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Chapter 18: Problem 47

Calculate \(\mathscr{E}^{\circ}\) values for the following cells. Which reactions are spontaneous as written (under standard conditions)? Balance the equations. Standard reduction potentials are found in Table 18.1. a. $\mathrm{MnO}_{4}^{-(a q)}+\mathrm{I}^{-}(a q) \longrightarrow \mathrm{I}_{2}(a q)+\mathrm{Mn}^{2+}(a q)$ b. $\mathrm{MnO}_{4}^{-}(a q)+\mathrm{F}^{-}(a q) \longrightarrow \mathrm{F}_{2}(g)+\mathrm{Mn}^{2+}(a q)$

Short Answer

Expert verified
The balanced half-reactions are: a. \( \mathrm{MnO}_{4}^{-}+\mathrm{I}^{-}+8 \mathrm{H}^{+}+2 \mathrm{e}^{-} \rightarrow \mathrm{I}_{2}+\mathrm{Mn}^{2+}+4 \mathrm{H}_{2} \mathrm{O} \) with E° = 0.97 V (spontaneous) b. \( \mathrm{MnO}_{4}^{-}+\mathrm{F}^{-}+8 \mathrm{H}^{+}+9 \mathrm{e}^{-} \rightarrow \mathrm{F}_{2}+\mathrm{Mn}^{2+}+4 \mathrm{H}_{2} \mathrm{O} \) with E° = -1.36 V (non-spontaneous)

Step by step solution

01

Balance the half-reactions

We will first balance the two given half-reactions in the acidic medium using the half-reaction method: a. MnO4- + I- → I2 + Mn2+ 1. Balance the atoms other than oxygen and hydrogen: As Mn and I are already balanced, move to the next step. 2. Balance the oxygen atoms: Add 4 H2O to the right side. MnO4- + I- → I2 + Mn2+ + 4H2O 3. Balance hydrogen atoms: Add 8 H+ to the left side. MnO4- + I- + 8H+ → I2 + Mn2+ + 4H2O 4. Balance the charge: 2 electrons are added to the left side. MnO4- + I- + 8H+ + 2e- → I2 + Mn2+ + 4H2O b. MnO4- + F- → F2 + Mn2+ 1. Balance the atoms other than oxygen and hydrogen: As Mn and F are already balanced, move to the next step. 2. Balance oxygen atoms: Add 4H2O to the right side. MnO4-+ F- → F2 + Mn2+ + 4H2O 3. Balance hydrogen atoms: Add 8 H+ on the left side. MnO4-+ F- + 8H+ → F2 + Mn2+ + 4H2O 4. Balance the charge: 9 electrons are added to the left side. MnO4-+ F- + 8H+ + 9e- → F2 + Mn2+ + 4H2O
02

Calculate the standard cell potential for each reaction

We can now calculate the E° using reduction potentials from Table 18.1. For both reactions, we have MnO4- (reduced) and Mn2+ (oxidized), so, the reduction potential for MnO4- is the same (1.51 V). We will calculate E° for both reactions: a. I2/I-: For this reaction, we have E°(I2) = 0.54 V. ∆E°(cell) = E°(cathode) - E°(anode) = E°(MnO4-/Mn2+) - E°(I2/I-) = 1.51 - 0.54 = 0.97 V b. F2/F-: For this reaction, we have E°(F2) = 2.87 V. ∆E°(cell) = E°(cathode) - E°(anode) = E°(MnO4-/Mn2+) - E°(F2/F-) = 1.51 - 2.87 = -1.36 V
03

Determine if reactions are spontaneous

To determine if a reaction is spontaneous, we will check the sign of E°. If E° is positive, the reaction is spontaneous (as written). a. ∆E°(cell) = 0.97 V; The reaction is spontaneous as E° > 0. b. ∆E°(cell) = -1.36 V; The reaction is not spontaneous as E° < 0.
04

Summary

The given half-reactions have been balanced, and their standard cell potentials have been calculated using the standard reduction potentials provided in Table 18.1. For reaction (a), the E° is 0.97 V, making it a spontaneous reaction. For reaction (b), the E° is -1.36 V, making it a non-spontaneous reaction.

