The standard reduction potential of \(\mathrm{Eu}^{2+}(a q)\) is \(-0.43 \mathrm{V}\) . Using Appendix E, which of the following substances is capable of reducing Eu' \((a q)\) to \(\mathrm{Eu}^{2+}(a q)\) under standard conditions: Al, Co, \(\mathrm{H}_{2} \mathrm{O}_{2}, \mathrm{N}_{2} \mathrm{H}_{5}^{+}, \mathrm{H}_{2} \mathrm{C}_{2} \mathrm{O}_{4} ?\)

Short Answer

Expert verified
The substances that can reduce Eu'(aq) to Eu²⁺(aq) under standard conditions are Hydrogen Peroxide (H₂O₂) and Oxalic Acid (H₂C₂O₄).

Step by step solution

01

Find the standard reduction potentials of the given substances

To know which substance can reduce Eu'(aq) to Eu²⁺(aq), we need to find the standard reduction potentials of all given substances: Al, Co, H₂O₂, N₂H₅⁺, and H₂C₂O₄. These values can be found in Appendix E. Let's list them down: - Al: Given as Al³⁺(aq) + 3e⁻ → Al, E° = -1.66 V - Co: Given as Co²⁺(aq) + 2e⁻ → Co, E° = -0.28 V - H₂O₂: Given as H₂O₂(aq) + 2H⁺ + 2e⁻ → 2H₂O(l), E° = 1.77 V - N₂H₅⁺: Given as N₂H₅⁺(aq) + 3H₂O(l) → N₂(g) + 8H⁺ + 8e⁻, E° = -0.60 V - H₂C₂O₄: Given as H₂C₂O₄(aq)+ 2H⁺ + 2e⁻ → 2CO₂(g) + 2H₂O(l), E° = 0.575 V We also have the standard reduction potential of Eu²⁺(aq) provided in the question: Eu²⁺(aq) + 2e⁻ → Eu, E° = -0.43 V. Now that we have the values, let's compare them to find the substance capable of reducing Eu'(aq) to Eu²⁺(aq).
02

Compare standard reduction potentials and determine the substances that can reduce Eu' to Eu²⁺

Let's compare the standard reduction potentials of the given substances with that of Eu²⁺/Eu': 1. Al/Al³⁺: E° = -1.66 V < -0.43 V (Eu²⁺/Eu') 2. Co/Co²⁺: E° = -0.28 V < -0.43 V (Eu²⁺/Eu') 3. H₂O₂/2H₂O: E° = 1.77 V > -0.43 V (Eu²⁺/Eu') 4. N₂H₅⁺/N₂: E° = -0.60 V < -0.43 V (Eu²⁺/Eu') 5. H₂C₂O₄/2CO₂ + 2H₂O: E° = 0.575 V > -0.43 V (Eu²⁺/Eu') From the comparisons above, we can see that only H₂O₂ and H₂C₂O₄ have higher standard reduction potentials than Eu²⁺/Eu', which means they are capable of reducing Eu'(aq) to Eu²⁺(aq) under standard conditions. The substances that can reduce Eu'(aq) to Eu²⁺(aq) under standard conditions are Hydrogen Peroxide (H₂O₂) and Oxalic Acid (H₂C₂O₄).

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Key Concepts

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

Electrochemistry
Electrochemistry is a branch of chemistry that deals with the chemical changes caused by the movement of electrons from one substance to another, a field at the heart of batteries, corrosion, and many industrial processes. At its core, it's the study of reduction and oxidation (redox) reactions that involve electron transfer.

In a redox reaction, one substance gains electrons, undergoing reduction, while another loses electrons, which is called oxidation. These reactions are often coupled, as the electrons lost by one substance are gained by another. In the context of the exercise, hydrogens peroxide (H₂O₂) and oxalic acid (H₂C₂O₄) can donate electrons to reduce europium ions (Eu') to europium metal (Eu), showing the practical applications of electrochemistry in determining reactivity and feasibility of such reactions.

Electrochemical cells, including galvanic or voltaic cells, are practical devices that harness these redox reactions to produce electrical energy. Understanding the standard reduction potential of different substances allows for the prediction of the direction of electron flow and the ability of chemicals to act as oxidizing or reducing agents in these cells.
Reduction and Oxidation Reactions
Reduction and oxidation reactions, often referred to as redox reactions, are processes where electrons are transferred between chemical species. The mnemonic 'OIL RIG' - oxidation is loss, reduction is gain - can help students remember the essence of these processes.

In the context of the exercise, we focus on reduction: the gain of electrons. A substance that gains electrons is said to be reduced, and its oxidation state decreases. Conversely, the oxidation state of a substance that loses electrons increases, and the process is known as oxidation. For example, when Europium ion (Eu') is reduced to form Europium metal (Eu²⁺), it gains electrons.

