Question

You have a concentration cell in which the cathode has a silver electrode with 0.10 \(M\) Ag \(^{+}\). The anode also has a silver electrode with \(\mathrm{Ag}^{+}(a q), 0.050 \space \mathrm{M}\space \mathrm{S}_{2} \mathrm{O}_{3}^{2-},\) 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

The concentration of Ag+ at the anode is \(1.51\times10^{-6} M\), and the equilibrium constant K for the formation of \(Ag(S_2O_3)_2^{3-}\) is 8.82.

Step-by-step solution

Write down the Nernst Equation

The Nernst Equation is given by: \(E_{cell}=E_{0}-\frac{RT}{nF}\ln Q\) where: - \(E_{cell}\) is the cell voltage - \(E_{0}\) is the standard cell voltage (since both electrodes are Ag, \(E_{0} = 0\)) - R is the gas constant (\(R = 8.314\space J\space mol^{-1}\space K^{-1}\)) - T is the temperature (assume standard temperature, 298 K) - n is the number of electrons involved in the redox reaction (n = 1) - F is the Faraday's constant (\(F = 9.648\times10^4\space C\space mol^{-1}\)) - Q is the reaction quotient

Determine the value of Q

The reaction quotient Q is given by the ratio of concentrations: \(Q = \frac{[Ag+]_{anode}}{[Ag+]_{cathode}}\) Since we know the concentration of Ag+ at the cathode, we have: \(Q = \frac{[Ag+]_{anode}}{0.10}\)

Use the Nernst Equation to find Ag+ concentration at the anode

Plug in the known values into the Nernst Equation and solve for [Ag+] at the anode: \(0.76 V = 0-\frac{8.314 J\space mol^{-1}\space K^{-1}\times 298 K}{1\times9.648\times10^4\space C\space mol^{-1}}\times\ln\frac{[Ag+]_{anode}}{0.10}\) Now, isolate [Ag+] and solve: \([Ag+]_{anode} = 0.10\times e^{\frac{0.76\times9.648\times10^4}{8.314\times298}}\) \([Ag+]_{anode} = 1.51\times10^{-6} M\) So the concentration of Ag+ at the anode is \(1.51\times10^{-6} M\).

Write the equilibrium expression for the reaction

The equilibrium expression for the reaction is given as: \(K = \frac{[Ag(S_2O_3)_2^{3-}]}{[Ag^+][S_2O_3^{2-}]^2}\)

Calculate the value of the equilibrium constant

Plug the known concentration values into the equilibrium expression, and solve for K: \(K = \frac{1.0\times10^{-3}}{(1.51\times10^{-6})(0.050)^2}\) \(K = 8.82\) So the equilibrium constant K for the formation of \(Ag(S_2O_3)_2^{3-}\) is 8.82.

Key concepts

Nernst Equation

The Nernst Equation is a crucial tool in electrochemistry, and it describes the relationship between the electrochemical cell potential, temperature, number of moles of electrons transferred, and reaction quotient. It is expressed as:\[ E_{cell} = E_{0} - \frac{RT}{nF}\ln Q \]In the context of concentration cells, the Nernst Equation allows us to calculate cell potentials based on concentration differences between two cells. Since a concentration cell's potential relies solely on the concentration gradient, the standard cell potential \(E_{0}\) is zero when identical electrodes, like silver electrodes in this example, are used.

• R represents the universal gas constant \(8.314\, J\, mol^{-1}\, K^{-1}\).
• T is the absolute temperature, typically in Kelvin, often assumed to be \(298\, K\).
• n is the number of electrons transferred in the redox process. For silver, it's usually \(1\).
• F stands for Faraday's constant, approximately \(9.648\times 10^4\, C\, mol^{-1}\).

Using this equation, we can determine how the concentration of ions like \(\text{Ag}^+\) affects the overall cell voltage, illustrating the power of chemical gradients in driving reactions.

Equilibrium Constant

The equilibrium constant, represented as \(K\), is an essential concept in chemistry that quantifies the extent of a chemical reaction at equilibrium. For the reaction:\[ \text{Ag}^{+}(aq) + 2 \text{S}_2\text{O}_3^{2-}(aq) \rightleftharpoons \text{Ag} (\text{S}_2\text{O}_3)_2^{3-}(aq) \]The equilibrium constant \(K\) can be expressed as:\[ K = \frac{[\text{Ag} (\text{S}_2\text{O}_3)_2^{3-}]}{[\text{Ag}^+][\text{S}_2\text{O}_3^{2-}]^2} \]The value of \(K\) tells us about the position of equilibrium:

• If \(K \gg 1\), the reaction favors the formation of products.
• If \(K \ll 1\), the reactants are favored.
• When \(K \approx 1\), there is a balance between products and reactants.

In the provided example, we calculated \(K = 8.82\), indicating a preference for product formation under the given conditions. Understanding \(K\) allows chemists to predict how changing reaction conditions, like concentration or pressure, will affect the balance of reactants and products.

Reaction Quotient

The reaction quotient, or \(Q\), is a measure used to determine the direction in which a reaction mixture will proceed to reach equilibrium. It's closely related to the equilibrium constant \(K\), but \(Q\) is calculated using the current concentrations of the reactants and products at any point in time, not just at equilibrium.For a reaction like the one involving silver and thiosulfate:\[ Q = \frac{[\text{Ag}(\text{S}_2\text{O}_3)_2^{3-}]}{[\text{Ag}^+][\text{S}_2\text{O}_3^{2-}]^2} \]The value of \(Q\) helps us understand whether a reaction will move forward to produce more products (if \(Q < K\)) or in reverse to form more reactants (if \(Q > K\)). If \(Q = K\), the system is already at equilibrium.In the cell described in the exercise, the Nernst Equation uses \(Q\) to show how cell potential is influenced by ion concentration differences. As the concentration of ions alters the reaction quotient, the direction and extent of the reaction shift correspondingly. Understanding \(Q\) empowers students to anticipate reaction dynamics in non-equilibrium states.

Silver Electrode

Silver electrodes are commonly used in electrochemical cells due to their stable and inert nature. In concentration cells, such as the one described in the exercise, silver electrodes help maintain uniform electron transfer reactions across the cell.Silver electrodes serve two important functions:

• As conductors for electron flow, maintaining an efficient and stable path for the redox reactions.
• Providing a constant reference point for measuring and comparing electrode potentials.

When used in conjunction with silver ions \(\text{Ag}^+\), they facilitate reliable redox reactions essential for generating electrical energy in the form of cell potential.In the described concentration cell:

• The electrode reactions involve the transfer of \(\text{Ag}^+\) ions at varying concentrations, creating a gradient that drives the electric current.
• The standard potential of silver electrodes is a well-documented zero under typical conditions, simplifying calculations of cell potential in the Nernst Equation.

Overall, silver electrodes play a vital role in ensuring that the rapid flow of electrons aligns with theoretical predictions, crucial for both experimental understanding and industrial applications.