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Most popular questions from this chapter

Give the balanced cell equation and determine \(\mathscr{E}^{\circ}\) for the galvanic cells based on the following half-reactions. Standard reduction potentials are found in Table 18.1. a. $\mathrm{Cr}_{2} \mathrm{O}_{7}^{2-}+14 \mathrm{H}^{+}+6 \mathrm{e}^{-} \rightarrow 2 \mathrm{Cr}^{3+}+7 \mathrm{H}_{2} \mathrm{O}$ $\mathrm{H}_{2} \mathrm{O}_{2}+2 \mathrm{H}^{+}+2 \mathrm{e}^{-} \rightarrow 2 \mathrm{H}_{2} \mathrm{O}$ b. \(2 \mathrm{H}^{+}+2 \mathrm{e}^{-} \rightarrow \mathrm{H}_{2}\) \(\mathrm{Al}^{3+}+3 \mathrm{e}^{-} \rightarrow \mathrm{Al}\)

Look up the reduction potential for \(\mathrm{Fe}^{3+}\) to \(\mathrm{Fe}^{2+} .\) Look up the reduction potential for \(\mathrm{Fe}^{2+}\) to Fe. Finally, look up the reduction potential for \(\mathrm{Fe}^{3+}\) to Fe. You should notice that adding the reduction potentials for the first two does not give the potential for the third. Why not? Show how you can use the first two potentials to calculate the third potential.

The ultimate electron acceptor in the respiration process is molecular oxygen. Electron transfer through the respiratory chain takes place through a complex series of oxidationreduction reactions. Some of the electron transport steps use iron-containing proteins called \(c y\) tochromes. All cytochromes transport electrons by converting the iron in the cytochromes from the +3 to the +2 oxidation state. Consider the following reduction potentials for three different cytochromes used in the transfer process of electrons to oxygen (the potentials have been corrected for \(\mathrm{pH}\) and for temperature): cytochrome $\mathrm{a}\left(\mathrm{Fe}^{3+}\right)+\mathrm{e}^{-} \longrightarrow\( cytochrome \)\mathrm{a}\left(\mathrm{Fe}^{2+}\right)$ $$ \begin{array}{c}\mathscr{C}=0.385 \mathrm{~V}\end{array} $$ cytochrome $\mathrm{b}\left(\mathrm{Fe}^{3+}\right)+\mathrm{e}^{-} \longrightarrow\( cytochrome \)\mathrm{b}\left(\mathrm{Fe}^{2+}\right)$ $$ \begin{array}{l}\mathscr{E}=0.030 \mathrm{~V}\end{array} $$ cytochrome $\mathrm{c}\left(\mathrm{Fe}^{3+}\right)+\mathrm{e}^{-} \longrightarrow \mathrm{cytochrome} \mathrm{c}\left(\mathrm{Fe}^{2+}\right)$ $$ \begin{array}{c}\mathscr{C}=0.254 \mathrm{~V}\end{array} $$ In the electron transfer series, electrons are transferred from one cytochrome to another. Using this information, determine the cytochrome order necessary for spontaneous transport of electrons from one cytochrome to another, which eventually will lead to electron transfer to \(\mathrm{O}_{2}\).

The following standard reduction potentials have been determined for the aqueous chemistry of indium: $$\operatorname{In}^{3+}(a q)+2 \mathrm{e}^{-} \longrightarrow \operatorname{In}^{+}(a q) \quad \mathscr{E}^{\circ}=-0.444 \mathrm{V}$$ $$\operatorname{In}^{+}(a q)+\mathrm{e}^{-} \longrightarrow \operatorname{In}(s) \qquad \quad \mathscr{E}^{\circ}=-0.126 \mathrm{V}$$ a. What is the equilibrium constant for the disproportionation reaction, where a species is both oxidized and reduced, shown below? $$3 \ln ^{+}(a q) \longrightarrow 2 \operatorname{In}(s)+\operatorname{In}^{3+}(a q)$$ b. What is \(\Delta G_{i}^{\circ}\) for \(\operatorname{In}^{+}(a q)\) if $\Delta G_{f}^{\circ}=-97.9 \mathrm{kJ} / \mathrm{mol}\( for \)\operatorname{In}^{3+}(a q) ?$

Sketch the galvanic cells based on the following overall reactions. Show the direction of electron flow, the direction of ion migration through the salt bridge, and identify the cathode and anode. Give the overall balanced equation. Assume that all concentrations are 1.0 \(M\) and that all partial pressures are 1.0 atm. a. $\mathrm{IO}_{3}^{-}(a q)+\mathrm{Fe}^{2+}(a q) \Longrightarrow \mathrm{Fe}^{3+}(a q)+\mathrm{I}_{2}(a q)$ b. $\mathrm{Zn}(s)+\mathrm{Ag}^{+}(a q) \rightleftharpoons \mathrm{Zn}^{2+}(a q)+\mathrm{Ag}(s)$
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