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Chapter 10: Problem 132

In a nickel-cadmium battery, the relevant redox reaction is: \(2 \mathrm{NiO}(\mathrm{OH})+\mathrm{Cd}+2 \mathrm{H}_{2} \mathrm{O} \rightarrow\) \(2 \mathrm{Ni}(\mathrm{OH})_{2}+\mathrm{Cd}(\mathrm{OH})_{2}\) Does this agree with the EMF series? What are the oxidation states of nickel before and after the reaction?

Short Answer

Expert verified
The given redox reaction agrees with the EMF series, as the total cell potential \(E_{cell}^{\circ} = 2.21 V\) is positive, indicating a spontaneous reaction. The oxidation states of nickel change from +1 in \(NiO(OH)\) to +3 in \(Ni(OH)_2\) during the reaction.

Step by step solution

01

Identify the redox reaction

The given redox reaction is: \[2 \mathrm{NiO}(\mathrm{OH})+\mathrm{Cd}+2 \mathrm{H}_{2} \mathrm{O} \rightarrow 2 \mathrm{Ni}(\mathrm{OH})_{2}+\mathrm{Cd}(\mathrm{OH})_{2}\]
02

Identify the half-reactions

To identify the half-reactions, let's look at the chemical changes involved in the reaction. The substances changing their oxidation states are: 1. Nickel - \(NiO(OH)\) turns into \(Ni(OH)_2\). 2. Cadmium - \(Cd\) turns into \(Cd(OH)_2\). Thus, the two half-reactions are: Oxidation: \(Cd \rightarrow Cd(OH)_2\) Reduction: \(2 NiO(OH) + 2 H_2O \rightarrow 2 Ni(OH)_2\)
03

Calculate standard electrode potential

To check if this reaction agrees with the EMF series, we need to determine the standard electrode potentials for the half-reactions. According to the EMF series: \(E^{\circ}_{Ni^{3+}/Ni^{2+}} = 1.81 V\) with the reduction half-reaction as: \[NiO(OH) + H_{2}O + e^{-} \rightarrow Ni(OH)_{2}\] \(E^{\circ}_{Cd^{2+}/Cd} = -0.40 V\) with the oxidation half-reaction as: \[Cd \rightarrow Cd^{2+} + 2e^{-}\] Now, let's find the total cell potential: \(E_{cell}^{\circ} = E^{\circ}_{Reduction} - E^{\circ}_{Oxidation}\) \(E_{cell}^{\circ} = 1.81 V - (-0.40 V) = 2.21 V\) Since the total cell potential is positive, the reaction is spontaneous and agrees with the EMF series.
04

Determine the oxidation states of nickel

We need to find the oxidation state of nickel before and after the reaction: Before the reaction, nickel is in \(NiO(OH)\), where oxygen has an oxidation state of -2 and hydrogen has an oxidation state of +1. Let \(x\) be the oxidation state of nickel, then: \[x + (-2) + 1 = 0\] Solving for \(x\), we get \(x = 1\), indicating that the initial oxidation state of nickel is +1. After the reaction, nickel is in \(Ni(OH)_2\), where oxygen has an oxidation state of -2 and hydrogen has an oxidation state of +1. Using the same approach to find the oxidation state of nickel: \[x + 2(-2) + 2(1) = 0\] Solving for \(x\), we get \(x = 3\), indicating the final oxidation state of nickel is +3.

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Most popular questions from this chapter

A piece of magnesium can be attached to the iron hull of a boat to prevent it from rusting. (a) Why is the magnesium called a sacrificial metal? (b) How does the Mg keep the iron hull from rusting? (c) Which can be thought of as the positive cathode, the \(\mathrm{Mg}\) or the steel hull? Explain. (d) Putting a block of \(\mathrm{Mg}\) on a steel boat hull does not create a battery. Why not?

Which of the following are electron-transfer reactions? (a) \(2 \mathrm{CrO}_{4}^{2-}+2 \mathrm{H}^{+} \rightarrow \mathrm{Cr}_{2} \mathrm{O}_{7}^{2-}+\mathrm{H}_{2} \mathrm{O}\) (b) \(\mathrm{Fe}+\mathrm{NO}_{3}^{-}+4 \mathrm{H}^{+} \rightarrow \mathrm{Fe}^{3+}+\mathrm{NO}+2 \mathrm{H}_{2} \mathrm{O}\) (c) \(2 \mathrm{C}_{2} \mathrm{H}_{6}+7 \mathrm{O}_{2} \rightarrow 4 \mathrm{CO}_{2}+6 \mathrm{H}_{2} \mathrm{O}\) (d) \(2 \mathrm{AgBr} \rightarrow 2 \mathrm{Ag}+\mathrm{Br}_{\dot{2}}\)

Assign an oxidation state to each carbon in: (a) \(\mathrm{H}_{3} \mathrm{CCH}_{3}\) (b) \(\mathrm{H}_{2} \mathrm{CCH}_{2}\) (c) HCCH

Which of the following are electron-transfer reactions? For those that are, indicate which reactant is the reducing agent and which reactant is the oxidizing agent. (a) \(3 \mathrm{H}_{2} \mathrm{SO}_{3}+2 \mathrm{HNO}_{3} \rightarrow 3 \mathrm{H}_{2} \mathrm{SO}_{4}+2 \mathrm{NO}+\mathrm{H}_{2} \mathrm{O}\) (b) \(\mathrm{Mg}+2 \mathrm{H}_{2} \mathrm{O} \rightarrow \mathrm{Mg}(\mathrm{OH})_{2}+\mathrm{H}_{2}\) (c) \(\mathrm{SO}_{3}{ }^{2-}+2 \mathrm{H}^{+} \rightarrow \mathrm{SO}_{2}+\mathrm{H}_{2} \mathrm{O}\) (d) \(\mathrm{PbO}+\mathrm{CO} \rightarrow \mathrm{Pb}+\mathrm{CO}_{2}\)

Consider a battery made from lead, \(\mathrm{Pb}\), and copper, \(\mathrm{Cu}\), along with \(\mathrm{Pb}^{2+}\) ions and \(\mathrm{Cu}^{2+}\) ions. In such a battery, what is the oxidizing agent, and what is the reducing agent? (Hint: Start by consulting the EMF series, and determine which metal oxidizes and which metal ion gets reduced.)
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