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Chapter 18: Problem 30

Balance the following oxidation-reduction reactions that occur in acidic solution using the half-reaction method. a. \(\mathrm{Cu}(s)+\mathrm{NO}_{3}^{-}(a q) \rightarrow \mathrm{Cu}^{2+}(a q)+\mathrm{NO}(g)\) b. \(\mathrm{Cr}_{2} \mathrm{O}_{7}^{2-}(a q)+\mathrm{Cl}^{-}(a q) \rightarrow \mathrm{Cr}^{3+}(a q)+\mathrm{Cl}_{2}(g)\) c. \(\mathrm{Pb}(s)+\mathrm{PbO}_{2}(s)+\mathrm{H}_{2} \mathrm{SO}_{4}(a q) \rightarrow \operatorname{PbSO}_{4}(s)\) d. \(\mathrm{Mn}^{2+}(a q)+\mathrm{NaBiO}_{3}(s) \rightarrow \mathrm{Bi}^{3+}(a q)+\mathrm{MnO}_{4}^{-}(a q)\) e. \(\mathrm{H}_{3} \mathrm{AsO}_{4}(a q)+\mathrm{Zn}(s) \rightarrow \mathrm{AsH}_{3}(g)+\mathrm{Zn}^{2+}(a q)\)

Short Answer

Expert verified
Balancing the other reactions: c. \(3\mathrm{Pb}(s) + 2\mathrm{PbO}_{2}(s) + 3\mathrm{H}_{2} \mathrm{SO}_{4}(a q) \rightarrow 5\operatorname{PbSO}_{4}(s) + 6\mathrm{H_2O}(l)\) d. \(\mathrm{Mn}^{2+}(a q)+ 2\mathrm{NaBiO}_{3}(s) + 2\mathrm{H}^+(a q) \rightarrow \mathrm{Bi}^{3+}(a q)+ 2\mathrm{MnO}_{4}^{-}(a q) + 2\mathrm{Na}^+(a q) + \mathrm{H_2O}(l)\) e. \(2\mathrm{H}_{3} \mathrm{AsO}_{4}(a q)+ 3\mathrm{Zn}(s) \rightarrow 2\mathrm{AsH}_{3}(g)+ 3\mathrm{Zn}^{2+}(a q) + 4\mathrm{H}_2\mathrm{O}(l)\)

Step by step solution

01

a. Balancing the reaction: Cu(s) + NO3−(aq) → Cu2+(aq) + NO(g) Step 1: Write the half-reactions

Oxidation half-reaction: \(\mathrm{Cu}(s) \rightarrow \mathrm{Cu}^{2+}(a q)\) Reduction half-reaction: \(\mathrm{NO}_{3}^{-}(a q) \rightarrow \mathrm{NO}(g)\) Step 2: Balance atoms and charges
02

Oxidation half-reaction: \[\mathrm{Cu}(s) \rightarrow \mathrm{Cu}^{2+}(a q) + 2e^-\] Reduction half-reaction: \[\mathrm{NO}_{3}^{-}(a q) + 4\mathrm{H}^+(a q) + 3e^- \rightarrow \mathrm{NO}(g) + 2\mathrm{H}_2\mathrm{O}(l)\] Step 3: Combine the half-reactions

Multiply the oxidation half-reaction by 3 and the reduction half-reaction by 2 to equalize the electrons: \[3\mathrm{Cu}(s) \rightarrow 3\mathrm{Cu}^{2+}(a q) + 6e^-\] \[2\mathrm{NO}_{3}^{-}(a q) + 8\mathrm{H}^+(a q) + 6e^- \rightarrow 2\mathrm{NO}(g) + 4\mathrm{H}_2\mathrm{O}(l)\] Then, add the two half-reactions together: \[3\mathrm{Cu}(s) + 2\mathrm{NO}_{3}^{-}(a q) + 8\mathrm{H}^+(a q) \rightarrow 3\mathrm{Cu}^{2+}(a q) + 2\mathrm{NO}(g) + 4\mathrm{H}_2\mathrm{O}(l)\] We've balanced the first reaction: Cu(s) + NO3−(aq) → Cu2+(aq) + NO(g).
03

b. Balancing the reaction: Cr2O72−(aq) + Cl−(aq) → Cr3+(aq) + Cl2(g) Step 1: Write the half-reactions

