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Chapter 24: Problem 36

24-36. Complete and balance the following equations: (a) \(\mathrm{Cr}_{2} \mathrm{O}_{7}^{2-}(a q)+\mathrm{I}^{-}(a q) \rightarrow \mathrm{Cr}^{3+}(a q)+\mathrm{I}_{2}(s) \quad\) (acidic) (b) \(\mathrm{CuS}(s)+\mathrm{NO}_{3}^{-}(a q) \rightarrow\) \(\mathrm{Cu}^{2+}(a q)+\mathrm{S}(s)+\mathrm{NO}(g) \quad\) (acidic)

Short Answer

Expert verified
(a) Balanced: \(14\mathrm{H}^{+} + \mathrm{Cr}_{2}\mathrm{O}_{7}^{2-} + 6\mathrm{I}^{-} \rightarrow 2\mathrm{Cr}^{3+} + 7\mathrm{H}_2\mathrm{O} + 3\mathrm{I}_{2}\). (b) Balanced: \(3\mathrm{CuS} + 2\mathrm{NO}_3^- + 8\mathrm{H}^{+} \rightarrow 3\mathrm{Cu}^{2+} + 3\mathrm{S} + 2\mathrm{NO} + 4\mathrm{H}_2\mathrm{O}\).

Step by step solution

01

Split Redox Reaction (a) into Half-Reactions

Identify the oxidation and reduction components. In reaction (a), the chromium is reduced from \(\mathrm{Cr}_{2}\mathrm{O}_{7}^{2-}\) to \(\mathrm{Cr}^{3+}\) and iodine is oxidized from \(\mathrm{I}^{-}\) to \(\mathrm{I}_{2}\). Write the half-reactions: \[ \mathrm{Cr}_{2}\mathrm{O}_{7}^{2-} \rightarrow \mathrm{Cr}^{3+} \]\[ \mathrm{I}^{-} \rightarrow \mathrm{I}_{2} \]
02

Balance Atom and Charge in Chromium Half-Reaction

Balance the atoms other than O and H first. For the chromium reaction, balance Cr:\[ \mathrm{Cr}_{2}\mathrm{O}_{7}^{2-} \rightarrow 2\mathrm{Cr}^{3+} \]Next, balance the oxygen atoms by adding water molecules.\[ \mathrm{Cr}_{2}\mathrm{O}_{7}^{2-} \rightarrow 2\mathrm{Cr}^{3+} + 7\mathrm{H}_2\mathrm{O} \]Then, balance hydrogen by adding \( \mathrm{H}^{+} \) ions:\[ 14\mathrm{H}^{+} + \mathrm{Cr}_{2}\mathrm{O}_{7}^{2-} \rightarrow 2\mathrm{Cr}^{3+} + 7\mathrm{H}_2\mathrm{O} \]Finally, balance charges by adding electrons.\[ 6e^- + 14\mathrm{H}^{+} + \mathrm{Cr}_{2}\mathrm{O}_{7}^{2-} \rightarrow 2\mathrm{Cr}^{3+} + 7\mathrm{H}_2\mathrm{O} \]
03

Balance Atom and Charge in Iodine Half-Reaction

For iodine, balance the iodine atoms:\[ 2\mathrm{I}^{-} \rightarrow \mathrm{I}_{2} \]Next, balance charge by adding electrons:\[ 2\mathrm{I}^{-} \rightarrow \mathrm{I}_{2} + 2e^- \]
04

Equalize Electron Transfer

Ensure both half-reactions transfer the same number of electrons. Multiply the iodine half-reaction by 3:\[ 6\mathrm{I}^{-} \rightarrow 3\mathrm{I}_{2} + 6e^- \]
05

Combine and Balance Equation (a)

Add the balanced half-reactions together:\[6e^- + 14\mathrm{H}^{+} + \mathrm{Cr}_{2}\mathrm{O}_{7}^{2-} + 6\mathrm{I}^{-} \rightarrow 2\mathrm{Cr}^{3+} + 7\mathrm{H}_2\mathrm{O} + 3\mathrm{I}_{2} + 6e^- \]Cancel electrons and simplify:\[ 14\mathrm{H}^{+} + \mathrm{Cr}_{2}\mathrm{O}_{7}^{2-} + 6\mathrm{I}^{-} \rightarrow 2\mathrm{Cr}^{3+} + 7\mathrm{H}_2\mathrm{O} + 3\mathrm{I}_{2} \]
06

Split Redox Reaction (b) into Half-Reactions

Identify the oxidation and reduction parts. Here, copper is oxidized from \(\mathrm{CuS}\) to \(\mathrm{Cu}^{2+}\) and nitrogen is reduced from \(\mathrm{NO}_3^-\) to \(\mathrm{NO}\). Write the half-reactions:\[ \mathrm{CuS} \rightarrow \mathrm{Cu}^{2+} + \mathrm{S} \]\[ \mathrm{NO}_3^- \rightarrow \mathrm{NO} \]
07

