Question

You have a concentration cell in which the cathode has a silver electrode with \(0.10 \mathrm{MAg}^{+}\). The anode also has a silver electrode with \(\mathrm{Ag}^{+}(a q), 0.050 \mathrm{M} \mathrm{S}_{2} \mathrm{O}_{3}{ }^{2-}\), and \(1.0 \times 10^{-3} \mathrm{M} \mathrm{Ag}\left(\mathrm{S}_{2} \mathrm{O}_{3}\right)_{2}{ }^{3-}\). You read the voltage to be \(0.76 \mathrm{~V}\). a. Calculate the concentration of \(\mathrm{Ag}^{+}\) at the anode. b. Determine the value of the equilibrium constant for the formation of \(\mathrm{Ag}\left(\mathrm{S}_{2} \mathrm{O}_{3}\right)_{2}{ }^{3-}\) \(\mathrm{Ag}^{+}(a q)+2 \mathrm{~S}_{2} \mathrm{O}_{3}{ }^{2-}(a q) \rightleftharpoons \mathrm{Ag}\left(\mathrm{S}_{2} \mathrm{O}_{3}\right)_{2}{ }^{3-}(a q) \quad K=?\)

Short answer

(a) The concentration of Ag⁺ at the anode is 2.996 M. (b) The value of the equilibrium constant for the formation of Ag(S₂O₃)₂³⁻ is 1333.33.

Step-by-step solution

Identify the cell reactions

The cell can be represented as: Ag+ (aq, cathode) | Ag(s) || Ag(s) | S₂O₃²⁻, Ag(S₂O₃)₂³⁻, Ag+ (aq, anode) At the cathode (reduction half-cell) where the Ag+ concentration is 0.10 M, the reaction is: \( Ag^{+}(aq) + e^{-} \rightarrow Ag(s) \) (reduction reaction) At the anode (oxidation half-cell) with Ag+(aq), S₂O₃²⁻, and Ag(S₂O₃)₂³⁻, the reaction is: \( Ag(s) \rightarrow Ag^{+}(aq) + e^{-} \) (oxidation reaction)

Calculate the Cell Potential

The Nernst equation relates the cell potential (E_cell) to the standard cell potential (E°_cell) and the concentrations of the species involved: \( E_{cell} = E°_{cell} - \dfrac{RT}{nF} \cdot ln (Q) \) Here, E°_cell = E°_cathode - E°_anode E°_cathode = E°_anode (since both are made of silver) E°_cell = 0 Given: E_cell = 0.76 V T = 298 K n = 1 (number of electrons involved in the reaction) R = 8.314 J/(K*mol) F = 96485 C/mol

Calculate the reaction quotient (Q)

For the cell reaction, \( Q_c = \dfrac{[M_{Ag+}^{(Anode)}]}{[M_{Ag+}^{(Cathode)}]} \) Substitute the given values into the Nernst equation and solve for Q: \( 0.76\,\text{V} = 0\,\text{V} - \dfrac{8.314 \text{ J/(K*mol)} × 298 \text{K}}{1 × 96485\, \text{C/mol}} \cdot ln(Q) \) First, find the value of the natural logarithm of Q (ln(Q)): \( ln(Q) = \dfrac{0.76\,\text{V} × 1 × 96485\, \text{C/mol}}{8.314 \text{ J/(K*mol)} × 298 \text{K}} = 3.40 \) Now, find the value of Q: \( Q = exp(3.40) = 29.96 \)

Calculate the Ag+ concentration at the anode

Since Q = [Ag⁺(Anode)]/[Ag⁺(Cathode)], we can find the concentration of Ag⁺ at the anode using the known concentration of Ag⁺ at the cathode (0.10 M): \( [M_{Ag^{+}}^{(Anode)}] = Q × [M_{Ag^{+}}^{(Cathode)}] = 29.96 × 0.10\,\text{M} \) \( [M_{Ag^{+}}^{(Anode)}] = 2.996\, \text{M} \) (a) The concentration of Ag⁺ at the anode is 2.996 M.

