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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 Eu虏鈦(aq) under standard conditions are Hydrogen Peroxide (H鈧侽鈧) and Oxalic Acid (H鈧侰鈧侽鈧).

Step by step solution

01

Find the standard reduction potentials of the given substances

To know which substance can reduce Eu'(aq) to Eu虏鈦(aq), we need to find the standard reduction potentials of all given substances: Al, Co, H鈧侽鈧, N鈧侶鈧呪伜, and H鈧侰鈧侽鈧. These values can be found in Appendix E. Let's list them down: - Al: Given as Al鲁鈦(aq) + 3e鈦 鈫 Al, E掳 = -1.66 V - Co: Given as Co虏鈦(aq) + 2e鈦 鈫 Co, E掳 = -0.28 V - H鈧侽鈧: Given as H鈧侽鈧(aq) + 2H鈦 + 2e鈦 鈫 2H鈧侽(l), E掳 = 1.77 V - N鈧侶鈧呪伜: Given as N鈧侶鈧呪伜(aq) + 3H鈧侽(l) 鈫 N鈧(g) + 8H鈦 + 8e鈦, E掳 = -0.60 V - H鈧侰鈧侽鈧: Given as H鈧侰鈧侽鈧(aq)+ 2H鈦 + 2e鈦 鈫 2CO鈧(g) + 2H鈧侽(l), E掳 = 0.575 V We also have the standard reduction potential of Eu虏鈦(aq) provided in the question: Eu虏鈦(aq) + 2e鈦 鈫 Eu, E掳 = -0.43 V. Now that we have the values, let's compare them to find the substance capable of reducing Eu'(aq) to Eu虏鈦(aq).
02

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

Let's compare the standard reduction potentials of the given substances with that of Eu虏鈦/Eu': 1. Al/Al鲁鈦: E掳 = -1.66 V < -0.43 V (Eu虏鈦/Eu') 2. Co/Co虏鈦: E掳 = -0.28 V < -0.43 V (Eu虏鈦/Eu') 3. H鈧侽鈧/2H鈧侽: E掳 = 1.77 V > -0.43 V (Eu虏鈦/Eu') 4. N鈧侶鈧呪伜/N鈧: E掳 = -0.60 V < -0.43 V (Eu虏鈦/Eu') 5. H鈧侰鈧侽鈧/2CO鈧 + 2H鈧侽: E掳 = 0.575 V > -0.43 V (Eu虏鈦/Eu') From the comparisons above, we can see that only H鈧侽鈧 and H鈧侰鈧侽鈧 have higher standard reduction potentials than Eu虏鈦/Eu', which means they are capable of reducing Eu'(aq) to Eu虏鈦(aq) under standard conditions. The substances that can reduce Eu'(aq) to Eu虏鈦(aq) under standard conditions are Hydrogen Peroxide (H鈧侽鈧) and Oxalic Acid (H鈧侰鈧侽鈧).

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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 (H鈧侽鈧) and oxalic acid (H鈧侰鈧侽鈧) 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 (Eu虏鈦), 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掳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 Eu虏鈦/Eu', like H鈧侽鈧 and H鈧侰鈧侽鈧, 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

A voltaic cell that uses the reaction $$ \operatorname{PdCl}_{4}^{2-}(a q)+\mathrm{Cd}(s) \longrightarrow \mathrm{Pd}(s)+4 \mathrm{Cl}^{-}(a q)+\mathrm{Cd}^{2+}(a q) $$ has a measured standard cell potential of \(+1.03 \mathrm{V}\) . (a) Write the two half-cell reactions. (b) By using data from Appendix E, determine \(E_{\text { red }}^{\circ}\) for the reaction involving Pd. (c) Sketch the voltaic cell, label the anode and cathode, and indicate the direction of electron flow.

A cell has a standard cell potential of \(+0.177 \mathrm{V}\) at 298 \(\mathrm{K}\) . What is the value of the equilibrium constant for the reaction ( a ) if \(n=1 ?(\mathbf{b})\) if \(n=2 ?(\mathbf{c})\) if \(n=3 ?\)

Gold exists in two common positive oxidation states, \(+1\) and \(+3 .\) The standard reduction potentials for these oxidation states are $$ \begin{array}{ll}{\mathrm{Au}^{+}(a q)+\mathrm{e}^{-}} \quad {\longrightarrow \mathrm{Au}(s) \quad E_{\mathrm{red}}^{\circ}=+1.69 \mathrm{V}} \\\ {\mathrm{Au}^{3+}(a q)+3 \mathrm{e}^{-} \longrightarrow \mathrm{Au}(s)} \quad {E_{\mathrm{red}}^{\circ}=+1.50 \mathrm{V}}\end{array} $$ (a) Can you use these data to explain why gold does not tarnish in the air? ( b) Suggest several substances that should be strong enough oxidizing agents to oxidize gold metal. (c) Miners obtain gold by soaking gold-containing ores in an aqueous solution of sodium cyanide. A very soluble complex ion of gold forms in the aqueous solution because of the redox reaction $$ \begin{array}{rl}{4 \mathrm{Au}(s)+8 \mathrm{NaCN}(a q)+2} & {\mathrm{H}_{2} \mathrm{O}(l)+\mathrm{O}_{2}(g)} \\ {\longrightarrow} & {4 \mathrm{Na}\left[\mathrm{Au}(\mathrm{CN})_{2}\right](a q)+4 \mathrm{NaOH}(a q)}\end{array} $$ What is being oxidized, and what is being reduced in this reaction? (d) Gold miners then react the basic aqueous product solution from part (c) with Zn dust to get gold metal. Write a balanced redox reaction for this process. What is being oxidized, and what is being reduced?

The electrodes in a silver oxide battery are silver oxide \(\left(\mathrm{Ag}_{2} \mathrm{O}\right)\) and zinc. (a) Which electrode acts as the anode? (b) Which battery do you think has an energy density most similar to the silver oxide battery: a Li-ion battery, a nickel-cadmium battery, or a lead-acid battery? [ Section 20.7]

For each of the following balanced oxidation-reduction reactions, (i) identify the oxidation numbers for all the elements in the reactants and products and (ii) state the total number of electrons transferred in each reaction. $$ \begin{array}{l}{\text { (a) } \mathrm{I}_{2} \mathrm{O}_{5}(s)+5 \mathrm{CO}(g) \longrightarrow \mathrm{I}_{2}(s)+5 \mathrm{CO}_{2}(g)} \\\ {\text { (b) } 2 \mathrm{Hg}^{2+}(a q)+\mathrm{N}_{2} \mathrm{H}_{4}(a q) \longrightarrow 2 \mathrm{Hg}(l)+\mathrm{N}_{2}(g)+4 \mathrm{H}^{+}(a q)} \\\ {\text { (c) } 3 \mathrm{H}_{2} \mathrm{S}(a q)+2 \mathrm{H}^{+}(a q)+2 \mathrm{NO}_{3}^{-}(a q) \longrightarrow 3 \mathrm{S}(s)+} \\\\{\quad\quad 2 \mathrm{NO}(g)+4 \mathrm{H}_{2} \mathrm{O}(l)}\end{array} $$

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