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Chapter 22: Problem 104

Hydrazine has been employed as a reducing agent for metals. Using standard reduction potentials, predict whether the following metals can be reduced to the metallic state by hydrazine under standard conditions in acidic solution: (a) \(\mathrm{Fe}^{2+}\), (b) \(\mathrm{Sn}^{2+}\), (c) \(\mathrm{Cu}^{2+}\), (d) \(\mathrm{Ag}^{+}\), (e) \(\mathrm{Cr}^{3+}\), (f) \(\mathrm{Co}^{3+}\).

Short Answer

Expert verified
Based on the comparison of standard reduction potentials, hydrazine can only reduce \(\mathrm{Cr}^{3+}\) to its metallic state under standard conditions in an acidic solution. The other metal ions \(\mathrm{Fe}^{2+}\), \(\mathrm{Sn}^{2+}\), \(\mathrm{Cu}^{2+}\), \(\mathrm{Ag}^{+}\), and \(\mathrm{Co}^{3+}\) cannot be reduced by hydrazine under these conditions.

Step by step solution

01

Find the reduction potential of hydrazine

In acidic solution, the reduction half-reaction for hydrazine (\(\mathrm{N}_{2}\mathrm{H}_{4}\)) is given by: \[\mathrm{N}_{2}\mathrm{H}_{4}(l) + 4\mathrm{H}^{+}(aq) + 4e^{-} \rightarrow 2\mathrm{NH}_{4}^{+}(aq)\] The standard reduction potential (\(E^{\circ}\)) of hydrazine in this reaction is -0.76 V.
02

Compare reduction potentials with given metal ions

Now let's compare the standard reduction potentials for each metal ion and determine whether they can be reduced by hydrazine: (a) \(\mathrm{Fe}^{2+}(aq) + 2e^{-} \rightarrow \mathrm{Fe}(s)\), \(E^{\circ} = -0.44\) V. Since the reduction potential of \(\mathrm{Fe}^{2+}\) is greater than hydrazine, hydrazine cannot reduce it. (b) \(\mathrm{Sn}^{2+}(aq) + 2e^{-} \rightarrow \mathrm{Sn}(s)\), \(E^{\circ} = -0.14\) V. The reduction potential of \(\mathrm{Sn}^{2+}\) is greater than hydrazine, so hydrazine cannot reduce it. (c) \(\mathrm{Cu}^{2+}(aq) + 2e^{-} \rightarrow \mathrm{Cu}(s)\), \(E^{\circ} = +0.34\) V. The reduction potential of \(\mathrm{Cu}^{2+}\) is greater than hydrazine, so hydrazine cannot reduce it. (d) \(\mathrm{Ag}^{+}(aq) + e^{-} \rightarrow \mathrm{Ag}(s)\), \(E^{\circ} = +0.80\) V. The reduction potential of \(\mathrm{Ag}^{+}\) is greater than hydrazine, so hydrazine cannot reduce it. (e) \(\mathrm{Cr}^{3+}(aq) + 3e^{-} \rightarrow \mathrm{Cr}(s)\), \(E^{\circ} = -0.74\) V. The reduction potential of \(\mathrm{Cr}^{3+}\) is less than hydrazine, so hydrazine can reduce it. (f) \(\mathrm{Co}^{3+}(aq) + 3e^{-} \rightarrow \mathrm{Co}(s)\), \(E^{\circ} = -0.28\) V. The reduction potential of \(\mathrm{Co}^{3+}\) is greater than hydrazine, so hydrazine cannot reduce it.
03

Compile the conclusions

By comparing the standard reduction potentials, we can conclude the following: (a) \(\mathrm{Fe}^{2+}\) cannot be reduced by hydrazine. (b) \(\mathrm{Sn}^{2+}\) cannot be reduced by hydrazine. (c) \(\mathrm{Cu}^{2+}\) cannot be reduced by hydrazine. (d) \(\mathrm{Ag}^{+}\) cannot be reduced by hydrazine. (e) \(\mathrm{Cr}^{3+}\) can be reduced by hydrazine. (f) \(\mathrm{Co}^{3+}\) cannot be reduced by hydrazine.

