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Balance the following redox equations. All occur in basic solution. (a) \(\mathrm{Fe}(\mathrm{OH})_{3}(\mathrm{s})+\mathrm{Cr}(\mathrm{s}) \rightarrow\) \(\mathrm{Cr}(\mathrm{OH})_{3}(\mathrm{s})+\mathrm{Fe}(\mathrm{OH})_{2}(\mathrm{s})\) (b) \(\mathrm{NiO}_{2}(\mathrm{s})+\mathrm{Zn}(\mathrm{s}) \rightarrow\) \(\mathrm{Ni}(\mathrm{OH})_{2}(\mathrm{s})+\mathrm{Zn}(\mathrm{OH})_{2}(\mathrm{s})\) (c) \(\mathrm{Fe}(\mathrm{OH})_{2}(\mathrm{s})+\mathrm{CrO}_{4}^{2-}(\mathrm{aq}) \rightarrow\) \(\mathrm{Fe}(\mathrm{OH})_{3}(\mathrm{s})+\left[\mathrm{Cr}(\mathrm{OH})_{4}\right]^{-}(\mathrm{aq})\) (d) \(\mathrm{N}_{2} \mathrm{H}_{4}(\mathrm{aq})+\mathrm{Ag}_{2} \mathrm{O}(\mathrm{s}) \rightarrow \mathrm{N}_{2}(\mathrm{g})+\mathrm{Ag}(\mathrm{s})\)

Short Answer

Expert verified
(a) 3Fe(OH)_3 + Cr → 3Fe(OH)_2 + Cr(OH)_3. (b) NiO_2 + Zn + 2H_2O → Ni(OH)_2 + Zn(OH)_2. (c) 2Fe(OH)_2 + CrO_4^2- + H_2O → 2Fe(OH)_3 + [Cr(OH)_4]^-. (d) N_2H_4 + 2Ag_2O → N_2 + 4Ag + 2H_2O.

Step by step solution

01

Identify Oxidation and Reduction

In the given redox reactions, determine which species are undergoing oxidation and which are undergoing reduction. For example, in reaction (a), Fe(OH)_3 contains Fe in the +3 oxidation state and is being reduced to Fe(OH)_2, where Fe is in the +2 oxidation state. Simultaneously, Cr is being oxidized into Cr(OH)_3 where Cr is in +3 oxidation state.
02

Write Half-Reactions

Split each of the provided redox reactions into their oxidation and reduction half-reactions. For reaction (a), the reduction half-reaction is Fe(OH)_3 + e^- → Fe(OH)_2. The oxidation half-reaction is Cr → Cr(OH)_3 + 3e^-.
03

Balance Atoms Other Than Oxygen and Hydrogen

Balance all atoms in each half-reaction other than oxygen and hydrogen. For well-balanced reactions, check that the number of metal and non-metal atoms are balanced. For instance, in both half-reactions of reaction (a), Fe, Cr are balanced.
04

Balance Oxygen Atoms

Add H2O molecules as needed to balance the oxygen atoms in each half-reaction. In reaction (a), add water molecules to the oxidation half-reaction to balance it: Cr + 3H2O → Cr(OH)_3.
05

Balance Hydrogen Atoms

Add OH^- ions to balance hydrogen atoms since the solution is basic. In reaction (a), add OH^- to the right side: Cr + 3H2O → Cr(OH)_3 + 3OH^-.
06

Balance Charge with Electrons

Balance each half-reaction's charge by adding electrons. For reaction (a), the reduction half-reaction is: Fe(OH)_3 + e^- → Fe(OH)_2 + OH^-. The oxidation half-reaction should be adjusted for charges as well: Cr + 3OH^- → Cr(OH)_3 + 3e^-.
07

Equalize Electrons and Combine

Make sure the number of electrons lost in oxidation equals electrons gained in reduction. Then add the half-reactions together. For reaction (a), multiply the reduction half-reaction by 3 and combine: 3[Fe(OH)_3 + e^- → Fe(OH)_2 + OH^-] and [Cr + 3OH^- → Cr(OH)_3 + 3e^-].
08

Simplify and Verify Final Equation

Cancel out terms common to both sides such as electrons, water, and hydroxide ions. Simplify the reaction to ensure mass and charge balance is achieved. For reaction (a), the final balanced equation is 3Fe(OH)_3 + Cr → 3Fe(OH)_2 + Cr(OH)_3.

