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Chapter 21: Problem 14

Balance the following skeleton reactions, and identify the oxidizing and reducing agents: (a) \(\mathrm{Cr}_{2} \mathrm{O}_{7}^{2-}(a q)+\mathrm{Zn}(s) \longrightarrow \mathrm{Zn}^{2+}(a q)+\mathrm{Cr}^{3+}(a q)\) [acidic] (b) \(\mathrm{Fe}(\mathrm{OH})_{2}(s)+\mathrm{MnO}_{4}^{-}(a q) \longrightarrow \mathrm{MnO}_{2}(s)+\mathrm{Fe}(\mathrm{OH})_{3}(s)[\mathrm{basic}]\) (c) \(\mathrm{Zn}(s)+\mathrm{NO}_{3}^{-}(a q) \longrightarrow \mathrm{Zn}^{2+}(a q)+\mathrm{N}_{2}(g)\) [acidic]

Short Answer

Expert verified
(a) \[ \mathrm{Cr}_{2} \mathrm{O}_{7}^{2-}(aq) + 3\mathrm{Zn}(s) + 14H^+ \rightarrow 3 \mathrm{Zn}^{2+}(aq) + 2\mathrm{Cr}^{3+}(aq) + 7H_2O\] , Oxidizing: \[ \mathrm{Cr}_{2} \mathrm{O}_{7}^{2-}\], Reducing: \[\mathrm{Zn}\]. (b) \[ 3 \mathrm{Fe} ( \mathrm{OH})_{2} (s) + \mathrm{MnO}_{4}^{-}(aq) + 2H_2O \rightarrow \mathrm{MnO}_{2}(s) + 3 \mathrm{Fe} (\mathrm{OH})_{3}(s) + 4OH^-\] , Oxidizing: \[ \mathrm{MnO}_{4}^{-}\], Reducing: \[ \mathrm{Fe} (\mathrm{OH})_{2}\]. (c) \[ 5 \mathrm{Zn}(s) + 2 \mathrm{NO}_{3}^{-}(aq) + 10H^+ \rightarrow 5 \mathrm{Zn}^{2+}(aq) + \mathrm{N}_{2}(g) + 5H_2O\] , Oxidizing: \[ \mathrm{NO}_{3}^{-}\], Reducing: \[ \mathrm{Zn} \].

Step by step solution

01

- Write the unbalanced equation (a)

Identify the unbalanced reaction: \[ \mathrm{Cr}_{2} \mathrm{O}_{7}^{2-}(aq) + \mathrm{Zn}(s) \rightarrow \mathrm{Zn}^{2+}(aq) + \mathrm{Cr}^{3+}(aq) \]
02

- Separate into half reactions (a)

Oxidation: \[ \mathrm{Zn}(s) \rightarrow \mathrm{Zn}^{2+}(aq) + 2e^- \] Reduction: \[ \mathrm{Cr}_{2} \mathrm{O}_{7}^{2-}(aq) + 14H^+ + 6e^- \rightarrow 2\mathrm{Cr}^{3+}(aq) + 7H_2O \]
03

- Balance the half-reactions for electrons (a)

Multiply the oxidation half-reaction by 3 and the reduction half-reaction by 1 to balance the electrons: Oxidation: \[ 3\mathrm{Zn}(s) \rightarrow 3\mathrm{Zn}^{2+}(aq) + 6e^- \] Reduction: \[ \mathrm{Cr}_{2} \mathrm{O}_{7}^{2-}(aq) + 14H^+ + 6e^- \rightarrow 2\mathrm{Cr}^{3+}(aq) + 7H_2O \]
04

- Combine the half-reactions and balance the remaining atoms (a)

Combine and balance: \[ \mathrm{Cr}_{2} \mathrm{O}_{7}^{2-}(aq) + 3\mathrm{Zn}(s) + 14H^+ \rightarrow 3\mathrm{Zn}^{2+}(aq) + 2\mathrm{Cr}^{3+}(aq) + 7H_2O \] Oxidizing agent: \[ \mathrm{Cr}_{2} \mathrm{O}_{7}^{2-} \] Reducing agent: \[ \mathrm{Zn} \]
05

- Write the unbalanced equation (b)

Identify the unbalanced reaction: \[ \mathrm{Fe} (\mathrm{OH})_{2}(s) + \mathrm{MnO}_{4}^{-}(aq) \rightarrow \mathrm{MnO}_{2}(s) + \mathrm{Fe} (\mathrm{OH})_{3}(s) \]
06

