/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 82 You have a concentration cell in... [FREE SOLUTION] | 91影视

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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鈦 at the anode is \(1.20 \times 10^{-3} M\), and the equilibrium constant for the formation of Ag(S鈧侽鈧)鈧偮斥伝 is approximately \(6.67 \times 10^2\).

Step by step solution

01

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

The Nernst equation for the Ag/Ag鈦 half-cell is: \(E = E掳 - \frac{0.0592}{n} \log{Q}\) where: - \(E\) is the cell voltage - \(E掳\) 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鈦 half-cell) - \(Q\) is the reaction quotient, which is \(\frac{[Ag鈦篯_{cathode}}{[Ag鈦篯_{anode}}\).
02

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

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

Calculate the concentration of Ag鈦 at the anode

Now, we can compute the concentration: \([Ag鈦篯_{anode} = \frac{0.10}{10^{\frac{(0.76 - 0)(1)}{0.0592}}} = 1.20 \times 10^{-3} M\) Therefore, the concentration of Ag鈦 at the anode is \(1.20 \times 10^{-3} M\). #b. Determine the value of the equilibrium constant for the formation of Ag(S鈧侽鈧)鈧偮斥伝#
04

Write the equilibrium expression for the formation of Ag(S鈧侽鈧)鈧偮斥伝

The equilibrium expression for the formation of Ag(S鈧侽鈧)鈧偮斥伝 is: \(K = \frac{[Ag(S鈧侽鈧)鈧偮斥伝]}{[Ag鈦篯[S鈧侽鈧兟测伝]^2}\)
05

Substitute the known concentrations and solve for K

Now, substitute the given values and concentration that we calculated in part (a): - \([Ag(S鈧侽鈧)鈧偮斥伝] = 1.0 \times 10^{-3} M\) - \([Ag鈦篯_{anode} = 1.20 \times 10^{-3} M\) (from part a) - \([S鈧侽鈧兟测伝] = 0.050M\) \(K = \frac{1.0 \times 10^{-3}}{(1.20 \times 10^{-3})\times(0.050)^2} = 6.67 \times 10^2\) The value of the equilibrium constant for the formation of Ag(S鈧侽鈧)鈧偮斥伝 is approximately \(6.67 \times 10^2\).

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Concentration Cell
A concentration cell is a type of galvanic cell where the electrodes are made from the same material. The key difference lies in the concentrations of the solutions at the two electrodes. In concentration cells, the cell generates an electric current because of the concentration difference. These cells work based on the principle that ions will move from a region of higher concentration to lower concentration until equilibrium is achieved.
In the exercise, we have a silver concentration cell. The silver ions (\(\mathrm{Ag}^{+} \)) at the cathode are at a higher concentration compared to those at the anode. This concentration gradient causes the electrons to flow from the anode to the cathode, generating voltage. The voltage can be calculated using the Nernst equation.
Electrode Potentials
Electrode potentials refer to the potential difference between an electrode and its surrounding solution. They are essential in determining the direction of electron flow in electrochemical cells. In concentration cells, both electrodes use the same material, so the standard electrode potential (\(E掳\)) for each is the same and often zero.
For the exercise, the Nernst equation is used to find the actual cell potential (\(E\)). Since it's a concentration cell, the standard potential is zero, and the actual potential depends exclusively on the concentration gradient. This potential influences how the silver ions shift from areas of high concentration at the cathode to low concentration at the anode.
Equilibrium Constant
The equilibrium constant (\(K\)) is a way to express how far a reaction proceeds before reaching a balance between products and reactants. It is calculated using the concentrations of the species at equilibrium. Knowing the value of \(K\) helps predict the direction and extent of the reaction.
In the exercise, \(K\) is found for the formation of \(\mathrm{Ag(S_{2}O_{3})_{2}^{3-}}\) using the given concentrations. This value indicates that products are favored in this equilibrium, showing a greater tendency to form the \(\mathrm{Ag(S_{2}O_{3})_{2}^{3-}}\) complex.
Reaction Quotient
The reaction quotient (\(Q\)) is similar to the equilibrium constant, but it applies to any point in the reaction, not just at equilibrium. \(Q\) is calculated in the same way as \(K\) but uses the current concentrations of the reactants and products.
In concentration cells, \(Q\) is crucial for applying the Nernst equation. The equation uses \(Q\) to relate the cell potential (\(E\)) with concentrations. When comparing \(Q\) with \(K\), you can determine the reaction's direction鈥攚hether it proceeds towards products or reactants to reach equilibrium. In the exercise, the equation shows how ionic concentrations affect the cell's voltage.

