You have a concentration cell in which the cathode has a silver electrode with 0.10\(M \mathrm{Ag}^{+} .\) The anode also has a silver electrode with \(\mathrm{Ag}^{+}(a q), 0.050 M \mathrm{S}_{2} \mathrm{O}_{3}^{2-},\) and $1.0 \times 10^{-3} 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

Expert verified
The concentration of Ag⁺ at the anode is \(1.20 \times 10^{-3} M\), and the equilibrium constant for the formation of Ag(S₂O₃)₂³⁻ is approximately \(6.67 \times 10^2\).

Step by step solution

01

Write the Nernst equation for Ag/Ag⁺ half-cell

The Nernst equation for the Ag/Ag⁺ half-cell is: \(E = E° - \frac{0.0592}{n} \log{Q}\) where: - \(E\) is the cell voltage - \(E°\) is the standard cell potential (which is 0 for a concentration cell) - \(n\) is the number of moles of electrons involved in the reaction (1 for Ag/Ag⁺ half-cell) - \(Q\) is the reaction quotient, which is \(\frac{[Ag⁺]_{cathode}}{[Ag⁺]_{anode}}\).
02

Rearrange the Nernst equation to solve for [Ag⁺]_{anode}

Rearrange the Nernst equation to make [Ag⁺]_{anode} the subject: \([Ag⁺]_{anode} = \frac{[Ag⁺]_{cathode}}{10^{\frac{(E - E°)n}{0.0592}}}\) Substitute the given values: - \([Ag⁺]_{cathode} = 0.10\,M\) - \(E = 0.76\,V\) - \(n = 1\) - \(E° = 0\,V\) \([Ag⁺]_{anode} = \frac{0.10}{10^{\frac{(0.76 - 0)(1)}{0.0592}}}\)
03

Calculate the concentration of Ag⁺ at the anode

Now, we can compute the concentration: \([Ag⁺]_{anode} = \frac{0.10}{10^{\frac{(0.76 - 0)(1)}{0.0592}}} = 1.20 \times 10^{-3} M\) Therefore, the concentration of Ag⁺ at the anode is \(1.20 \times 10^{-3} M\). #b. Determine the value of the equilibrium constant for the formation of Ag(S₂O₃)₂³⁻#
04

Write the equilibrium expression for the formation of Ag(S₂O₃)₂³⁻

The equilibrium expression for the formation of Ag(S₂O₃)₂³⁻ is: \(K = \frac{[Ag(S₂O₃)₂³⁻]}{[Ag⁺][S₂O₃²⁻]^2}\)
05

Substitute the known concentrations and solve for K

Now, substitute the given values and concentration that we calculated in part (a): - \([Ag(S₂O₃)₂³⁻] = 1.0 \times 10^{-3} M\) - \([Ag⁺]_{anode} = 1.20 \times 10^{-3} M\) (from part a) - \([S₂O₃²⁻] = 0.050M\) \(K = \frac{1.0 \times 10^{-3}}{(1.20 \times 10^{-3})\times(0.050)^2} = 6.67 \times 10^2\) The value of the equilibrium constant for the formation of Ag(S₂O₃)₂³⁻ is approximately \(6.67 \times 10^2\).

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Most popular questions from this chapter

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