Balance Atom and Charge in Copper Half-Reaction

For the copper half-reaction, balance sulfur and copper initially:\[ \mathrm{CuS} \rightarrow \mathrm{Cu}^{2+} + \mathrm{S} \]Add electrons to balance charge:\[ \mathrm{CuS} \rightarrow \mathrm{Cu}^{2+} + \mathrm{S} + 2e^- \]
08

Balance Atom and Charge in Nitrogen Half-Reaction

For the nitrogen half-reaction, balance nitrogen atom and oxygen atoms by adding water:\[ \mathrm{NO}_3^- \rightarrow \mathrm{NO} + 2\mathrm{H}_2\mathrm{O} \]Balance hydrogen by adding \(\mathrm{H}^{+}\):\[ \mathrm{NO}_3^- + 4\mathrm{H}^{+} \rightarrow \mathrm{NO} + 2\mathrm{H}_2\mathrm{O} \]Add electrons to balance charge:\[ \mathrm{NO}_3^- + 4\mathrm{H}^{+} + 3e^- \rightarrow \mathrm{NO} + 2\mathrm{H}_2\mathrm{O} \]
09

Equalize the Electron Transfer

Ensure both half-reactions transfer the same number of electrons. Multiply the copper half-reaction by 3 and the nitrogen half-reaction by 2:\[ 3(\mathrm{CuS} \rightarrow \mathrm{Cu}^{2+} + \mathrm{S} + 2e^-) \]\[ 2(\mathrm{NO}_3^- + 4\mathrm{H}^{+} + 3e^- \rightarrow \mathrm{NO} + 2\mathrm{H}_2\mathrm{O}) \]
10

Combine and Balance Equation (b)

Add the scaled half-reactions together and cancel electrons:\[3\mathrm{CuS} + 2\mathrm{NO}_3^- + 8\mathrm{H}^{+} \rightarrow 3\mathrm{Cu}^{2+} + 3\mathrm{S} + 2\mathrm{NO} + 4\mathrm{H}_2\mathrm{O}\]
11

Conclusion

Both redox reactions are now balanced. Equation (a) is now\[ 14\mathrm{H}^{+} + \mathrm{Cr}_{2}\mathrm{O}_{7}^{2-} + 6\mathrm{I}^{-} \rightarrow 2\mathrm{Cr}^{3+} + 7\mathrm{H}_2\mathrm{O} + 3\mathrm{I}_{2} \]. Equation (b) is\[ 3\mathrm{CuS} + 2\mathrm{NO}_3^- + 8\mathrm{H}^{+} \rightarrow 3\mathrm{Cu}^{2+} + 3\mathrm{S} + 2\mathrm{NO} + 4\mathrm{H}_2\mathrm{O} \].

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Half-Reactions
Redox reactions can seem complex, but breaking them down into simpler components called half-reactions helps to understand the process. A half-reaction is a way of separating the oxidation process from the reduction process. In each redox reaction, there are two half-reactions: one for oxidation and one for reduction.
  • Oxidation half-reaction: This involves the loss of electrons. For example, iodine is oxidized from \( \mathrm{I}^- \) to \( \mathrm{I}_2 \).
  • Reduction half-reaction: This involves the gain of electrons. For chromium, \( \mathrm{Cr}_{2}\mathrm{O}_{7}^{2-} \) reduces to \( \mathrm{Cr}^{3+} \).
For a reaction to be complete, we need both parts to occur simultaneously so that the electrons lost in oxidation are gained in reduction. By writing out half-reactions, we isolate these changes to see the electron flow clearly.
Balancing Chemical Equations
Balancing chemical equations ensures that atoms and charge are conserved. In redox reactions, this becomes a bit tricky because we have to balance both the atoms and the electrons.
For half-reactions, you begin by balancing all the atoms except oxygen and hydrogen. Then:
  • Balance Oxygen: Add water molecules \( \mathrm{H}_2\mathrm{O} \) to balance oxygen atoms.
  • Balance Hydrogen: Add hydrogen ions \( \mathrm{H}^{+} \) to balance hydrogen atoms.
  • Balance Charge: Add electrons to balance the charge on both sides of the reaction.
For example, in the chromium reaction, we add water and hydrogen ions to balance oxygens and hydrogens, and then add electrons to balance the charge. Balancing equations keeps them in line with the law of conservation of mass and charge.
Oxidation and Reduction
Oxidation and reduction are two sides of the same coin in chemical reactions. They are complementary processes where one species loses electrons (oxidation) and another gains electrons (reduction).
  • Oxidation: Often remembered by the mnemonic 'OIL' - Oxidation Is Loss of electrons. For instance, iodine \( \mathrm{I}^{-} \) loses electrons becoming \( \mathrm{I}_2 \).
  • Reduction: Remembered by 'RIG' - Reduction Is Gain of electrons. For example, chromium in \( \mathrm{Cr}_{2}\mathrm{O}_{7}^{2-} \) gains electrons to form \( \mathrm{Cr}^{3+} \).
In any redox reaction, oxidation and reduction occur simultaneously, which is why they are also called "redox reactions." Understanding these concepts is crucial for balancing the reaction equations and ensures proper calculation of electron transfer.

