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The standard potential for the reduction of \(\mathrm{AgSCN}(s)\) is \(+0.0895 \mathrm{~V}\) $$ \mathrm{AgSCN}(s)+\mathrm{e}^{-\ldots} \mathrm{Ag}(s)+\mathrm{SCN}^{-}(a q) $$ Using this value and the electrode potential for \(\mathrm{Ag}^{+}(a q)\), calculate the \(K_{s p}\) for AgSCN.

Short Answer

Expert verified
The standard potential for the complete redox reaction can be found by subtracting the standard potential of the $\mathrm{Ag^{+}}$ half-cell from the given standard potential for the reduction of $\mathrm{AgSCN}$: \(E^{\circ} = E_{\mathrm{red}} - E_{\mathrm{red}}^{\mathrm{Ag^{+}}}\). Then, use the Nernst equation to find the equilibrium constant (K): \(E^{\circ} = -\frac{RT}{nF} \ln K\). Finally, calculate the solubility product constant (Ksp) for $\mathrm{AgSCN}$: \(K_{s p} = K\).

Step by step solution

01

The given half-cell reaction is the reduction of AgSCN: $$\mathrm{AgSCN}(s)+\mathrm{e}^{-} \longrightarrow \mathrm{Ag}(s)+\mathrm{SCN}^{-}(a q)$$ For this reaction, we are given the standard reduction potential: \(E_{\mathrm{red}} = +0.0895~\mathrm{V}\). We also need the half-cell reaction involving \(\mathrm{Ag^{+}}\)(aq), which is the following: $$\mathrm{Ag}^{+}(a q)+\mathrm{e}^{-} \longrightarrow \mathrm{Ag}(s)$$ For this reaction, the standard reduction potential is given: \(E_{\mathrm{red}}^{\mathrm{Ag^{+}}}\). We will assume the information is provided or found in a standard reduction potential table. #Step 2: Determine the standard potential for the complete redox reaction#

To find the standard potential for the complete redox reaction, we need to subtract the standard potential of the Ag+ half-cell from the given standard potential for the reduction of AgSCN: $$E^{\circ} = E_{\mathrm{red}} - E_{\mathrm{red}}^{\mathrm{Ag^{+}}}$$ #Step 3: Use the Nernst equation to find the equilibrium constant (K)#
02

The Nernst equation relates the standard potential to the equilibrium constant of a redox reaction: $$E^{\circ} = -\frac{RT}{nF} \ln K$$ In this case, n = 1 since there is only 1 electron transfer in both half-cell reactions, and the complete redox reaction is $$\mathrm{AgSCN}(s) \longrightarrow \mathrm{Ag}^{+}(a q)+\mathrm{SCN}^{-}(a q)$$ Using values for R (8.314 J/mol·K) and F (96485 C/mol), along with the value of \(E^{\circ}\) from Step 2, we can solve for K. #Step 4: Calculate Ksp#

The value of K we found in Step 3 is the solubility equilibrium constant for the redox reaction: $$K = [\mathrm{Ag}^{+}(a q)][\mathrm{SCN}^{-}(a q)]$$ This value is equal to the solubility product constant (Ksp) for AgSCN: $$K_{s p} = K$$

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Key Concepts

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

Standard Reduction Potential
Understanding the standard reduction potential is vital for analyzing and predicting the outcome of redox reactions in electrochemistry. Imagine you are looking at a species that can either gain or lose an electron; knowing the likelihood of either scenario is crucial. The standard reduction potential (often abbreviated as E° or SRP) measures how readily a chemical species gains electrons in its most stable form at standard conditions, which are usually set at 25°C, 1 atm pressure, and 1 M concentration for all reactants and products.

The greater the positive value of E°, the stronger the species acts as an oxidizing agent, meaning it has a greater tendency to gain electrons. Hence, a high positive E° value indicates a strong drive for electron uptake. Conversely, a more negative E° implies a species is more likely to donate electrons and is, therefore, a stronger reducing agent. These values are often tabulated, allowing us to compare different half-reactions and predict the direction in which a reaction will proceed.
Nernst Equation
The Nernst equation bridges our understanding of chemical and electrical forces in a reaction. It connects the standard reduction potential to the actual reduction potential under non-standard conditions, accounting for temperature, concentration, and the number of moles of electrons transferred in the reaction. The Nernst equation is expressed as:
\[ E = E^\circ - \frac{RT}{nF} \ln Q \]
where E is the cell potential under non-standard conditions, is the standard cell potential, R is the universal gas constant, T is the temperature in Kelvin, n is the number of moles of electrons transferred, F is Faraday’s constant, and Q is the reaction quotient. Through this equation, we can find the equilibrium constant K for the reaction when the cell potential is at equilibrium, meaning there is no net reaction occurring.
Equilibrium Constant
At the heart of chemical reactions lies the concept of dynamic equilibrium, where the rates of the forward and reverse reactions are equal, resulting in no net change in the concentration of reactants and products over time. The equilibrium constant (K) quantifies this balance, representing the ratio of product concentrations to reactant concentrations at equilibrium, raised to the power of their respective coefficients in the balanced equation.

