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 \(\left[\mathrm{Ce}^{4+}\right]=3.0 \mathrm{M}\), \(\left[\mathrm{Ce}^{3+}\right]=0.10 \mathrm{M}\), 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

(a) The emf of the cell under standard conditions is 2.35 V. (b) The emf of the cell with given concentrations is 2.332 V. (c) The emf of the cell with different given concentrations is 2.24 V.

Step-by-step solution

(1) Write Half-Reactions and Find Standard Reduction Potentials

Split the given reaction into two half-reactions: Half-Reaction 1: Ce4+(aq) + e- → Ce3+(aq) Half-Reaction 2: Cr(s) → Cr3+(aq) + 3e- Next, we need to find the standard reduction potentials (E°) for these half-reactions. We can find them in a standard reduction potential table: E°(Ce4+/Ce3+) = +1.61 V E°(Cr3+/Cr) = -0.74 V (Keep in mind that the reaction is reversed, so we change the sign.) Now we can determine the overall cell potential (E°cell).

(2) Calculate E°cell (Standard Cell Potential)

Use the following formula to calculate E°cell: E°cell = E°(cathode) - E°(anode) E°cell = E°(Ce4+/Ce3+) - E°(Cr3+/Cr) E°cell = (+1.61 V) - (-0.74 V) E°cell = 2.35 V So, the cell potential under standard conditions is 2.35 V.

(3) Calculate the emf of the cell at given concentrations using the Nernst equation

Now, we need to calculate the emf of the cell at the given concentrations using the Nernst equation: E = E° - (RT/nF) * ln(Q) where E is the emf of the cell, R is the gas constant (8.314 J/mol·K), T is the temperature in Kelvin (298K), n is the number of electrons transferred (3), F is the Faraday constant (9.6485 × 10^4 C/mol), and Q is the reaction quotient. For part (b), the concentrations are as follows: [Ce4+] = 3M [Ce3+] = 0.1M [Cr3+] = 0.01M First, calculate the reaction quotient Q: Q = ([Ce3+]^3 * [Cr3+]) / [Ce4+]^3 Q = ((0.1)^3 * 0.01) / (3)^3 Q = 1.85185185 × 10^-6 Now, use the Nernst equation to find the emf (E): E = 2.35 V - ((8.314 × 298 K) / (3 * 9.6485 × 10^4 C/mol)) * ln(1.85185185 × 10^-6) E = 2.35 V - 0.017707 V E = 2.332 V For part (b), emf of the cell is 2.332 V. For part (c), the concentrations are as follows: [Ce4+] = 0.1 M [Ce3+] = 1.75 M [Cr3+] = 2.5 M Calculate the reaction quotient Q: Q = ([Ce3+]^3 * [Cr3+]) / [Ce4+]^3 Q = ((1.75)^3 * 2.5) / (0.1)^3 Q = 13.515625 Now, use the Nernst equation to find the emf (E): E = 2.35 V - ((8.314 × 298 K) / (3 * 9.6485 × 10^4 C/mol)) * ln(13.515625) E = 2.35 V - 0.11297 V E = 2.24 V For part (c), emf of the cell is 2.24 V. To summarize the answers: (a) The emf of the cell under standard conditions is 2.35 V. (b) The emf of the cell with given concentrations is 2.332 V. (c) The emf of the cell with different given concentrations is 2.24 V.

Key concepts

Standard Reduction Potentials

Standard reduction potentials (E°) are intrinsic values indicating the tendency of a chemical species to gain electrons and get reduced. They are measured under standard conditions, which includes a temperature of 298 K, a pressure of 1 atmosphere, and 1 M concentration for each ion. These values are found in a standard reduction potential table and are crucial for predicting the direction of redox reactions and calculating the electromotive force (emf) of electrochemical cells, such as the voltaic cell in our exercise.

To calculate the overall cell potential (E°cell), identify each half-reaction involved. The half-reaction with the higher reduction potential acts as the cathode—the site where reduction occurs. The other half-reaction becomes the anode—the site of oxidation. The difference between the reduction potential of the cathode and the anode yields the standard cell potential. For example, in the provided exercise, the redox pair Ce⁴⁺/Ce³⁺ has a higher potential than Cr³⁺/Cr, hence it acts as the cathode.