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Most popular questions from this chapter

Solid phosphorus, \(\mathrm{P}_{4}(s)\), reacts with \(\mathrm{BaSO}_{4}(s)\) under oxygen-free, anhydrous conditions to produce \(\mathrm{P}_{4} \mathrm{O}_{10}(s)\) and \(\mathrm{BaS}(s) ;\) write a balanced equation for this process. How much phosphorus is required to react completely with \(2.16\) grams of \(\mathrm{BaSO}_{4}(s) ?\)

Iodine pentoxide is a reagent for the quantitative determination of carbon monoxide. The equation for the reaction is $$ 5 \mathrm{CO}(g)+\mathrm{I}_{2} \mathrm{O}_{5}(s) \rightarrow \mathrm{I}_{2}(s)+5 \mathrm{CO}_{2}(g) $$ The iodine produced is dissolved in \(\mathrm{KI}(a q)\) to form \(\mathrm{I}_{3}^{-}(a q)\) and then determined by reaction with \(\mathrm{Na}_{2} \mathrm{~S}_{2} \mathrm{O}_{3}(a q)\) according to $$ 2 \mathrm{~S}_{2} \mathrm{O}_{3}^{2-}(a q)+\mathrm{I}_{3}^{-}(a q) \rightarrow 3 \mathrm{I}^{-}(a q)+\mathrm{S}_{4} \mathrm{O}_{6}^{2-}(a q) $$ Calculate the mass percentage of \(\mathrm{CO}(g)\) in a \(56.04\) -gram sample of air if it requires \(10.0 \mathrm{~mL}\) of \(0.0350 \mathrm{M} \mathrm{Na}_{2} \mathrm{~S}_{2} \mathrm{O}_{3}(a q)\) to react completely with the \(\mathrm{I}_{3}^{-}(a q)\) produced.

Use the method of half reactions to balance the following equations: (a) \(\mathrm{IO}_{4}^{-}(a q)+\mathrm{I}^{-}(a q) \rightarrow \mathrm{IO}_{3}^{-}(a q)+\mathrm{I}_{3}^{-}(a q)\) (basic) (b) \(\mathrm{H}_{2} \mathrm{MoO}_{4}(a q)+\mathrm{Cr}^{2+}(a q) \rightarrow\) \(\operatorname{Mo}(s)+\mathrm{Cr}^{3+}(a q) \quad\) (acidic)

The concentration of \(\mathrm{H}_{2} \mathrm{~S}(g)\) in air can be determined by the following method. A sample of air is passed through a \(\mathrm{Cd}^{2+}(a q)\) solution, where the sulfur is precipitated as \(\operatorname{CdS}(s)\). This precipitate is then treated with an excess of \(\mathrm{I}_{2}(a q)\), which oxidizes the sulfide ion to elemental sulfur, \(\mathrm{S}(s)\). The amount of excess \(\mathrm{I}_{2}(a q)\) is determined by titration with \(\mathrm{Na}_{2} \mathrm{~S}_{2} \mathrm{O}_{3}(a q)\) as described in the previous problem. Suppose that a \(10.75\) -gram sample of air is passed through a \(\mathrm{Cd}^{2+}(a q)\) solution. A \(30.00-\mathrm{mL}\) sample of \(0.0115\) M \(\mathrm{I}_{2}(a q)\) is then added to the \(\operatorname{CdS}(s)\) The unreacted \(\mathrm{I}_{2}(a q)\) required \(7.65 \mathrm{~mL}\) of \(0.0750 \mathrm{M}\) \(\mathrm{Na}_{2} \mathrm{~S}_{2} \mathrm{O}_{3}(a q)\) for reduction to \(\mathrm{I}^{-}(a q) .\) Calculate the mass percentage of \(\mathrm{H}_{2} \mathrm{~S}(s)\) in the air.

Write Lewis formulas for the covalent titanium ions (a) \(\mathrm{TiF}_{6}^{2-}(a q)\) (b) \(\mathrm{TiBr}_{5}^{-}(a q)\)
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