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Key Concepts

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

Oxidation-Reduction
Oxidation-reduction, often referred to as redox, involves chemical reactions where electrons are transferred between substances. This process consists of two key parts: oxidation, where a substance loses electrons, and reduction, where a substance gains electrons.
In oxidation-reduction reactions, each element has a specific oxidation state. When an element undergoes oxidation, it will increase its oxidation state by losing electrons. Conversely, reduction will lower the oxidation state as the element gains electrons.
  • When hydrazine (\(\mathrm{N}_{2}\mathrm{H}_{4}\)) interacts with metal ions, it donates its electrons to reduce them.
  • The efficiency of a reducing agent like hydrazine is decided by its standard reduction potential. A negative potential means it can donate electrons to metals with more positive potentials.
  • Knowing which metals can be reduced by hydrazine is determined by comparing their respective standard reduction potentials.
Understanding these concepts is crucial for predicting the outcome of redox reactions, especially in determining which metals can be reduced using specific reducing agents like hydrazine.
Metal Ion Reduction
The process of metal ion reduction is a fundamental concept in chemistry, often involved in obtaining pure metals from their ionic form. It occurs when metal ions gain electrons and transform into solid metals.
Metal ion reduction is significantly influenced by the reduction potential. Metals with a more positive standard reduction potential are more easily reduced.
  • Reduction potentials are measured in volts (V) which helps predict their likelihood to gain electrons.
  • Metals with a higher reduction potential compared to the reducing agent are less likely to be reduced by that agent.
  • In our case, hydrazine鈥檚 potential at -0.76 V has to be lower than that of the metal to successfully reduce it.
The comparison of potentials helps determine the feasibility of reducing each specific metal, allowing predictions on whether hydrazine can transform these metal ions into their metallic state.
Electrochemistry
Electrochemistry is the branch of chemistry that examines chemical reactions which take place at the interface of an electrode, typically involving the movement of electrons.
Central to electrochemistry are concepts like electrodes, electrochemical cells, and importantly, the standard electrode, or reduction potential. This potential indicates how likely a substance is to be reduced.
  • Standard reduction potentials provide groundwork for predicting reactions in an electrochemical cell.
  • Unlike typical reactions, electrochemical cells harness the energy from redox reactions to do electrical work.
  • To determine whether hydrazine can reduce a metal ion, one must evaluate the cell potential. If it鈥檚 positive, reduction of the metal ion is possible in the electrochemical environment.
Thus, understanding these principles is vital not only in theoretical chemistry but also in practical applications, such as designing batteries and electroplating. The comparison of standard potentials ensures successful predictions in electrochemical applications.

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Most popular questions from this chapter

Ultrapure germanium, like silicon, is used in semiconductors. Germanium of "ordinary" purity is prepared by the high-temperature reduction of \(\mathrm{GeO}_{2}\) with carbon. The Ge is converted to \(\mathrm{GeCl}_{4}\) by treatment with \(\mathrm{Cl}_{2}\) and then purified by distillation; \(\mathrm{GeCl}_{4}\) is then hydrolyzed in water to \(\mathrm{GeO}_{2}\) and reduced to the elemental form with \(\mathrm{H}_{2}\). The element is then zone refined. Write a balanced chemical equation for each of the chemical transformations in the course of forming ultrapure Ge from \(\mathrm{GeO}_{2}\).

a) Determine the number of sodium ions in the chemical formula of albite, \(\mathrm{Na}_{x} \mathrm{AlSi}_{3} \mathrm{O}_{g}\). (b) Determine the number of hydroxide ions in the chemical formula of tremolite, \(\mathrm{Ca}_{2} \mathrm{Mg}_{5}\left(\mathrm{Si}_{4} \mathrm{O}_{11}\right)_{2}(\mathrm{OH})_{\mathrm{x}}\) -

Select the member of group \(4 \mathrm{~A}\) that best fits each description: (a) has the lowest first ionization energy, (b) is found in oxidation states ranging from \(-4\) to \(+4\), (c) is most abundant in Earth's crust.

You have two glass bottles, one containing oxygen and one filled with nitrogen. How could you determine which one is which? [Sections \(22.5\) and \(22.7\) ]

The standard heats of formation of \(\mathrm{H}_{2} \mathrm{O}(g), \mathrm{H}_{2} \mathrm{~S}(g), \mathrm{H}_{2} \mathrm{Se}(g)\), and \(\mathrm{H}_{2} \mathrm{Te}(g)\) are \(-241.8,-20.17,+29.7\), and \(+99.6 \mathrm{~kJ} /\) mol, respectively. The enthalpies necessary to convert the elements in their standard states to one mole of gaseous atoms are \(248,277,227\), and \(197 \mathrm{~kJ} / \mathrm{mol}\) of atoms for \(\mathrm{O}, \mathrm{S}, \mathrm{Se}\), and Te, respectively. The enthalpy for dissociation of \(\mathrm{H}_{2}\) is \(436 \mathrm{~kJ} / \mathrm{mol}\). Calculate the average \(\mathrm{H}-\mathrm{O}, \mathrm{H}-\mathrm{S}, \mathrm{H}-\mathrm{Se}\), and \(\mathrm{H}\) - Te bond enthalpies, and comment on their trend.
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