Write the expression for the equilibrium constant (K)

For the given equilibrium reaction, the expression for K is: \( K = \dfrac{[Ag(S_{2}O_{3})_{2}^{3-}]}{[Ag^{+}] \cdot [S_{2}O_{3}^{2-}]^{2}} \)

Calculate the equilibrium constant

Plug in the given and calculated concentrations into the expression for K: \( K = \dfrac{(1.0 \times 10^{-3}\,\mathrm{M})}{(2.996\,\mathrm{M}) \cdot (0.050\,\mathrm{M})^{2}} = 1333.33 \) (b) The value of the equilibrium constant for the formation of Ag(S₂O₃)₂³⁻ is 1333.33.

Key concepts

Understanding the Nernst Equation

The Nernst equation plays a pivotal role in electrochemistry by relating the cell potential to its standard potential and the activities or concentrations of the reactants and products. It is an indispensable tool for chemists to predict the behavior of electrochemical cells under non-standard conditions.

At its core, the Nernst equation is expressed as \( E_{cell} = E°_{cell} - \frac{RT}{nF} \cdot ln(Q) \). Here, \(E_{cell}\) is the actual cell potential measured under certain conditions, and \(E°_{cell}\) is the standard cell potential when the reactants and products are under standard conditions (1M concentration for solutions, 1 bar pressure for gases). The term \(RT/nF\) accounts for the temperature (T), the number of moles of electrons transferred in the reaction (n), the gas constant (R), and the Faraday's constant (F). The reaction quotient (Q) represents the ratio of the product activities to the reactant activities at a given instant.

One crucial aspect to remember is that for a concentration cell where the electrodes are the same material but the concentrations differ, the standard cell potential, \(E°_{cell}\), is zero. This is because the reduction potential for both the cathode and anode are identical, thus they cancel out. Understanding how to manipulate the Nernst equation to find unknowns such as concentration or cell potential is vital in solving electrochemistry problems.

Calculation of Cell Potential

Cell potential, denoted as \(E_{cell}\), represents the driving force of an electrochemical cell and is measured in volts (V). In a concentration cell, where the two electrodes have different concentrations, the cell potential is directly linked to the concentration gradient of the electrodes. It tells us how inclined a cell is to push electrons from the anode to the cathode.

For the concentration cell described in the exercise, the measured cell potential was 0.76V. This information, along with the concentrations of species at the electrodes, allows the determination of the cell potentials at non-standard conditions using the Nernst equation. The equation effectively bridges the gap between measurable electrical energy and chemical concentration.

The provided step-by-step solution appropriately calculated the cell potential, emphasizing the direct relationship between cell potential and the tendency for a chemical reaction to occur. The positive cell potential in our exercise signals a spontaneous reaction, and as the concentrations of reactants and products change, so too will the cell potential, a concept that should be grasped to predict the direction of electrochemical reactions.

Equilibrium Constant Calculation

The equilibrium constant, denoted by K, is a dimensionless value that reflects the ratio of the concentration of products to reactants for a reaction at equilibrium. Each reaction has a characteristic equilibrium constant that depends on temperature but is independent of the concentrations of reactants and products. In electrochemistry, the equilibrium constant gives insight into the extent of a reaction.

For the redox reaction in the concentration cell, the equilibrium constant is calculated using the concentrations of the ions at equilibrium. From the Nernst equation, we can derive the concentrations of these ions under given conditions and then determine K. As demonstrated in the solution, the equilibrium constant for the formation of the complex ion \( Ag(S_2O_3)_2^{3-} \) was found to be 1333.33.

The high value of K reflects a reaction heavily favoring the formation of the product, indicating that under the given conditions, the complex ion is predominantly formed. Students should understand that the equilibrium constant provides information on the position of equilibrium and helps predict whether a reaction will proceed to form more products or reactants.