To predict whether a substance will be oxidized or reduced in a reaction, we use standard reduction potentials. These are measured under standard conditions: solute concentrations of 1 M, a pressure of 1 atm for gases, and a temperature of 25°C (298 K). Substances with higher reduction potentials will likely gain electrons and be reduced, whereas those with lower potentials tend to lose electrons and be oxidized.
Standard Electrode Potentials
Standard electrode potentials, also known as standard reduction potentials, are a measure of the tendency of a chemical species to acquire electrons and be reduced, set under standard conditions (1 M concentration, 1 atm pressure, and 298 K temperature). The values are measured against the standard hydrogen electrode, which has a potential of 0 V.

In the exercise, different substances were compared based on their standard reduction potentials to determine if they could reduce Eu'. Substances with a more positive standard reduction potential than Eu²⁺/Eu', like H₂O₂ and H₂C₂O₄, have a greater tendency to gain electrons and can act as reducing agents for Eu'. This comparison is crucial for understanding the flow of electrons in electrochemical cells and for predicting the outcome of redox reactions.

The use of standard reduction potentials simplifies the analysis of electrochemical reactions by providing a comparative scale. However, it's important to remember that actual conditions, such as non-standard concentrations and temperatures, can shift these potentials, affecting the chemical behavior of the species involved.

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Most popular questions from this chapter

Hydrogen gas has the potential for use as a clean fuel in reaction with oxygen. The relevant reaction is $$ 2 \mathrm{H}_{2}(g)+\mathrm{O}_{2}(g) \longrightarrow 2 \mathrm{H}_{2} \mathrm{O}(l) $$ Consider two possible ways of utilizing this reaction as an electrical energy source: (i) Hydrogen and oxygen gases are combusted and used to drive a generator, much as coal is currently used in the electric power industry; (ii) hydrogen and oxygen gases are used to generate electricity directly by using fuel cells that operate at \(85^{\circ} \mathrm{C}\) . (a) Use data in Appendix C to calculate \(\Delta H^{\circ}\) and \(\Delta S^{\circ}\) for the reaction. We will assume that these values do not change appreciably with temperature. (b) Based on the values from part (a), what trend would you expect for the magnitude of \(\Delta G\) for the reaction as the temperature increases? (c) What is the significance of the change in the magnitude of \(\Delta G\) with temperature with respect to the utility of hydrogen as a fuel? (d) Based on the analysis here, would it be more efficient to use the combustion method or the fuel-cell method to generate electrical energy from hydrogen?

A mixture of copper and gold metals that is subjected to electrorefining contains tellurium as an impurity. The standard reduction potential between tellurium and its lowest common oxidation state, \(\mathrm{Te}^{4+},\) is $$ \mathrm{Te}^{4+}(a q)+4 \mathrm{e}^{-} \longrightarrow \mathrm{Te}(s) \quad E_{\mathrm{red}}^{\circ}=0.57 \mathrm{V} $$ Given this information, describe the probable fate of tellurium impurities during electrorefining. Do the impurities fall to the bottom of the refining bath, unchanged, as copper is oxidized, or do they go into solution as ions? If they go into solution, do they plate out on the cathode?

Copper corrodes to cuprous oxide, \(\mathrm{Cu}_{2} \mathrm{O},\) or cupric oxide, \(\mathrm{CuO},\) depending on environmental conditions. (a) What is the oxidation state of copper in cuprous oxide? (b) What is the oxidation state of copper in cupric oxide? (c) Copper peroxide is another oxidation product of elemental copper. Suggest a formula for copper peroxide based on its name. (d) Copper(III) oxide is another unusual oxidation product of elemental copper. Suggest a chemical formula for copper(II) oxide.

If the equilibrium constant for a one-electron redox reaction at 298 \(\mathrm{K}\) is \(8.7 \times 10^{4}\) , calculate the corresponding \(\Delta G^{\circ}\) and \(E^{\circ} .\)

(a) Write the reactions for the discharge and charge of a nickel-cadmium (nicad) rechargeable battery. (b) Given the following reduction potentials, calculate the standard emf of the cell: $$ \begin{array}{r}{\operatorname{Cd}(\mathrm{OH})_{2}(s)+2 \mathrm{e}^{-} \longrightarrow \mathrm{Cd}(s)+2 \mathrm{OH}^{-}(a q)} \\\ {E_{\mathrm{red}}^{\circ}=-0.76 \mathrm{V}}\end{array} $$ $$ \begin{array}{r}{\mathrm{NiO}(\mathrm{OH})(s)+\mathrm{H}_{2} \mathrm{O}(l)+\mathrm{e}^{-} \longrightarrow \mathrm{Ni}(\mathrm{OH})_{2}(s)+\mathrm{OH}^{-}(a q)} \\\ {E_{\mathrm{red}}^{\circ}=+0.49 \mathrm{V}}\end{array} $$ (c) A typical nicad voltaic cell generates an emf of \(+1.30 \mathrm{V}\) . Why is there a difference between this value and the one you calculated in part (b)? (d) Calculate the equilibrium constant for the overall nicad reaction based on this typical emf value.

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