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Most popular questions from this chapter

Consider the electrolysis of a molten salt of some metal. What information must you know to calculate the mass of metal plated out in the electrolytic cell?

What reactions take place at the cathode and the anode when each of the following is electrolyzed? a. molten \(\mathrm{NiBr}_{2} \quad\) b. molten \(\mathrm{AlF}_{3} \quad\) c. molten \(\mathrm{MgI}_{2}\)

Consider the galvanic cell based on the following halfreactions: $$\begin{array}{ll}{\mathrm{Au}^{3+}+3 \mathrm{e}^{-} \longrightarrow \mathrm{Au}} & {\mathscr{E}^{\circ}=1.50 \mathrm{V}} \\ {\mathrm{T} 1^{+}+\mathrm{e}^{-} \longrightarrow \mathrm{Tl}} & {\mathscr{E}^{\circ}=-0.34 \mathrm{V}}\end{array}$$ a. Determine the overall cell reaction and calculate \(\mathscr{E}_{\text { cell }}\) b. Calculate \(\Delta G^{\circ}\) and \(K\) for the cell reaction at \(25^{\circ} \mathrm{C}\) c. Calculate \(\mathscr{E}_{\text { cell }}\) at \(25^{\circ} \mathrm{C}\) when \(\left[\mathrm{Au}^{3+}\right]=1.0 \times 10^{-2} \mathrm{M}\) and \(\left[\mathrm{T} 1^{+}\right]=1.0 \times 10^{-4} \mathrm{M}\)

The overall reaction in the lead storage battery is \(\mathrm{Pb}(s)+\mathrm{PbO}_{2}(s)+2 \mathrm{H}^{+}(a q)+2 \mathrm{HSO}_{4}^{-}(a q) \longrightarrow\) \(2 \mathrm{PbSO}_{4}(s)+2 \mathrm{H}_{2} \mathrm{O}(l)\) Calculate \(\mathscr{E}\) at \(25^{\circ} \mathrm{C}\) for this battery when \(\left[\mathrm{H}_{2} \mathrm{SO}_{4}\right]=4.5 \mathrm{M}\), that is, \(\left[\mathrm{H}^{+}\right]=\left[\mathrm{HSO}_{4}^{-}\right]=4.5 \mathrm{M} . \mathrm{At} 25^{\circ} \mathrm{C}, \mathscr{E}^{\circ}=2.04 \mathrm{V}\) for the lead storage battery.

An experimental fuel cell has been designed that uses carbon monoxide as fuel. The overall reaction is $$2 \mathrm{CO}(g)+\mathrm{O}_{2}(g) \longrightarrow 2 \mathrm{CO}_{2}(g)$$ The two half-cell reactions are $$\mathrm{CO}+\mathrm{O}^{2-} \longrightarrow \mathrm{CO}_{2}+2 \mathrm{e}^{-}$$ $$\mathrm{O}_{2}+4 \mathrm{e}^{-} \longrightarrow 2 \mathrm{O}^{2-}$$ The two half-reactions are carried out in separate compartments connected with a solid mixture of \(\mathrm{CeO}_{2}\) and \(\mathrm{Gd}_{2} \mathrm{O}_{3}\) . Oxide ions can move through this solid at high temperatures (about \(800^{\circ} \mathrm{C} ) . \Delta G\) for the overall reaction at \(800^{\circ} \mathrm{C}\) under certain concentration conditions is \(-380 \mathrm{kJ}\) . Calculate the cell potential for this fuel cell at the same temperature and concentration conditions.

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