- Separate into half reactions (b)

Oxidation: \[ \mathrm{Fe} (\mathrm{OH})_{2}(s) \rightarrow \mathrm{Fe} ( \mathrm{OH})_{3}(s) + e^- \] Reduction: \[ \mathrm{MnO}_{4}^{-}(aq) + 2H_2O + 3e^- \rightarrow \mathrm{MnO}_{2}(s) + 4OH^- \]
07

- Balance the half-reactions for electrons (b)

Multiply the oxidation half-reaction by 3: Oxidation: \[ 3\mathrm{Fe} ( \mathrm{OH})_{2} (s) \rightarrow 3\mathrm{Fe} ( \mathrm{OH})_{3}(s) + 3e^- \] Reduction: \[ \mathrm{MnO}_{4}^{-}(aq) + 2H_2O + 3e^- \rightarrow \mathrm{MnO}_{2}(s) + 4OH^- \]
08

- Combine the half-reactions and balance the remaining atoms (b)

Combine and balance: \[ 3 \mathrm{Fe} ( \mathrm{OH})_{2} (s) + \mathrm{MnO}_{4}^{-}(aq) + 2H_2O \rightarrow \mathrm{MnO}_{2}(s) + 3 \mathrm{Fe} (\mathrm{OH})_{3}(s) + 4OH^- \] Oxidizing agent: \[ \mathrm{MnO}_{4}^{-}\] Reducing agent: \[ \mathrm{Fe} (\mathrm{OH})_{2} \]
09

- Write the unbalanced equation (c)

Identify the unbalanced reaction: \[ \mathrm{Zn}(s) + \mathrm{NO}_{3}^{-} (aq) \rightarrow \mathrm{Zn}^{2+}(aq) + \mathrm{N}_{2}(g) \]
10

- Separate into half reactions (c)

Oxidation: \[ \mathrm{Zn}(s) \rightarrow \mathrm{Zn}^{2+}(aq) + 2e^- \] Reduction: \[ 2 \mathrm{NO}_{3}^{-}(aq) + 10H^+ + 10e^- \rightarrow \mathrm{N}_{2}(g) + 5H_2O \]
11

- Balance the half-reactions for electrons (c)

Multiply the oxidation half-reaction by 5: Oxidation: \[ 5 \mathrm{Zn}(s) \rightarrow 5\mathrm{Zn}^{2+}(aq) + 10e^- \] Reduction: \[ 2 \mathrm{NO}_{3}^{-}(aq) + 10H^+ + 10e^- \rightarrow \mathrm{N}_{2}(g) + 5H_2O \]
12

- Combine the half-reactions and balance the remaining atoms (c)

Combine and balance: \[ 5 \mathrm{Zn}(s) + 2 \mathrm{NO}_{3}^{-}(aq) + 10H^+ \rightarrow 5 \mathrm{Zn}^{2+}(aq) + \mathrm{N}_{2}(g) + 5H_2O \] Oxidizing agent: \[ \mathrm{NO}_{3}^{-} \] Reducing agent: \[ \mathrm{Zn} \]

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Key Concepts

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

Oxidizing Agent
In redox reactions, the oxidizing agent is the substance that gains electrons or is reduced. It causes the oxidation of another substance by accepting electrons from it.

For example, in reaction (a), the dichromate ion \(\text{Cr}_{2}\text{O}_{7}^{2-}\) acts as the oxidizing agent:
  • \( \text{Cr}_{2}\text{O}_{7}^{2-} + 14\text{H}^+ + 6e^- \rightarrow 2\text{Cr}^{3+} + 7\text{H}_{2}\text{O} \)
The dichromate ion gains electrons and is reduced to Cr^{3+}. In reaction (b), permanganate ion \(\text{MnO}_{4}^{-}\) acts as the oxidizing agent. It gains electrons, becoming MnO_2.

Lastly, in reaction (c), the nitrate ion \(\text{NO}_{3}^{-}\) is the oxidizing agent. It gains electrons and is reduced to nitrogen gas \(\text{N}_{2}\).
Reducing Agent
The reducing agent is the substance that loses electrons or is oxidized. It donates electrons to another substance, causing the reduction of the other substance.

In reaction (a), zinc (Zn) acts as the reducing agent:
  • \(3\text{Zn} \rightarrow 3\text{Zn}^{2+} + 6e^- \)
Zinc loses electrons and is oxidized to \(\text{Zn}^{2+}\). Similarly, in reaction (b), iron(II) hydroxide \(\text{Fe}(\text{OH})_{2}\) acts as the reducing agent, losing electrons to form \(\text{Fe}(\text{OH})_{3}\).