When it comes to redox reactions, K reflects the extent to which the reaction favors the formation of products. A large K indicates a reaction strongly favors products at equilibrium, while a small K shows a reaction that leans towards the reactants. In the context of solubility, the Ksp, or solubility product constant, specifically represents the extent to which a salt will dissolve to produce ions in solution. By calculating K or Ksp, we gain insight into the solubility and stability of compounds in different chemical environments.
Electrochemistry
Electrochemistry is the branch of chemistry that deals with the interconversion of chemical energy and electrical energy. It describes processes that involve the movement of electrons, typically in redox reactions that occur in electrochemical cells. There are two main types of electrochemical cells: galvanic (or voltaic) cells, which generate electrical energy from spontaneous redox reactions, and electrolytic cells, which drive non-spontaneous chemical reactions using electrical energy.

This fascinating field covers batteries, corrosion, electroplating, and many other applications rooted in daily life. By understanding concepts like reduction potentials, equilibrium constants, and the Nernst equation, we can design and optimize batteries, prevent metal corrosion, refine metals, and perform electro-synthesis of valuable compounds. Electrochemistry not only aids in solving practical problems but also deepens our knowledge of chemical reactions and their driving forces.

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Most popular questions from this chapter

In a galvanic cell the cathode is an \(\mathrm{Ag}^{+}(1.00 \mathrm{M}) / \mathrm{Ag}(\mathrm{s})\) half-cell. The anode is a standard hydrogen electrode immersed in a buffer solution containing \(0.10 \mathrm{M}\) benzoic acid \(\left(\mathrm{C}_{6} \mathrm{H}_{5} \mathrm{COOH}\right)\) and \(0.050 \mathrm{M}\) sodium benzoate \(\left(\mathrm{C}_{6} \mathrm{H}_{5} \mathrm{COO}^{-} \mathrm{Na}^{+}\right)\). The measured cell voltage is \(1.030 \mathrm{~V}\). What is the \(\mathrm{pK}_{a}\) of benzoic acid?

(a) The nonrechargeable lithium batteries used for photography use lithium metal as the anode. What advantages might be realized by using lithium rather than zinc, cadmium, lead, or nickel? (b) The rechargeable lithiumion battery does not use lithium metal as an electrode material. Nevertheless, it still has a substantial advantage over nickel-based batteries. Suggest an explanation.

Derive an equation that directly relates the standard emf of a redox reaction to its equilibrium constant.

Complete and balance the following equations, and identify the oxidizing and reducing agents. Recall that the \(\mathrm{O}\) atoms in hydrogen peroxide, \(\mathrm{H}_{2} \mathrm{O}_{2}\), have an atypical oxidation state. (a) \(\mathrm{NO}_{2}^{-}(a q)+\mathrm{Cr}_{2} \mathrm{O}_{7}^{2-}(a q)-\cdots\) \(\mathrm{Cr}^{3+}(a q)+\mathrm{NO}_{3}^{-}(a q)\) (acidic solution) (b) \(\mathrm{S}(s)+\mathrm{HNO}_{3}(a q) \rightarrow-\rightarrow \mathrm{H}_{2} \mathrm{SO}_{3}(a q)+\mathrm{N}_{2} \mathrm{O}(g)\) (acidic solution) (c) \(\mathrm{Cr}_{2} \mathrm{O}_{7}^{2-}(a q)+\mathrm{CH}_{3} \mathrm{OH}(a q) \longrightarrow\) \(\mathrm{HCO}_{2} \mathrm{H}(a q)+\mathrm{Cr}^{3+}(a q)\) (acidic solution) (d) \(\mathrm{MnO}_{4}^{-}(a q)+\mathrm{Cl}^{-}(a q) \longrightarrow \mathrm{Mn}^{2+}(a q)+\mathrm{Cl}_{2}(a q)\) (acidic solution) (e) \(\mathrm{NO}_{2}^{-}(a q)+\mathrm{Al}(s) \longrightarrow \mathrm{NH}_{4}{ }^{+}(a q)+\mathrm{AlO}_{2}^{-}(a q)\) (basic solution) (f) \(\mathrm{H}_{2} \mathrm{O}_{2}(a q)+\mathrm{ClO}_{2}(a q) \longrightarrow \mathrm{ClO}_{2}^{-}(a q)+\mathrm{O}_{2}(g)\) (basic solution)

A voltaic cell utilizes the following reaction: \(4 \mathrm{Fe}^{2+}(a q)+\mathrm{O}_{2}(g)+4 \mathrm{H}^{+}(a q)-\cdots 4 \mathrm{Fe}^{3+}(a q)+2 \mathrm{H}_{2} \mathrm{O}(l)\) (a) What is the emf of this cell under standard conditions? (b) What is the emf of this cell when \(\left[\mathrm{Fe}^{2+}\right]=1.3 \mathrm{M}\) \(\left[\mathrm{Fe}^{3+}\right]=0.010 \mathrm{M}, \mathrm{P}_{\mathrm{O}_{2}}=0.50 \mathrm{~atm}\), and the \(\mathrm{pH}\) of the so- lution in the cathode compartment is \(3.50 ?\)

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