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Chapter 18: Problem 109

What reactions take place at the cathode and the anode when each of the following is electrolyzed? (Assume standard conditions.) a. 1.0 \(M \mathrm{NiBr}_{2}\) solution b. 1.0 \(M \mathrm{AlF}_{3}\) solution c. 1.0 \(M \mathrm{MnI}_{2}\) solution

Short Answer

Expert verified
For the electrolysis of the given solutions under standard conditions, the following half-reactions take place: a. 1.0 M NiBr₂ solution: Cathode: Ni²⁺ + 2e⁻ → Ni Anode: 2Br⁻ → Br₂ + 2e⁻ b. 1.0 M AlF₃ solution: Cathode: Al³⁺ + 3e⁻ → Al Anode: 2F⁻ → F₂ + 2e⁻ c. 1.0 M MnI₂ solution: Cathode: Mn²⁺ + 2e⁻ → Mn Anode: 2I⁻ → I₂ + 2e⁻

Step by step solution

01

Identify the ions in the solution

In a 1.0 M NiBr₂ solution, the ions present are Ni²⁺ and Br⁻.
02

Determine the reactions at the cathode and anode using reduction potentials

The standard reduction potentials for the relevant half-reactions are: Ni²⁺ + 2e⁻ → Ni (E° = -0.25 V) 2Br⁻ → Br₂ + 2e⁻ (E° = +1.09 V) Since the reaction with the more positive reduction potential occurs at the cathode, Ni²⁺ will be reduced to Ni, and Br⁻ will be oxidized to Br₂ at the anode.
03

Write balanced half-reactions

The balanced half-reactions for the electrolysis of 1.0 M NiBr₂ are: Cathode: Ni²⁺ + 2e⁻ → Ni Anode: 2Br⁻ → Br₂ + 2e⁻ #1.0 M AlF₃ solution#
04

Identify the ions in the solution

In a 1.0 M AlF₃ solution, the ions present are Al³⁺ and F⁻.
05

Determine the reactions at the cathode and anode using reduction potentials

The standard reduction potentials for the relevant half-reactions are: Al³⁺ + 3e⁻ → Al (E° = -1.66 V) 2F⁻ → F₂ + 2e⁻ (E° = +2.87 V) Since the reaction with the more positive reduction potential occurs at the cathode, Al³⁺ will be reduced to Al, and F⁻ will be oxidized to F₂ at the anode.
06

Write balanced half-reactions

The balanced half-reactions for the electrolysis of 1.0 M AlF₃ are: Cathode: Al³⁺ + 3e⁻ → Al Anode: 2F⁻ → F₂ + 2e⁻ #1.0 M MnI₂ solution#
07

Identify the ions in the solution

In a 1.0 M MnI₂ solution, the ions present are Mn²⁺ and I⁻.
08

Determine the reactions at the cathode and anode using reduction potentials

The standard reduction potentials for the relevant half-reactions are: Mn²⁺ + 2e⁻ → Mn (E° = -1.18 V) 2I⁻ → I₂ + 2e⁻ (E° = +0.54 V) Since the reaction with the more positive reduction potential occurs at the cathode, Mn²⁺ will be reduced to Mn, and I⁻ will be oxidized to I₂ at the anode.
09

Write balanced half-reactions

The balanced half-reactions for the electrolysis of 1.0 M MnI₂ are: Cathode: Mn²⁺ + 2e⁻ → Mn Anode: 2I⁻ → I₂ + 2e⁻

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Most popular questions from this chapter

It takes 15 kWh (kilowatt-hours) of electrical energy to produce 1.0 kg aluminum metal from aluminum oxide by the Hall-Heroult process. Compare this to the amount of energy necessary to melt 1.0 kg aluminum metal. Why is it economically feasible to recycle aluminum cans? [The enthalpy of fusion for aluminum metal is 10.7 $\mathrm{kJ} / \mathrm{mol}(1 \text { watt }=1 \mathrm{J} / \mathrm{s}) . ]$

A disproportionation reaction involves a substance that acts as both an oxidizing and a reducing agent, producing higher and lower oxidation states of the same element in the products. Which of the following disproportionation reactions are spontaneous under standard conditions? Calculate $\Delta G^{\circ}\( and \)K\( at \)25^{\circ} \mathrm{C}$ for those reactions that are spontaneous under standard conditions. a. $2 \mathrm{Cu}^{+}(a q) \longrightarrow \mathrm{Cu}^{2+}(a q)+\mathrm{Cu}(s)$ b. $3 \mathrm{Fe}^{2+}(a q) \longrightarrow 2 \mathrm{Fe}^{3+}(a q)+\mathrm{Fe}(s)$ c. $\mathrm{HClO}_{2}(a q) \longrightarrow \mathrm{ClO}_{3}^{-}(a q)+\mathrm{HClO}(a q) \quad$ (unbalanced) Use the half-reactions: $\mathrm{ClO}_{3}^{-}+3 \mathrm{H}^{+}+2 \mathrm{e}^{-} \longrightarrow \mathrm{HClO}_{2}+\mathrm{H}_{2} \mathrm{O} \quad \mathscr{E}^{\circ}=1.21 \mathrm{V}$ $\mathrm{HClO}_{2}+2 \mathrm{H}^{+}+2 \mathrm{e}^{-} \longrightarrow \mathrm{HClO}+\mathrm{H}_{2} \mathrm{O} \quad \mathscr{E}^{\circ}=1.65 \mathrm{V}$

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 and standard cell potential at \(25^{\circ} \mathrm{C}\) for the rechargeable nickel-cadmium alkaline battery is $\mathrm{Cd}(s)+\mathrm{NiO}_{2}(s)+2 \mathrm{H}_{2} \mathrm{O}(l) \longrightarrow$ $\mathrm{Ni}(\mathrm{OH})_{2}(s)+\mathrm{Cd}(\mathrm{OH})_{2}(s) \qquad \mathscr{E}^{\circ}=1.10 \mathrm{V}$ For every mole of Cd consumed in the cell, what is the maximum useful work that can be obtained at standard conditions?

A solution containing \(\mathrm{Pt}^{4+}\) is electrolyzed with a current of 4.00 \(\mathrm{A} .\) How long will it take to plate out 99\(\%\) of the platinum in 0.50 \(\mathrm{L}\) of a \(0.010-M\) solution of \(\mathrm{Pt}^{4+} ?\)
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