You have a concentration cell in which the cathode has a silver electrode with \(0.10 \mathrm{M} \mathrm{Ag}^{+}\). 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 \(\mathrm{Ag}\left(\mathrm{S}_{2} \mathrm{O}_{3}\right)_{2}{ }^{3-}\). \(\mathrm{Ag}^{+}(a q)+2 \mathrm{~S}_{2} \mathrm{O}_{3}{ }^{2-}(a q) \rightleftharpoons \mathrm{Ag}\left(\mathrm{S}_{2} \mathrm{O}_{3}\right)_{2}^{3-}(a q) \quad K=?\)

Step by step solution

01

Understanding the Nernst Equation

The Nernst Equation relates the cell potential to the concentrations of the reactants and products. It can be written as: \[E_{cell} = E^{0}_{cell} - \frac{RT}{nF} \ln Q\] Where: - \(E_{cell}\) is the cell potential (voltage) - \(E^{0}_{cell}\) is the standard cell potential - \(R\) is the gas constant - \(T\) is the temperature in Kelvin - \(n\) is the number of moles of electrons transferred - \(F\) is the Faraday constant - \(Q\) is the reaction quotient Since we have a concentration cell with identical electrodes, the standard cell potential, \(E^{0}_{cell}\), will be 0. Thus, we can simplify the Nernst Equation to: \[E_{cell} = - \frac{RT}{nF} \ln Q\]

02

Determine the reaction quotient Q

For our given reaction: \(\text{Ag}^{+}(aq) \rightleftharpoons \text{Ag}\left(\text{S}_{2}\text{O}_{3}\right)_{2}{ }^{3-}(aq)\), Q can be calculated as: \[Q = \frac{[\text{Ag}\left(\text{S}_{2}\text{O}_{3}\right)_{2}{ }^{3-}]}{[\text{Ag}^{+}][\text{S}_{2}\text{O}_{3}{ }^{2-}]^2}\]

03

Calculate the concentration of Ag+ at the anode

Now we can substitute the known values into the Nernst Equation: \[0.76 = -\frac{(8.314)(298)}{(2)(96485)} \ln Q\] We are given \([\text{S}_{2}\text{O}_{3}{ }^{2-}] = 0.050 M\) and \([\text{Ag}\left(\text{S}_{2}\text{O}_{3}\right)_{2}{ }^{3-}] = 1.0 \times 10^{-3} M\), and we need to find \([\text{Ag}^{+}]\) at the anode. Plug these values into the equation for Q and solve for \([\text{Ag}^{+}]\): \[\frac{1.0 \times 10^{-3}}{(0.050)^2[\text{Ag}^{+}]} = Q\] Now, substitute the value of Q back into the Nernst Equation and solve for \([\text{Ag}^{+}]\). The resulting value is the concentration of \(\text{Ag}^{+}\) at the anode.

04

Calculate the equilibrium constant K

Now that we have the concentration of \(\text{Ag}^{+}\) at the anode, we can calculate the equilibrium constant (K) for the given reaction. We know: \[K = \frac{[\text{Ag}\left(\text{S}_{2}\text{O}_{3}\right)_{2}{ }^{3-}]}{[\text{Ag}^{+}][\text{S}_{2}\text{O}_{3}{ }^{2-}]^2}\] Substitute the values for the concentrations into the equation and solve for K. The resulting value is the equilibrium constant for the formation of \(\text{Ag}\left(\text{S}_{2}\text{O}_{3}\right)_{2}{ }^{3-}\).

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