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Which species in each pair is a better reducing agent under standard-state conditions? (a) Na or \(\mathrm{Li},\) (b) \(\mathrm{H}_{2}\) or \(\mathrm{I}_{2},\) (c) \(\mathrm{Fe}^{2+}\) or \(\mathrm{Ag},\) (d) \(\mathrm{Br}^{-}\) or \(\mathrm{Co}^{2+}\)

Short Answer

Expert verified
Li, H鈧, Fe, and Co虏鈦 are the better reducing agents in each respective pair.

Step by step solution

01

Understand the Concept of Reducing Agents

A reducing agent is a substance that loses electrons in a redox chemical reaction and gets oxidized. The effectiveness of a reducing agent is inversely related to its reduction potential. A lower or more negative standard electrode potential indicates a stronger reducing agent.
02

Identify Standard Reduction Potentials

Consult a table of standard reduction potentials to find the values for each species: (a) Na and Li, (b) H鈧 and I鈧, (c) Fe虏鈦 and Ag, and (d) Br鈦 and Co虏鈦.
03

Compare Reduction Potentials for (a) Na vs Li

Standard reduction potential of Na is -2.71 V and that of Li is -3.04 V. Since Li has a more negative potential, Li is a better reducing agent than Na.
04

Compare Reduction Potentials for (b) H鈧 vs I鈧

H鈧 has a reduction potential of 0.00 V and I鈧 has a potential of +0.54 V. Since H鈧 has a more negative (or lower) reduction potential, H鈧 is the better reducing agent.
05

Compare Reduction Potentials for (c) Fe虏鈦 vs Ag

Fe from Fe虏鈦/Fe has a reduction potential of -0.44 V, while Ag from Ag鈦/Ag has a reduction potential of +0.80 V. Fe虏鈦 has a more negative reduction potential; thus, Fe is a better reducing agent.
06

Compare Reduction Potentials for (d) Br鈦 vs Co虏鈦

Br鈦/Br鈧 has a reduction potential of +1.07 V, while Co虏鈦/Co has a reduction potential of -0.28 V. Co虏鈦 with a more negative reduction potential is a better reducing agent than Br鈦.

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Key Concepts

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

Standard Electrode Potential
Standard electrode potential, often denoted as E掳, is a measurement that allows us to compare how easily a chemical species gives up electrons during a redox reaction.
Electrode potentials are measured under standard conditions: a solute concentration of 1 M, pressure of 1 bar, and a temperature of 25掳C. These conditions ensure consistency and reliability in results. Knowing the standard electrode potential of a species helps in determining its tendency to be oxidized or reduced.
Lower or more negative electrode potentials indicate a higher tendency to donate electrons, making the substance a better reducing agent. Conversely, higher or more positive potentials suggest a tendency to gain electrons.
Redox Reactions
Redox reactions, or reduction-oxidation reactions, are chemical processes in which electrons are transferred between substances. These reactions are crucial in numerous biological and industrial processes.
Each redox reaction involves two halves: oxidation and reduction. In one reactant, oxidation occurs where it loses electrons; in the other, reduction takes place where it gains electrons.
The essence of a redox reaction is in balancing electron transfer between the oxidized and reduced species. By examining redox reactions, we can identify which substances are acting as oxidizing or reducing agents.
Oxidation
Oxidation is the process where a substance loses electrons during a chemical reaction. This loss is often accompanied by an increase in the oxidation state.
The term "oxidation" originally referred to reactions involving oxygen; however, it now encompasses any reaction where electrons are lost.
To easily remember, oxidation can be thought of as "loss of electrons" or using the acronym OIL (Oxidation Is Loss). This concept helps in identifying substances that diminish in electron density, affecting their chemical properties.
Reduction Potentials
Reduction potential is a measure of the tendency of a chemical species to gain electrons and undergo reduction. A species with a higher reduction potential is more likely to gain electrons.
Reduction potentials are often thought of in the context of galvanic cells, where the flow of electrons from one half-cell to another occurs.
When comparing reduction potentials, the more negative the value, the greater the ability of the species to donate electrons and act as a reducing agent. Conversely, more positive reduction potentials indicate a greater tendency to act as an oxidizing agent.

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Most popular questions from this chapter

The \(E^{\circ}\) value of one cell reaction is positive and that of another cell reaction is negative. Which cell reaction will proceed toward the formation of more products at equilibrium?

Oxalic acid \(\left(\mathrm{H}_{2} \mathrm{C}_{2} \mathrm{O}_{4}\right)\) is present in many plants and vegetables. (a) Balance the following equation in acid solution: $$\mathrm{MnO}_{4}^{-}+\mathrm{C}_{2} \mathrm{O}_{4}^{2-} \longrightarrow \mathrm{Mn}^{2+}+\mathrm{CO}_{2}$$ (b) If a \(1.00-\mathrm{g}\) sample of \(\mathrm{H}_{2} \mathrm{C}_{2} \mathrm{O}_{4}\) requires \(24.0 \mathrm{~mL}\) of \(0.0100 \mathrm{M} \mathrm{KMnO}_{4}\) solution to reach the equivalence point, what is the percent by mass of \(\mathrm{H}_{2} \mathrm{C}_{2} \mathrm{O}_{4}\) in the sample?

The \(\mathrm{SO}_{2}\) present in air is mainly responsible for the phenomenon of acid rain. The concentration of \(\mathrm{SO}_{2}\) can be determined by titrating against a standard permanganate solution as follows: $$5 \mathrm{SO}_{2}+2 \mathrm{MnO}_{4}^{-}+2 \mathrm{H}_{2} \mathrm{O} \longrightarrow \underset{5 \mathrm{SO}_{4}^{2-}}{\longrightarrow}+2\mathrm{Mn}^{2+}+4 \mathrm{H}^{+}$$ Calculate the number of grams of \(\mathrm{SO}_{2}\) in a sample of air if \(7.37 \mathrm{~mL}\) of \(0.00800 \mathrm{M} \mathrm{KMnO}_{4}\) solution are required for the titration.

Consider a galvanic cell consisting of a magnesium electrode in contact with \(1.0 \mathrm{M} \mathrm{Mg}\left(\mathrm{NO}_{3}\right)_{2}\) and a cadmium electrode in contact with \(1.0 \mathrm{M} \mathrm{Cd}\left(\mathrm{NO}_{3}\right)_{2},\) Calculate \(E^{\circ}\) for the cell, and draw a diagram showing the cathode, anode, and direction of electron flow.

Given the following standard reduction potentials, calculate the ion-product, \(K_{w},\) for water at \(25^{\circ} \mathrm{C}\) $$\begin{aligned}2 \mathrm{H}^{+}(a q)+2 e^{-} \longrightarrow \mathrm{H}_{2}(g) & E^{\circ} &=0.00 \mathrm{~V} \\\2 \mathrm{H}_{2} \mathrm{O}(l)+2 e^{-} \longrightarrow \mathrm{H}_{2}(g)+2 \mathrm{OH}^{-}(&a q) & \\\& & E^{\circ}=-0.83 \mathrm{~V}\end{aligned}$$

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