Nernst Equation

The Nernst equation enables us to calculate the emf of an electrochemical cell under non-standard conditions. The equation is:\[\begin{equation}E = E° - \frac{RT}{nF} \ln(Q)ewlineewlineewlineewlineewlineewlineewlineewlineewlineewlineewlineewlineewlineewlineewlineewlineewlineewlineewlineewlineewlineewlineewlineewlineewlineewlineewlineewlineewlineewlineewlineewlineewlineewlineewlineewlineewlineewlineewlineewlineewlineewlineewlineewlineewlineewlineewlineewlineewlineewlineewlineewlineewlineewlineewlineewlineewlineewlineewlineewlineewlineewlineewlineewlineewlineewlineewlineewlineewlineewlineewlineewlineewlineewlineewlineewlineewlineewlineewlineewlineewlineewlineewlineewlineewlineewlineewlineewlineewlineewlineewlineewlineewlineewlineewlineewlineewlineewlineewlineewlineewlineewlineewlineewlineewlineewlineewlineewlineewlineewlineewlineewlineewlineewlineewlineewlineewlineewlineewlineewlineewlineewlineewlineewlineewlineewlineewlineewlineewlineewlineewlineewlineewlineewlineewlineewlineewlineewlineewlineewlineewlineewlineewlineewlineewlineewlineewlineewlineewlineewlineewlineewlineewlineewlineewlineewlineewlineewlineewlineewlineewlineewlineewlineewlineewlineewlineewlineewlineewlineewlineewlineewlineewlineewlineewlineewlineewlineewlineewlineewlineewlineewlineewlineewlineewlineewlineewlineewlineewlineewlineewlineewlineewlineewlineewlineewlineewlineewlineewlineewlineewlineewlineewlineewlineewlineewlineewlineewlineewlineewlineewlineewlineewlineewlineewlineewlineewlineewlineewlineewlineewlineewlineewlineewlineewlineewlineewlineewlineewlineewlineewlineewlineewlineewlineewlineewlineewlineewlineewlineewlineewlineewlineewlineewlineewlineewlineewlineewlineewlineewlineewlineewlineewlineewlineewlineewlineewlineewlineewlineewlineewlineewlineewlineewlineewlineewlineewlineewlineewlineewlineewlineewlineewlineewlineewlineewlineewlineewlineewlineewlineewlineewlineewlineewlineewlineewlineewlineewlineewlineewlineewlineewlineewlineewlineewlineewlineewlineewlineewlineewlineewlineewlineewlineewlineewlineewlineewlineewlineewlineewlineewlineewlineewlineewlineewlineewlineewlineewlineewlineewlineewlineewlineewlineewlineewlineewlineewlineewlineewlineewlineewlineewlineewlineewlineewlineewlineewlineewlineewlineewlineewlineewlineewlineewlineewlineewlineewlineewlineewlineewlineewlineewlineewlineewlineewlineewlineewlineewlineewlineewlineewlineewlineewlineewlineewlineewlineewlineewlineewlineewlineewlineewlineewlineewlineewlineewlineewlineewlineewlineewlineewline\end{equation}\]where E is the emf of the cell, R is the gas constant, T is the temperature in Kelvin, n is the number of electrons transferred in the cell reaction, F is the Faraday constant, and Q is the reaction quotient. This equation adjusts the standard cell potential to account for the effect of ion concentrations on the cell's emf. It is a vital tool for predicting cell behavior under various chemical environments.

Reaction Quotient (Q)

The reaction quotient (Q) is a calculation that expresses the ratio of products to reactants at any moment in time during a chemical reaction, raised to the power of their stoichiometric coefficients. For the example in our exercise, the reaction quotient is calculated based on the concentrations of the ions involved in the cell reaction. It represents how far the reaction has proceeded towards equilibrium. The reaction quotient is used in the Nernst equation to account for non-standard conditions and calculate the emf when the reactant and product concentrations are not at their standard states. Essentially, Q lets us find out the driving force of the reaction under the prevalent conditions.

Galvanic Cell Electrochemistry

Galvanic cell electrochemistry involves the study of chemical reactions that convert chemical energy into electrical energy through spontaneous redox reactions. By separating the oxidation and reduction half-reactions into two different compartments connected by a salt bridge, galvanic cells allow for electron flow through an external circuit. This electron flow is what we measure as electric current.

Each half-cell consists of an electrode and an electrolyte where either the oxidation or reduction half-reaction takes place. The difference in potential between the two electrodes under standard conditions is the cell's standard emf, while the Nernst equation allows for emf calculation under non-standard conditions. Understanding galvanic cells is fundamental for both theoretical analyses and practical applications, such as designing batteries and fuel cells.