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Chapter 20: Problem 37

Using standard reduction potentials (Appendix E), calculate the standard emf for each of the following reactions: (a) $\mathrm{Cl}_{2}(g)+2 \mathrm{I}^{-}(a q) \longrightarrow 2 \mathrm{Cl}^{-}(a q)+\mathrm{I}_{2}(s)$ (b) $\mathrm{Ni}(s)+2 \mathrm{Ce}^{4+}(a q) \longrightarrow \mathrm{Ni}^{2+}(a q)+2 \mathrm{Ce}^{3+}(a q)$ (c) $\mathrm{Fe}(s)+2 \mathrm{Fe}^{3+}(a q) \longrightarrow 3 \mathrm{Fe}^{2+}(a q)$ (d) $2 \mathrm{NO}_{3}^{-}(a q)+8 \mathrm{H}^{+}(a q)+3 \mathrm{Cu}(s) \longrightarrow 2 \mathrm{NO}(g)+ 4 \mathrm{H}_{2} \mathrm{O}(l)+3 \mathrm{Cu}^{2+}(a q)$

Short Answer

Expert verified
The standard emf values for the given reactions are: (a) 0.82 V (b) -1.46 V (c) 1.21 V (d) 0.62 V

Step by step solution

01

Half-Reactions and Standard Reduction Potentials for Reaction (a)

For this reaction, we have: Oxidation half-reaction: 2 I⁻(aq) → I₂(s) + 2 e⁻ Reduction half-reaction: Cl₂(g) + 2 e⁻ → 2 Cl⁻(aq) From Appendix E: E°(I⁻/I₂) = +0.54 V E°(Cl₂/Cl⁻) = +1.36 V #Step 2: Calculate the Standard emf#
02

Standard emf for Reaction (a)

Using the Nernst equation: \[E°_{cell} = E°_{cathode} - E°_{anode}\] For this reaction, emf is: \[E°_{cell} = E°(Cl₂/Cl⁻) - E°(I⁻/I₂) = 1.36 - 0.54 = 0.82 \ V\] #b) Reaction: Ni(s) + 2 Ce⁴⁺(aq) → Ni²⁺(aq) + 2 Ce³⁺(aq)# #Step 1: Identify Half-Reactions and Standard Reduction Potentials#
03

Half-Reactions and Standard Reduction Potentials for Reaction (b)

For this reaction, we have: Oxidation half-reaction: Ni(s) → Ni²⁺(aq) + 2 e⁻ Reduction half-reaction: 2 Ce⁴⁺(aq) + 2 e⁻ → 2 Ce³⁺(aq) From Appendix E: E°(Ni²⁺/Ni) = -0.26 V E°(Ce⁴⁺/Ce³⁺) = -1.72 V #Step 2: Calculate the Standard emf#
04

Standard emf for Reaction (b)

Using the Nernst equation, the standard emf is: \[E°_{cell} = E°(Ce⁴⁺/Ce³⁺) - E°(Ni²⁺/Ni) = -1.72 + 0.26 = -1.46 \ V\] #c) Reaction: Fe(s) + 2 Fe³⁺(aq) → 3 Fe²⁺(aq)# #Step 1: Identify Half-Reactions and Standard Reduction Potentials#
05

Half-Reactions and Standard Reduction Potentials for Reaction (c)

For this reaction, we have: Oxidation half-reaction: Fe(s) → Fe²⁺(aq) + 2 e⁻ Reduction half-reaction: 2 Fe³⁺(aq) + 2 e⁻ → 2 Fe²⁺(aq) From Appendix E: E°(Fe²⁺/Fe) = -0.44 V E°(Fe³⁺/Fe²⁺) = +0.77 V #Step 2: Calculate the Standard emf#
06

Standard emf for Reaction (c)

Using the Nernst equation, the standard emf is: \[E°_{cell} = E°(Fe³⁺/Fe²⁺) - E°(Fe²⁺/Fe) = 0.77 + 0.44 = 1.21 \ V\] #d) Reaction: 2 NO₃⁻(aq) + 8 H⁺(aq) + 3 Cu(s) → 2 NO(g) + 4 H₂O(l) + 3 Cu²⁺(aq)# #Step 1: Identify Half-Reactions and Standard Reduction Potentials#
07

Half-Reactions and Standard Reduction Potentials for Reaction (d)

