aq) + Cr(s) \rightarrow 3\mathrm{Ce}^{3+}(aq) + Cr^{3+}(aq)\] Now, we can calculate the emf.

Calculate the emf for part (a)

Under standard conditions, the emf can be calculated as: \(E_{cell}^0 = E_{\mathrm{red}}^0 - E_{\mathrm{ox}}^0\) where \(E_{\mathrm{red}}^0\) is the standard reduction potential for the reduction half reaction (\(Ce^{4+}\) to \(Ce^{3+}\)) and \(E_{\mathrm{ox}}^0\) is the standard reduction potential for the oxidation half reaction (reversed reaction of \(Cr^{3+}\) to \(Cr(s)\)). Hence, we have: \(E_{cell}^0 = 1.61~V - (-0.74~V) = 2.35 V\) So, the emf under standard conditions is \(2.35~V\).

Calculate the emf for parts (b) and (c) using the Nernst Equation

To calculate the cell potentials for parts (b) and (c), we use the Nernst Equation: \[E_{cell} = E_{cell}^0 - \frac{RT}{nF} \ln Q\] where: - \(E_{cell}\) is the cell potential - \(E_{cell}^0\) is the standard cell potential - \(R\) is the gas constant \((8.314~JK^{-1}mol^{-1})\) - \(T\) is the temperature in Kelvin \(298 K\) - \(n\) is the number of transferred electrons \(3\) - \(F\) is Faraday's constant \((96,485~C mol^{-1})\) - \(Q\) is the reaction quotient For part (b), we have: \([\mathrm{Ce}^{4+}]=3.0 M\), \([\mathrm{Ce}^{3+}]=0.10 M\), and \([\mathrm{Cr}^{3+}]=0.010 M\). We need to calculate the reaction quotient \(Q\). \(Q = \frac{[\mathrm{Ce}^{3+}]^3 [\mathrm{Cr}^{3+}]}{[\mathrm{Ce}^{4+}]^3 [\mathrm{Cr}]} = \frac{(0.10)^3 (0.010)}{(3.0)^3}\) Now, we can plug the values into the Nernst Equation: \(E_{cell} = 2.35 - \frac{8.314 × 298}{3 × 96,485} \ln Q\) Solving this, we get \(E_{cell} \approx 2.25V\). For part (c), we have: \([\mathrm{Ce}^{4+}]=0.10 M\), \([\mathrm{Ce}^{3+}]=1.75 M\), and \([\mathrm{Cr}^{3+}]=2.5 M\). We need to calculate the reaction quotient \(Q\). \(Q = \frac{[(1.75)^3 (2.5)]}{(0.10)^3}\) Now, we can plug the values into the Nernst Equation: \(E_{cell} = 2.35 - \frac{8.314 × 298}{3 × 96,485} \ln Q\) Solving this, we get \(E_{cell} \approx -0.11V\). The answers for the three parts are: (a) \(2.35 V\) (b) \(2.25 V\) (c) \(-0.11 V\)

Key concepts

Standard Reduction Potential

The standard reduction potential, often denoted as \( E^0 \), is a measure of the tendency of a chemical species to gain electrons and be reduced. It is measured in volts and compared against the standard hydrogen electrode, which is arbitrarily set at 0 volts. In a voltaic cell, the species with the higher reduction potential will be reduced to gain electrons, while the species with the lower reduction potential will be oxidized to lose electrons.

In the example voltaic cell under consideration, we need to look at two separate half-reactions:

• The reduction of \( ext{Ce}^{4+} \) to \( ext{Ce}^{3+} \), which has a standard reduction potential \( E_{Ce^{4+}/Ce^{3+}}^0 = 1.61 \text{ V} \).
• The reverse oxidation reaction of \( ext{Cr}(s) \) to \( ext{Cr}^{3+} \), where the reduction potential is \( E_{Cr^{3+}/Cr(s)}^0 = -0.74 \text{ V} \).

Recognizing the need to reverse the second reaction for the oxidation half, we use these potentials to calculate the overall cell potential. Understanding these potentials is crucial because they determine the direction of electron flow and the cell's ability to perform electrical work under standard conditions.

Nernst Equation

The Nernst equation is pivotal when calculating the electromotive force (emf) of a cell under non-standard conditions. This equation accounts for the effects of concentration, pressure, and temperature on the cell potential, providing insights beyond what standard potentials can offer.

The Nernst equation is given by:\[ E_{cell} = E_{cell}^0 - \frac{RT}{nF} \ln Q \]where:

• \( E_{cell} \) is the emf under non-standard conditions.
• \( E_{cell}^0 \) is the standard cell potential.
• \( R \) is the universal gas constant, \( 8.314 \space JK^{-1}mol^{-1} \).
• \( T \) is the temperature in Kelvin.
• \( n \) is the number of moles of electrons exchanged in the reaction (3 in our case).
• \( F \) is Faraday's constant, \( 96485 \space Cmol^{-1} \).
• \( Q \) is the reaction quotient.

Through this equation, you can adjust the expected voltage output of a cell, acknowledging how real-world conditions deviate from standards, an essential consideration for practical applications.

EMF Calculation

Calculating the electromotive force (emf) under standard conditions is relatively straightforward. You subtract the standard reduction potential of the oxidation half-reaction from that of the reduction half-reaction:

\[ E_{cell}^0 = E_{red}^0 - E_{ox}^0 \]

For the voltaic cell mentioned, we have:

• The reduction of \( ext{Ce}^{4+} \) to \( ext{Ce}^{3+} \) contributes \( 1.61 \text{ V} \).
• The "oxidation" side for chromium (where we reverse the reduction potential for \( ext{Cr}^{3+} \) to \( ext{Cr}(s) \), making it \( +0.74 \text{ V} \)).
• Hence, \( E_{cell}^0 = 1.61 \text{ V} - (-0.74 \text{ V}) = 2.35 \text{ V} \).

This initial emf provides a baseline for what the cell can achieve under optimal, standard conditions before applying the real-world adjustments given by the Nernst equation.

Reaction Quotient

The reaction quotient \( Q \) provides a snapshot of a reaction's progress and is pivotal in the Nernst equation. It compares the concentrations (or, in some cases, pressures) of products and reactants not at equilibrium.

The general form for \( Q \) is:\[ Q = \frac{[ ext{products}]}{[ ext{reactants}]} \]

For our voltaic cell:

• The products are \( ext{Ce}^{3+} \) and \( ext{Cr}^{3+} \).
• The reactants are \( ext{Ce}^{4+} \) and \( ext{Cr}(s) \), though solids like chromium do not appear in \( Q \).

For example:- For part (b), \( Q = \frac{(0.10)^3 (0.010)}{(3.0)^3} \).- For part (c), \( Q = \frac{(1.75)^3 (2.5)}{(0.10)^3} \).Calculating \( Q \) allows you to assess whether a reaction will shift toward the products or reactants, influencing both the direction and magnitude of the cell's voltage. Understanding \( Q \) in the Nernst equation helps predict how far the cell's potential will differ from standard conditions.