Oxidation half-reaction: \(\mathrm{Cl}^-(a q) \rightarrow \mathrm{Cl}_{2}(g)\) Reduction half-reaction: \(\mathrm{Cr}_{2} \mathrm{O}_{7}^{2-}(a q) \rightarrow \mathrm{Cr}^{3+}(a q)\) Step 2: Balance atoms and charges
04

Oxidation half-reaction: \[\mathrm{2Cl}^-(a q) \rightarrow \mathrm{Cl}_{2}(g) + 2e^-\] Reduction half-reaction: \[\mathrm{Cr}_{2} \mathrm{O}_{7}^{2-}(a q) + 14\mathrm{H}^+(a q) + 6e^- \rightarrow 2\mathrm{Cr}^{3+}(a q) + 7\mathrm{H}_2\mathrm{O}(l)\] Step 3: Combine the half-reactions

Multiply the oxidation half-reaction by 3 and the reduction half-reaction by 2 to equalize the electrons: \[6\mathrm{Cl}^-(a q) \rightarrow 3\mathrm{Cl}_{2}(g) + 6e^-\] \[\mathrm{Cr}_{2} \mathrm{O}_{7}^{2-}(a q) + 14\mathrm{H}^+(a q) + 6e^- \rightarrow 2\mathrm{Cr}^{3+}(a q) + 7\mathrm{H}_2\mathrm{O}(l)\] Then, add the two half-reactions together: \[6\mathrm{Cl}^-(a q) + \mathrm{Cr}_{2} \mathrm{O}_{7}^{2-}(a q) + 14\mathrm{H}^+(a q) \rightarrow 3\mathrm{Cl}_{2}(g) + 2\mathrm{Cr}^{3+}(a q) + 7\mathrm{H}_2\mathrm{O}(l)\] We've balanced the second reaction: Cr2O72−(aq) + Cl−(aq) → Cr3+(aq) + Cl2(g). Note: We'll follow similar procedure for the reactions c, d, and e.

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Most popular questions from this chapter

A fuel cell designed to react grain alcohol with oxygen has the following net reaction: $$ \mathrm{C}_{2} \mathrm{H}_{5} \mathrm{OH}(l)+3 \mathrm{O}_{2}(g) \longrightarrow 2 \mathrm{CO}_{2}(g)+3 \mathrm{H}_{2} \mathrm{O}(l) $$ The maximum work that 1 mole of alcohol can do is \(1.32 \times\) \(10^{3} \mathrm{~kJ}\). What is the theoretical maximum voltage this cell can achieve at \(25^{\circ} \mathrm{C} ?\)

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 \(\mathrm{Pt}^{4+}\) ?

Consider the following galvanic cell at \(25^{\circ} \mathrm{C}\) : $$ \mathrm{Pt}\left|\mathrm{Cr}^{2+}(0.30 \mathrm{M}), \mathrm{Cr}^{3+}(2.0 \mathrm{M})\right|\left|\mathrm{Co}^{2+}(0.20 \mathrm{M})\right| \mathrm{Co} $$ 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.

A zinc-copper battery is constructed as follows at \(25^{\circ} \mathrm{C}\) : $$ \mathrm{Zn}\left|\mathrm{Zn}^{2+}(0.10 M)\right|\left|\mathrm{Cu}^{2+}(2.50 M)\right| \mathrm{Cu} $$ The mass of each electrode is \(200 . \mathrm{g}\). a. Calculate the cell potential when this battery is first connected. b. Calculate the cell potential after \(10.0 \mathrm{~A}\) of current has flowed for \(10.0 \mathrm{~h}\). (Assume each half-cell contains \(1.00 \mathrm{~L}\) of solution.) c. Calculate the mass of each electrode after \(10.0 \mathrm{~h}\). d. How long can this battery deliver a current of \(10.0 \mathrm{~A}\) before it goes dead?

When jump-starting a car with a dead battery, the ground jumper should be attached to a remote part of the engine block. Why?
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