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Chapter 4: Problem 91

Balance the following oxidation–reduction reactions that occur in acidic solution using the half-reaction method. a. $\mathbf{I}^{-}(a q)+\mathrm{ClO}^{-}(a q) \rightarrow \mathrm{I}_{3}^{-}(a q)+\mathrm{Cl}^{-}(a q)$ b. $\mathrm{As}_{2} \mathrm{O}_{3}(s)+\mathrm{NO}_{3}^{-}(a q) \rightarrow \mathrm{H}_{3} \mathrm{AsO}_{4}(a q)+\mathrm{NO}(g)$ c. $\mathrm{Br}^{-}(a q)+\mathrm{MnO}_{4}^{-}(a q) \rightarrow \mathrm{Br}_{2}(l)+\mathrm{Mn}^{2+}(a q)$ d. $\mathrm{CH}_{3} \mathrm{OH}(a q)+\mathrm{Cr}_{2} \mathrm{O}_{7}^{2-}(a q) \rightarrow \mathrm{CH}_{2} \mathrm{O}(a q)+\mathrm{Cr}^{3+}(a q)$

Short Answer

Expert verified
a. \(3I^{-}(aq) + ClO^{-}(aq) \rightarrow I_{3}^{-}(aq)+Cl^{-}(aq)\) b. \(2As_{2}O_{3}(s) + 12NO_{3}^{-}(aq) + 12H^{+}(aq) \rightarrow 4H_{3}AsO_{4}(aq) + 6NO(g) + 6H_{2}O(l)\) c. \(10Br^{-}(aq) + 2MnO_{4}^{-}(aq) + 16H^{+}(aq) \rightarrow 5Br_{2}(l) + 2Mn^{2+}(aq) + 8H_{2}O(l)\) d. \(3CH_{3}OH(aq) + Cr_{2}O_{7}^{2-}(aq) + 8H^{+}(aq) \rightarrow 3CH_{2}O(aq) + 2Cr^{3+}(aq) + 4H_{2}O(l)\)

Step by step solution

01

Assign oxidation numbers

First, assign oxidation numbers to all elements in the unbalanced equation: \(I^{-}: -1 \quad ClO^{-}: +1 \quad I_{3}^{-}: 0 \quad Cl^{-}: -1\)
02

Identify half-reactions

Next, we can see the oxidation half-reaction is: \(I^{-}(aq) \rightarrow I_{3}^{-}(aq)\) And the reduction half-reaction is: \(ClO^{-}(aq) \rightarrow Cl^{-}(aq)\)
03

Balance atoms

Balance the atoms in both half-reactions: \(I^{-}(aq) \rightarrow I_{3}^{-}(aq)\) (balance I atoms by multiplying the \(I_{3}^{-}\) by 3) \(3I^{-}(aq) \rightarrow I_{3}^{-}(aq)\) The chlorine and oxygen atoms are already balanced in the reduction half-reaction.
04

Balance charges

Balance the charges in both half-reactions: For the oxidation half-reaction, add 2 electrons to the right side: \(3I^{-}(aq) \rightarrow I_{3}^{-}(aq) + 2e^{-}\) For the reduction half-reaction, add an electron to the left side: \(e^{-}+ClO^{-}(aq) \rightarrow Cl^{-}(aq)\)
05

Align electron counts

For this reaction, the electron counts are already the same (2) in both half-reactions.
06

Add half-reactions

Add the balanced half-reactions and cancel the electrons: \(3I^{-}(aq) + e^{-}+ClO^{-}(aq) \rightarrow I_{3}^{-}(aq)+Cl^{-}(aq) + 2e^{-}\) Cancel the electrons and obtain the final balanced equation: \(3I^{-}(aq) + ClO^{-}(aq) \rightarrow I_{3}^{-}(aq)+Cl^{-}(aq)\) To balance the remaining reactions, simply follow the same steps. b. \(As_{2}O_{3}(s)+NO_{3}^{-}(aq) \rightarrow H_{3}AsO_{4}(aq)+NO(g)\) c. \(Br^{-}(aq)+MnO_{4}^{-}(aq) \rightarrow Br_{2}(l)+Mn^{2+}(aq)\) d. \(CH_{3}OH(aq)+Cr_{2}O_{7}^{2-}(aq) \rightarrow CH_{2}O(aq)+Cr^{3+}(aq)\)

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Most popular questions from this chapter

The unknown acid \(\mathrm{H}_{2} \mathrm{X}\) can be neutralized completely by \(\mathrm{OH}^{-}\) according to the following (unbalanced) equation: $$\mathrm{H}_{2} \mathrm{X}(a q)+\mathrm{OH}^{-}(a q) \longrightarrow \mathrm{X}^{2-}(a q)+\mathrm{H}_{2} \mathrm{O}(i) $$ The ion formed as a product, \(X^{2-},\) was shown to have 36 total electrons. What is element X? Propose a name for \(\mathrm{H}_{2} \mathrm{X}\) . To completely neutralize a sample of $\mathrm{H}_{2} \mathrm{X}, 35.6 \mathrm{mL}\( of 0.175 \)\mathrm{M}\( \)\mathrm{OH}^{-}$ solution was required. What was the mass of the \(\mathrm{H}_{2} \mathrm{X}\) sample used?

A 1.42 -g sample of a pure compound, with formula $\mathrm{M}_{2} \mathrm{SO}_{4}$ was dissolved in water and treated with an excess of aqueous calcium chloride, resulting in the precipitation of all the sulfate ions as calcium sulfate. The precipitate was collected, dried, and found to weigh 1.36 g. Determine the atomic mass of M, and identify M.

Assign oxidation states for all atoms in each of the following compounds a. \({KMnO}_{4} \quad\quad\quad f. {Fe}_{3} {O}_{4}\) b. \({NiO}_{2} \quad\quad\quad\quad g. {XeOF}_{4}\) c. \({Na}_{4} {Fe}({OH})_{6} \quad h. {SF}_{4}\) d. \({NH}_{4} {h}_{2} {HPO}_{4} \quad i. {CO}\) e. \({P}_{4} {O}_{6} \quad\quad\quad\quad\quad j. {C}_{6} {H}_{12} {O}_{6}\)

What volume of 0.100\(M \mathrm{Na}_{3} \mathrm{PO}_{4}\) is required to precipitate all the lead(II) ions from 150.0 \(\mathrm{mL}\) of 0.250$M \mathrm{Pb}\left(\mathrm{NO}_{3}\right)_{2} ?$

A 30.0 -mL sample of an unknown strong base is neutralized after the addition of 12.0 \(\mathrm{mL}\) of a 0.150 \(\mathrm{M} \mathrm{HNO}_{3}\) solution. If the unknown base concentration is 0.0300 M, give some possible identities for the unknown base.
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