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Chapter 18: Problem 150

Given the following two standard reduction potentials, $$\begin{array}{ll}{\mathrm{M}^{3+}+3 \mathrm{e}^{-} \longrightarrow \mathrm{M}} & {\mathscr{E}^{\circ}=-0.10 \mathrm{V}} \\ {\mathrm{M}^{2+}+2 \mathrm{e}^{-} \longrightarrow \mathrm{M}} & {\mathscr{E}^{\circ}=-0.50 \mathrm{V}}\end{array}$$ solve for the standard reduction potential of the half-reaction $$\mathrm{M}^{3+}+\mathrm{e}^{-} \longrightarrow \mathrm{M}^{2+}$$ (Hint: You must use the extensive property \(\Delta G^{\circ}\) to determine the standard reduction potential.)

Short Answer

Expert verified
The standard reduction potential of the half-reaction M³⁺ + e⁻ → M²⁺ is -0.699 V.

Step by step solution

01

Find the ∆G° for M³⁺ → M and M²⁺ → M half-reactions

First, let's calculate the ∆G° values for both given reduction reactions using the equation: ∆G° = -nFE°. For M³⁺ → M: E° = -0.10 V; n = 3 ∆G° = -(3)(96,485 C/mol)(-0.10 V) ∆G° = 29,045.5 J/mol For M²⁺ → M: E° = -0.50 V; n = 2 ∆G° = -(2)(96,485 C/mol)(-0.50 V) ∆G° = 96,485 J/mol
02

Find the ∆G° for M³⁺ → M²⁺ half-reaction using given ∆G° values

We can relate the ∆G° of the reactions with the following equation: ∆G°(M³⁺ → M²⁺) = ∆G°(M²⁺ → M) - ∆G°(M³⁺ → M) ∆G°(M³⁺ → M²⁺) = 96,485 J/mol - 29,045.5 J/mol ∆G°(M³⁺ → M²⁺) = 67,439.5 J/mol
03

Find the standard reduction potential of M³⁺ → M²⁺ using the calculated ∆G°

Now we can find the standard reduction potential of the required half-reaction using the equation: E° = -∆G°/(nF) In this reaction (M³⁺ → M²⁺), only one electron is involved: n = 1. E°(M³⁺ → M²⁺) = -67,439.5 J/mol / (1 * 96,485 C/mol) E°(M³⁺ → M²⁺) = -0.699 V So, the standard reduction potential of the half-reaction M³⁺ + e⁻ → M²⁺ is -0.699 V.

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Most popular questions from this chapter

Consider a concentration cell that has both electrodes made of some metal M. Solution A in one compartment of the cell contains 1.0\(M \mathrm{M}^{2+} .\) Solution \(\mathrm{B}\) in the other cell compartment has a volume of 1.00 L. At the beginning of the experiment 0.0100 mole of \(\mathrm{M}\left(\mathrm{NO}_{3}\right)_{2}\) and 0.0100 mole of \(\mathrm{Na}_{2} \mathrm{SO}_{4}\) are dissolved in solution \(\mathrm{B}\) (ignore volume changes), where the reaction $$\mathrm{M}^{2+}(a q)+\mathrm{SO}_{4}^{2-}(a q) \rightleftharpoons \mathrm{MSO}_{4}(s)$$ occurs. For this reaction equilibrium is rapidly established, whereupon the cell potential is found to be 0.44 \(\mathrm{V}\) at \(25^{\circ} \mathrm{C} .\) Assume that the process $$\mathrm{M}^{2+}+2 \mathrm{e}^{-} \longrightarrow \mathrm{M}$$ has a standard reduction potential of \(-0.31 \mathrm{V}\) and that no other redox process occurs in the cell. Calculate the value of \(K_{\mathrm{sp}}\) for \(\mathrm{MSO}_{4}(s)\) at \(25^{\circ} \mathrm{C} .\)

One of the few industrial-scale processes that produce organic compounds electrochemically is used by the Monsanto Company to produce 1,4-dicyanobutane. The reduction reaction is $$2 \mathrm{CH}_{2}=\mathrm{CHCN}+2 \mathrm{H}^{+}+2 \mathrm{e}^{-} \longrightarrow \mathrm{NC}-\left(\mathrm{CH}_{2}\right)_{4}-\mathrm{CN}$$ The \(\mathrm{NC}-\left(\mathrm{CH}_{2}\right)_{4}-\mathrm{CN}\) is then chemically reduced using hydrogen gas to $\mathrm{H}_{2} \mathrm{N}-\left(\mathrm{CH}_{2}\right)_{6}-\mathrm{NH}_{2},$ which is used to produce production of nylon. What current must be used to produce \(150 . \mathrm{kg} \mathrm{NC}-\left(\mathrm{CH}_{2}\right)_{4}-\mathrm{CN}\) per hour?

The general rule for salt bridges is that anions flow to the anode and cations flow to the cathode. Explain why this is true.

Sketch the galvanic cells based on the following half- reactions. Show the direction of electron flow, show the direction of ion migration through the salt bridge, and identify the cathode and anode. Give the overall balanced equation, and determine \(\mathscr{E}^{\circ}\) for the galvanic cells. Assume that all concentrations are 1.0 \(M\) and that all partial pressures are 1.0 atm. a. $\mathrm{Cl}_{2}+2 \mathrm{e}^{-} \rightarrow 2 \mathrm{Cl}^{-} \quad \mathscr{E}^{\circ}=1.36 \mathrm{V}$ $\mathrm{Br}_{2}+2 \mathrm{e}^{-} \rightarrow 2 \mathrm{Br}^{-} \quad \mathscr{E}^{\circ}=1.09 \mathrm{V}$ b. $\mathrm{MnO}_{4}^{-}+8 \mathrm{H}^{+}+5 \mathrm{e}^{-} \rightarrow \mathrm{Mn}^{2+}+4 \mathrm{H}_{2} \mathrm{O} \quad \mathscr{E}^{\circ}=1.51 \mathrm{V}$ $\mathrm{IO}_{4}^{-}+2 \mathrm{H}^{+}+2 \mathrm{e}^{-} \rightarrow \mathrm{IO}_{3}^{-}+\mathrm{H}_{2} \mathrm{O} \quad \quad \mathscr{E}^{\circ}=1.60 \mathrm{V}$

An electrochemical cell consists of a standard hydrogen electrode and a copper metal electrode. If the copper electrode is placed in a solution of 0.10$M \mathrm{NaOH}\( that is saturated with \)\mathrm{Cu}(\mathrm{OH})_{2},$ what is the cell potential at $25^{\circ} \mathrm{C} ?\left[\text { For } \mathrm{Cu}(\mathrm{OH})_{2}\right.\( \)K_{\mathrm{sp}}=1.6 \times 10^{-19} . ]$
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