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Chapter 17: Problem 112

What reaction will take place at the cathode and the anode when each of the following is electrolyzed? (Assume standard conditions.) a. \(1.0 M\space \mathrm {KF}\) solution b. \(1.0 M\space \mathrm {CuCl}_{2}\) solution c. \(1.0 M \space \mathrm{MgI}_{2}\) solution

Short Answer

Expert verified
a. For a \(1.0 M\space KF\) solution, the reactions at the cathode and anode are: Cathode: \(K^{+} + e^{-} \rightarrow K\) Anode: \(2F^{-} \rightarrow F_{2} + 2e^{-}\) b. For a \(1.0 M\space CuCl_{2}\) solution, the reactions at the cathode and anode are: Cathode: \(Cu^{2+} + 2e^{-} \rightarrow Cu\) Anode: \(2Cl^{-} \rightarrow Cl_{2} + 2e^{-}\) c. For a \(1.0 M\space MgI_{2}\) solution, the reactions at the cathode and anode are: Cathode: \(Mg^{2+} + 2e^{-} \rightarrow Mg\) Anode: \(2I^{-} \rightarrow I_{2} + 2e^{-}\)

Step by step solution

01

Identify the ions present

For a 1.0 M KF solution, the dissociation in water will produce K+ and F- ions.
02

Determine the possible reduction and oxidation reactions

At the cathode (reduction): K+ + e- -> K At the anode (oxidation): 2F- -> F2 + 2e- b. Electrolysis of a 1.0 M CuCl2 solution
03

Identify the ions present

For a 1.0 M CuCl2 solution, the dissociation in water will produce Cu(2+) and 2Cl- ions.
04

Determine the possible reduction and oxidation reactions

At the cathode (reduction): Cu(2+) + 2e- -> Cu At the anode (oxidation): 2Cl- -> Cl2 + 2e- c. Electrolysis of a 1.0 M MgI2 solution
05

Identify the ions present

For a 1.0 M MgI2 solution, the dissociation in water will produce Mg(2+) and 2I- ions.
06

Determine the possible reduction and oxidation reactions

At the cathode (reduction): Mg(2+) + 2e- -> Mg At the anode (oxidation): 2I- -> I2 + 2e-

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Most popular questions from this chapter

The overall reaction in the lead storage battery is $$ \begin{array}{r} \mathrm{Pb}(s)+\mathrm{PbO}_{2}(s)+2 \mathrm{H}^{+}(a q)+2 \mathrm{HSO}_{4}^{-}(a q) \longrightarrow \\ 2 \mathrm{PbSO}_{4}(s)+2 \mathrm{H}_{2} \mathrm{O}(l) \end{array} $$ Calculate \(\mathscr{E}\) at \(25^{\circ} \mathrm{C}\) for this battery when \(\left[\mathrm{H}_{2} \mathrm{SO}_{4}\right]=4.5 \mathrm{M}\) that is, \(\left[\mathrm{H}^{+}\right]=\left[\mathrm{HSO}_{4}^{-}\right]=4.5 \mathrm{M} .\) At \(25^{\circ} \mathrm{C}, 8^{\circ}=2.04 \mathrm{V}\) for the lead storage battery.

Consider the following galvanic cell at \(25^{\circ} \mathrm{C}\) : $$ \mathrm{Pt}\left|\mathrm{Cr}^{2+}(0.30 M), \mathrm{Cr}^{3+}(2.0 M)\right|\left|\mathrm{Co}^{2+}(0.20 M)\right| \mathrm{Co} $$ The overall reaction and equilibrium constant value are $$\begin{aligned} 2 \mathrm{Cr}^{2+}(a q)+\mathrm{Co}^{2+}(a q) & \longrightarrow \\ 2 \mathrm{Cr}^{3+}(a q)+\mathrm{Co}(s) & K=2.79 \times 10^{7} \end{aligned}$$ Calculate the cell potential, \(\mathscr{E}\), for this galvanic cell and \(\Delta G\) for the cell reaction at these conditions.

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.

Consider the following galvanic cell: a. Label the reducing agent and the oxidizing agent, and describe the direction of the electron flow. b. Determine the standard cell potential. c. Which electrode increases in mass as the reaction proceeds, and which electrode decreases in mass?

When copper reacts with nitric acid, a mixture of \(\mathrm{NO}(g)\) and \(\mathrm{NO}_{2}(g)\) is evolved. The volume ratio of the two product gases depends on the concentration of the nitric acid according to the equilibrium $$2 \mathrm{H}^{+}(a q)+2 \mathrm{NO}_{3}^{-}(a q)+\mathrm{NO}(g) \rightleftharpoons 3 \mathrm{NO}_{2}(g)+\mathrm{H}_{2} \mathrm{O}(l)$$ Consider the following standard reduction potentials at \(25^{\circ} \mathrm{C}:\) $$\begin{aligned} &3 \mathrm{e}^{-}+4 \mathrm{H}^{+}(a q)+\mathrm{NO}_{3}^{-}(a q) \longrightarrow \mathrm{NO}(g)+2 \mathrm{H}_{2} \mathrm{O}(l)\ & \mathscr{E}^{\circ}=0.957 \mathrm{V} \end{aligned}$$ $$\begin{aligned} &\mathrm{e}^{-}+2 \mathrm{H}^{+}(a q)+\mathrm{NO}_{3}^{-}(a q) \longrightarrow \mathrm{NO}_{2}(g)+2 \mathrm{H}_{2} \mathrm{O}(l) &\mathscr{E}^{\circ}=0.775 \mathrm{V} \end{aligned}$$ a. Calculate the equilibrium constant for the above reaction. b. What concentration of nitric acid will produce a NO and \(\mathrm{NO}_{2}\) mixture with only \(0.20 \% \mathrm{NO}_{2}\) (by moles) at \(25^{\circ} \mathrm{C}\) and 1.00 atm? Assume that no other gases are present and that the change in acid concentration can be neglected.
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