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Key Concepts

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

Redox Chemistry
Redox chemistry is an essential aspect of chemical reactions involving the transfer of electrons between reactants. These reactions include both oxidation, the loss of electrons, and reduction, the gain of electrons. This dual process makes up what we call a redox reaction. The substance that donates electrons is known as the reducing agent, and the one that accepts electrons is the oxidizing agent.
Understanding redox reactions is crucial as they are fundamental to many natural and industrial processes, from cellular respiration in biology to the functioning of batteries in technology.
  • Oxidation involves an increase in oxidation state.
  • Reduction involves a decrease in oxidation state.
  • In basic solutions, hydroxide ions play a key role in balancing these reactions.
In summary, redox reactions are key to understanding electron flow and energy transfer in a variety of chemical processes.
Half-Reaction Method
The half-reaction method is an effective way to balance redox equations by separating the oxidation and reduction processes. This method involves writing two separate half-reactions, one for oxidation and one for reduction.
Each half-reaction is balanced independently, taking into account the elements involved and the changes in oxidation states. This method also ensures that the electrons lost in oxidation are equal to the electrons gained in reduction:
  • First, identify the species being oxidized and reduced by examining changes in oxidation states.
  • Write the oxidation and reduction reactions separately.
  • For each half-reaction, balance atoms other than oxygen and hydrogen first.
  • Balance oxygen by adding water molecules and hydrogen by adding hydroxide ions for basic solutions.
  • Balance the charge by adding electrons as necessary.
This meticulous approach helps ensure that all mass and charge considerations are satisfied in the final balanced equation.
Oxidation-Reduction Balance
Balancing oxidation-reduction (redox) reactions is crucial for accurately representing chemical changes that involve electron transfer. The goal is to ensure that both mass and charge are balanced on both sides of the equation.
Achieving this balance involves aligning the electron flow between oxidation, which involves electron loss, and reduction, which involves electron gain. Here are some key steps you might follow:
  • After writing and balancing half-reactions, ensure that the electrons lost in the oxidation half-reaction match those gained in the reduction half-reaction.
  • If necessary, multiply the half-reactions by appropriate coefficients to equalize the number of electrons exchanged.
  • Combine the balanced half-reactions by adding them together and eliminate any common terms such as electrons on both sides.
  • Simplify to get the smallest whole-number coefficients and verify that all elements and charges are balanced in the final equation.
This careful balance between oxidation and reduction mechanisms is essential for consistency in chemical equations, reflecting the conservation of mass and energy.

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Most popular questions from this chapter

Write equations for the half-reactions that occur at the anode and cathode in the electrolysis of molten KBr. What are the products formed at the anode and cathode in the electrolysis of aqueous KBr?

Balance each of the following unbalanced equations; then calculate the standard potential, \(E^{\circ}\) and decide whether each is product-favored at equilibrium as written. (All reactions are carried out in acid solution. (a) \(\mathrm{I}_{2}(\mathrm{s})+\mathrm{Br}^{-}(\mathrm{aq}) \rightarrow \mathrm{I}^{-}(\mathrm{aq})+\mathrm{Br}_{2}(\ell)\) (b) \(\mathrm{Fe}^{2+}(\mathrm{aq})+\mathrm{Cu}^{2+}(\mathrm{aq}) \rightarrow \mathrm{Cu}(\mathrm{s})+\mathrm{Fe}^{3+}(\mathrm{aq})\) (c) \(\mathrm{Fe}^{2+}(\mathrm{aq})+\mathrm{Cr}_{2} \mathrm{O}_{7}^{2-}(\mathrm{aq}) \rightarrow\) \(\mathrm{Fe}^{3+}(\mathrm{aq})+\mathrm{Cr}^{3+}(\mathrm{aq})\) (d) \(\mathrm{MnO}_{4}-(\mathrm{aq})+\mathrm{HNO}_{2}(\mathrm{aq}) \rightarrow\) \(\mathrm{Mn}^{2+}(\mathrm{aq})+\mathrm{NO}_{3}^{-}(\mathrm{aq})\)

Magnesium metal is oxidized, and silver ions are reduced in a voltaic cell using \(\mathrm{Mg}^{2+}(\mathrm{aq}, 1 \mathrm{M}) | \mathrm{Mg}\) and \(\mathrm{Ag}^{+}(\text {aq, } 1 \mathrm{M}) |\) Ag half-cells. (a) Label each part of the cell. (b) Write equations for the half-reactions occurring at the anode and the cathode, and write an equation for the net reaction in the cell. (c) Trace the movement of electrons in the exter. nal circuit. Assuming the salt bridge contains NaNO_, trace the movement of the Nat and \(\mathrm{NO}_{3}^{-}\) ions in the salt bridge that occurs when a voltaic cell produces current. Why is a salt bridge required in a cell?

In the electrolysis of a solution containing \(\mathrm{Ag}^{+}(\mathrm{aq}),\) metallic \(\mathrm{Ag}(\mathrm{s})\) deposits on the cathode Using a current of 1.12 A for 2.40 hours, what mass of silver forms?

A (a) Is it easier to reduce water in acid or base? To evaluate this, consider the half-reaction \(2 \mathrm{H}_{2} \mathrm{O}(\ell)+2 \mathrm{e}^{-} \rightarrow 2 \mathrm{OH}^{-}(\mathrm{aq})+\mathrm{H}_{2}(\mathrm{g})\) \(E^{*}=-0.83 \mathrm{V}\) (b) What is the reduction potential for water for solutions at \(\mathrm{pH}=7\) (neutral) and \(\mathrm{pH}=1\) (acid)? Comment on the value of \(E^{\circ}\) at \(\mathrm{pH}=1\)

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