For this reaction, we have: Oxidation half-reaction: 3 Cu(s) → 3 Cu²⁺(aq) + 6 e⁻ Reduction half-reaction: 2 NO₃⁻(aq) + 8 H⁺(aq) + 6 e⁻ → 2 NO(g) + 4 H₂O(l) From Appendix E: E°(Cu²⁺/Cu) = +0.34 V E°(NO₃⁻/NO) = +0.96 V #Step 2: Calculate the Standard emf#
08

Standard emf for Reaction (d)

Using the Nernst equation, the standard emf is: \[E°_{cell} = E°(NO₃⁻/NO) - E°(Cu²⁺/Cu) = 0.96 - 0.34 = 0.62 \ V\] In summary, the standard emf values for the given reactions are: (a) 0.82 V (b) -1.46 V (c) 1.21 V (d) 0.62 V

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Most popular questions from this chapter

(a) What is the definition of the volt? (b) Do all voltaic cells produce a positive cell potential?

Heart pacemakers are often powered by lithium-silver chromate "button" batteries. The overall cell reaction is $$ 2 \mathrm{Li}(s)+\mathrm{Ag}_{2} \mathrm{CrO}_{4}(s) \longrightarrow \mathrm{Li}_{2} \mathrm{CrO}_{4}(s)+2 \mathrm{Ag}(s) $$ (a) Lithium metal is the reactant at one of the electrodes of the battery. Is it the anode or the cathode? (b) Choose the two half-reactions from Appendix \(\mathrm{E}\) that most closely approximate the reactions that occur in the battery. What standard emf would be generated by a voltaic cell based on these half-reactions? (c) The battery generates an emf of \(+3.5 \mathrm{~V}\). How close is this value to the one calculated in part (b)? (d) Calculate the emf that would be generated at body temperature, \(37^{\circ} \mathrm{C}\). How does this compare to the emf you calculated in part (b)?

Indicate whether the following balanced equations involve oxidation-reduction. If they do, identify the elements that undergo changes in oxidation number. (a) $2 \mathrm{AgNO}_{3}(a q)+\mathrm{CoCl}_{2}(a q) \longrightarrow 2 \mathrm{AgCl}(s)+ \mathrm{Co}\left(\mathrm{NO}_{3}\right)_{2}(a q)$ (b) $2 \mathrm{PbO}_{2}(s) \longrightarrow 2 \mathrm{PbO}(s)+\mathrm{O}_{2}(g)$ (c) $2 \mathrm{H}_{2} \mathrm{SO}_{4}(a q)+2 \mathrm{NaBr}(s) \longrightarrow \mathrm{Br}_{2}(l)+\mathrm{SO}_{2}(g)+ \mathrm{Na}_{2} \mathrm{SO}_{4}(a q)+2 \mathrm{H}_{2} \mathrm{O}(l)$

Disulfides are compounds that have \(S-S\) bonds, like peroxides have \(\mathrm{O}-\mathrm{O}\) bonds. Thiols are organic compounds that have the general formula \(\mathrm{R}-\mathrm{SH}\), where \(\mathrm{R}\) is a generic hydrocarbon. The \(\mathrm{SH}^{-}\) ion is the sulfur counterpart of hydroxide, \(\mathrm{OH}^{-}\). Two thiols can react to make a disulfide, \(\mathrm{R}-\mathrm{S}-\mathrm{S}-\mathrm{R} .\) (a) What is the oxidation state of sulfur in a thiol? (b) What is the oxidation state of sulfur in a disulfide? (c) If you react two thiols to make a disulfide, are you oxidizing or reducing the thiols? (d) If you wanted to convert a disulfide to two thiols, should you add a reducing agent or oxidizing agent to the solution? (e) Suggest what happens to the H's in the thiols when they form disulfides.

From each of the following pairs of substances, use data in Appendix \(\mathrm{E}\) to choose the one that is the stronger reducing agent: (a) \(\mathrm{Al}(s)\) or \(\mathrm{Mg}(s)\) (b) \(\mathrm{Fe}(s)\) or \(\mathrm{Ni}(s)\) (c) \(\mathrm{H}_{2}(g\), acidic solution) or \(\operatorname{Sn}(s)\) (d) \(\mathrm{I}^{-}(a q)\) or \(\mathrm{Br}^{-}(a q)\)
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