For each of the following redox reactions, (i) write the half-reactions; (ii) write a balanced equation for the whole reaction, (iii) determine in which direction the reaction will proceed spontaneously under standard-state conditions: (a) \(\mathrm{H}_{2}(g)+\mathrm{Ni}^{2+}(a q) \longrightarrow \mathrm{H}^{+}(a q)+\mathrm{Ni}(s)\) (b) \(\mathrm{MnO}_{4}^{-}(a q)+\mathrm{Cl}^{-}(a q) \longrightarrow\) \(\mathrm{Mn}^{2+}(a q)+\mathrm{Cl}_{2}(g)\) (in acid solution) (c) \(\mathrm{Cr}(s)+\mathrm{Zn}^{2+}(a q) \longrightarrow \mathrm{Cr}^{3+}(a q)+\mathrm{Zn}(s)\)

Short Answer

Expert verified
The balanced equations are: (a) \(H2 + Ni^{2+} \rightarrow 2H^{+} + Ni\), (b) \(2MnO4^{-} + 16H^{+} + 10Cl^{-} \rightarrow 2Mn^{2+} + 5Cl2 + 8H2O\), and (c) \(2Cr + 3Zn^{2+} \rightarrow 2Cr^{3+} + 3Zn\). Predicted directions for each reaction under standard conditions respectively are: forward, reverse, forward.

Step by step solution

01

Half-Reaction formulation and Balancing

For reaction (a), the reduction half-reaction is: \(Ni^{2+}\) \(+\) 2e- \(\rightarrow\) \(Ni\), and the oxidation half-reaction is: \(H2\) \(\rightarrow\) 2\(H^{+}\) + 2e-. For reaction (b), the reduction half-reaction is: \(MnO4^{-}\) \(+\) 8H+ \(+\) e- \(\rightarrow\) \(Mn^{2+}\) + 4\(H2O\), and the oxidation half-reaction is: \(Cl^{-}\) \(\rightarrow\) \(Cl2\) + 2e-. For reaction (c), the reduction half-reaction is: \(Zn^{2+}\) \(+\) 2e- \(\rightarrow\) \(Zn\), and the oxidation half-reaction is: \(Cr\) \(\rightarrow\) \(Cr^{3+}\) + 3e-.
02

Writing Balanced Equations

The balanced equations for the reactions are calculated by insuring that the number of electrons in the oxidation and reduction half-reactions are equal. Thus, for reaction (a), the balanced equation is: \(H2 + Ni^{2+} \rightarrow 2H^{+} + Ni\). For reaction (b), the balanced equation is: \(2MnO4^{-} + 16H^{+} + 10Cl^{-} \rightarrow 2Mn^{2+} + 5Cl2 + 8H2O\). For reaction (c), the balanced equation is: \(2Cr + 3Zn^{2+} \rightarrow 2Cr^{3+} + 3Zn\).
03

Predicting the direction of the reaction

The direction of a redox reaction under standard conditions can be predicted from the relative strengths of the oxidizing and reducing agents, or from the standard reduction potentials of the half-reactions. For simplification, consider following reaction spontaneity rules: A reaction is spontaneous if reduction potential of reactant which is getting reduced is more than reduction potential of other reactant which is getting oxidised. As per these rules, reaction (a) and (c) will proceed spontaneously as \(Ni^{2+}\) and \(Zn^{2+}\) ions have more reduction potential than \(H2\) and \(Cr\) respectively while reaction (b) will not proceed spontaneously as \(Cl^{-}\) ions have less reduction potential than \(MnO4^{-}\) ions.

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Key Concepts

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

Half-Reaction Formulation
Understanding the half-reaction formulation is essential in grasping redox processes. A redox reaction involves both reduction (gain of electrons) and oxidation (loss of electrons). To break it down, we separate the overall reaction into two half-reactions: one for reduction and one for oxidation. This separation allows us to balance the reaction in complex environments, such as acidic or basic solutions.

For example, in reaction (a) from the exercise, we split the original reaction into two parts: the reduction of nickel ions to metallic nickel and the oxidation of hydrogen gas into hydrogen ions. Each half-reaction is balanced individually by ensuring the number of atoms of each element and the charges are equal on both sides. When we combine these half-reactions, we get a clearer insight into the electron transfer occurring in the overall redox process.
Balancing Redox Equations
After we have outlined the half-reactions, the next step is to balance the overall redox equation. Balancing redox equations is a deliberate process that entails equating the number of electrons lost in the oxidation half-reaction to the number of electrons gained in the reduction half-reaction. This ensures that the electrons - which are the essence of redox reactions - are conserved.