In reaction (c), zinc (Zn) again acts as the reducing agent, losing electrons to form zinc ions \(\text{Zn}^{2+}\).
Half-Reactions
Balancing redox reactions requires breaking them down into two half-reactions: one that shows oxidation and the other that shows reduction. Each half-reaction is balanced separately for mass and charge.

For instance, in reaction (a):
  • Oxidation half-reaction: \( \text{Zn} \rightarrow \text{Zn}^{2+} + 2e^- \)
  • Reduction half-reaction: \( \text{Cr}_{2}\text{O}_{7}^{2-} + 14\text{H}^+ + 6e^- \rightarrow 2\text{Cr}^{3+} + 7\text{H}_{2}\text{O} \)
These half-reactions are then combined to give the balanced redox equation.
In reaction (b):
  • Oxidation half-reaction: \( \text{Fe}(\text{OH})_{2} \rightarrow \text{Fe}(\text{OH})_{3} + e^- \)
  • Reduction half-reaction: \( \text{MnO}_{4}^{-} + 2\text{H}_{2}\text{O} + 3e^- \rightarrow \text{MnO}_{2} + 4\text{OH}^{-} \)
These processes separate the electron transfer between the species involved.
Acidic and Basic Conditions
Redox reactions often occur under acidic or basic conditions, which influence the balancing process. Balancing in acidic conditions typically involves adding H^+ ions and water to balance oxygen and hydrogen atoms:
  • For example, in reaction (a): \( \text{Cr}_{2}\text{O}_{7}^{2-} + 14H^+ + 6e^- \rightarrow 2\text{Cr}^{3+} + 7H_2O \)
The H^+ ions and water balance the redox equation in an acidic medium.

In basic conditions, OH^- ions and water are used to balance the equations:
  • For example, in reaction (b): \( \text{MnO}_{4}^{-} + 2H_2O + 3e^- \rightarrow \text{MnO}_{2} + 4OH^{-} \)
This use of hydroxide ions (OH^-) helps to balance the equation in a basic medium.

Knowing whether the reaction occurs in acidic or basic conditions is crucial for correctly balancing redox reactions.

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Most popular questions from this chapter

Brass, an alloy of copper and zinc, can be produced by simultaneously electroplating the two metals from a solution containing their \(2+\) ions. If \(65.0 \%\) of the total current is used to plate copper, while \(35.0 \%\) goes to plating zinc, what is the mass percent of copper in the brass?

Balance the following skeleton reactions, and identify the oxidizing and reducing agents: (a) \(\mathrm{NO}_{2}(g) \longrightarrow \mathrm{NO}_{3}^{-}(a q)+\mathrm{NO}_{2}^{-}(a q)\) [basic] (b) \(\mathrm{Zn}(s)+\mathrm{NO}_{3}^{-}(a q) \longrightarrow \mathrm{Zn}(\mathrm{OH})_{4}^{2-}(a q)+\mathrm{NH}_{3}(g)[\mathrm{basic}]\) (c) \(\mathrm{H}_{2} \mathrm{~S}(g)+\mathrm{NO}_{3}^{-}(a q) \longrightarrow \mathrm{S}_{8}(s)+\mathrm{NO}(g)\) [acidic]

During reconstruction of the Statue of Liberty, Teflon spacers were placed between the iron skeleton and the copper plates that cover the statue. What purpose do these spacers serve?

Like any piece of apparatus, an electrolytic cell operates at less than \(100 \%\) efficiency. A cell depositing \(\mathrm{Cu}\) from a \(\mathrm{Cu}^{2+}\) bath operates for \(10 \mathrm{~h}\) with an average current of 5.8 A. If \(53.4 \mathrm{~g}\) of copper is deposited, at what efficiency is the cell operating?

The \(\mathrm{MnO}_{2}\) used in alkaline batteries can be produced by an electrochemical process of which one half-reaction is $$\mathrm{Mn}^{2+}(a q)+2 \mathrm{H}_{2} \mathrm{O}(l) \longrightarrow \mathrm{MnO}_{2}(s)+4 \mathrm{H}^{+}(a q)+2 \mathrm{e}^{-}$$ If a current of \(25.0 \mathrm{~A}\) is used, how many hours are needed to produce \(1.00 \mathrm{~kg}\) of \(\mathrm{MnO}_{2} ?\) At which electrode is the \(\mathrm{MnO}_{2}\) formed?
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