In the exercise solution step 2, electron balance was achieved by multiplying the half-reactions by appropriate factors so that the electrons cancel out when the half-reactions are added together. This is a critical puzzle—it is only when the electrons balance that we can say the equation itself is balanced. Besides the electron count, we also need to check that atoms are balanced, which may require adding water molecules, hydrogen ions, or hydroxide ions, depending on whether the reaction occurs in an acidic or basic solution.
Standard Reduction Potentials
The standard reduction potentials (SRP) are a series of values that provide us a way to predict the behavior of redox reactions. These potentials, usually referenced to the standard hydrogen electrode, define the tendency of a species to gain electrons and thus be reduced.

In simple terms, the more positive the SRP value, the greater the species' affinity for electrons. We use these values to determine which substance in the reaction will act as the oxidizing agent (will be reduced) and which will act as the reducing agent (will be oxidized). When comparing two half-reactions, as we see in the exercise, the one with the higher SRP will predictably undergo reduction and the other oxidation. These standard potentials are not only useful for predicting the outcome of reactions but also for understanding the driving force behind the reaction's direction.
Reaction Spontaneity
How do we know if a redox reaction will proceed on its own, without the input of external energy? This is where the idea of spontaneity comes in, which can be deduced from the standard reduction potentials we discussed earlier. A reaction is considered spontaneous if it leads to a decrease in free energy.

Using the rules mentioned in step 3 of the solution, we determine the spontaneity of a redox reaction by comparing the standard reduction potentials: if the SRP of the reactant being reduced is more positive than that of the reactant being oxidized, the reaction is likely spontaneous under standard conditions. This principle helps us predict whether a redox reaction can occur as written or if we need to apply energy to drive it. In the exercise, this is applied to conclude that reactions (a) and (c) are spontaneous, while reaction (b) is not, under standard-state conditions.

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Most popular questions from this chapter

An acidified solution was electrolyzed using copper electrodes. A constant current of 1.18 A caused the anode to lose \(0.584 \mathrm{~g}\) after \(1.52 \times 10^{3} \mathrm{~s}\). (a) What is the gas produced at the cathode and what is its volume at STP? (b) Given that the charge of an electron is \(1.6022 \times 10^{-19} \mathrm{C}\), calculate Avogadro's number. Assume that copper is oxidized to \(\mathrm{Cu}^{2+}\) ions.

Discuss the advantages and disadvantages of fuel cells over conventional power plants in producing electricity.

Balance the following redox equations by the halfreaction method: (a) \(\mathrm{H}_{2} \mathrm{O}_{2}+\mathrm{Fe}^{2+} \longrightarrow \mathrm{Fe}^{3+}+\mathrm{H}_{2} \mathrm{O}\) (in acidic solution) (b) \(\mathrm{Cu}+\mathrm{HNO}_{3} \longrightarrow \mathrm{Cu}^{2+}+\mathrm{NO}+\mathrm{H}_{2} \mathrm{O}\) (in acidic solution) (c) \(\mathrm{CN}^{-}+\mathrm{MnO}_{4}^{-} \longrightarrow \mathrm{CNO}^{-}+\mathrm{MnO}_{2}\) (in basic solution) (d) \(\mathrm{Br}_{2} \longrightarrow \mathrm{BrO}_{3}^{-}+\mathrm{Br}^{-}\) (in basic solution) (e) \(\mathrm{S}_{2} \mathrm{O}_{3}^{2-}+\mathrm{I}_{2} \longrightarrow \mathrm{I}^{-}+\mathrm{S}_{4} \mathrm{O}_{6}^{2-}\) (in acidic solution)

The concentration of a hydrogen peroxide solution can be conveniently determined by titration against a standardized potassium permanganate solution in an acidic medium according to the following unbalanced equation: $$ \mathrm{MnO}_{4}^{-}+\mathrm{H}_{2} \mathrm{O}_{2} \longrightarrow \mathrm{O}_{2}+\mathrm{Mn}^{2+} $$ (a) Balance the above equation. (b) If \(36.44 \mathrm{~mL}\) of a \(0.01652 \mathrm{M} \mathrm{KMnO}_{4}\) solution are required to completely oxidize \(25.00 \mathrm{~mL}\) of a \(\mathrm{H}_{2} \mathrm{O}_{2}\) solution, calculate the molarity of the \(\mathrm{H}_{2} \mathrm{O}_{2}\) solution.

Explain why chlorine gas can be prepared by electrolyzing an aqueous solution of \(\mathrm{NaCl}\) but fluorine gas cannot be prepared by electrolyzing an aqueous solution of NaF.

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