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Chapter 20: Problem 7

Consider the half reaction \(\mathrm{Ag}^{+}(a q)+\mathrm{e}^{-} \longrightarrow \mathrm{Ag}(s)\) (a) Which of the lines in the following diagram indicates how the reduction potential varies with the concentration of \(\mathrm{Ag}^{+} ?\) (b) What is the value of \(E_{\text {red }}\) when \(\log \left[\mathrm{Ag}^{+}\right]=0 ?[\) Section \(20.6]\)

Short Answer

Expert verified
The reduction potential for the given reaction can be obtained using the Nernst equation: \[E = E^0 - \frac{0.05916}{n}\log_{10} Q\]. When \(\log [\mathrm{Ag}^{+}] = 0\), the reduction potential is \(E = 0.80 V\).

Step by step solution

01

Write down the Nernst equation for reduction potential

The Nernst equation gives the relationship between the reduction potential (E_red) and the concentration of species involved in a redox reaction. The Nernst equation is given by: \[E = E^0 - \frac{0.05916}{n}\log_{10} Q\] where \(E\) is the reduction potential, \(E^0\) is the standard reduction potential (constant), \(n\) is the number of moles of electrons in the half reaction, and \(Q\) is the reaction quotient.
02

Calculate the reduction potential based on the Nernst equation

For the given reaction, the standard reduction potential of silver is \( E^0 = 0.80 V \) and since it involves one electron, \(n = 1\). The reaction quotient \(Q\) is given by: \[Q = [\mathrm{Ag} ^+]\] Substitute the values in the Nernst equation: \[E = 0.80 - \frac{0.05916}{1}\log_{10} [\mathrm{Ag} ^{+} ]\]
03

Find the reduction potential when the logarithm of the Ag+ concentration is 0

As given, \(\log [\mathrm{Ag} ^{+}] = 0\). Now substitute this value in the above equation: \[E = 0.80 - \frac{0.05916}{1} \times 0\]
04

Simplify the equation to find the value of reduction potential

Simplify the equation: \[E = 0.80 - 0\] \[E = 0.80 V\] Summary: For the given reaction, the reduction potential (\(E\)) can be obtained by using the Nernst equation. The value of the reduction potential when the logarithm of the silver ion concentration is 0 is \(E = 0.80 V\).

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Key Concepts

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

Reduction Potential
Understanding the concept of reduction potential is key to grasping how electrochemical cells work. In chemistry, the reduction potential, often expressed as E, indicates the tendency of a chemical species to gain electrons and thereby be reduced. Imagine that each half-cell in an electrochemical cell is competing in a tug-of-war, wanting to snatch the electron; the reduction potential tells you how strong each participant is.

It's a bit like how a larger gravitational force indicates a stronger pull toward the Earth, a higher reduction potential means a stronger pull for electrons.

This concept is central to predicting which way a redox reaction will proceed; the species with higher reduction potential will be reduced, while the other gets oxidized. Basically, the substance with the 'will to gain electrons' wins. This concept is applied in a wide range of practical scenarios, from batteries to biosensors.
Standard Reduction Potential
Diving a bit deeper, we encounter the standard reduction potential, symbolized as 贰掳. This is the reduction potential of a half-reaction under standard conditions, which means all reactants and products are at 1 M concentration, the gas pressure is 1 atm, and the temperature is 298 K (25掳C).

The standard reduction potential is a way to set a common playing field for all half-reactions. It's like giving all runners in a race the same type of shoes to see who truly runs the fastest.

In the educational example, the standard reduction potential of the silver ion reduction reaction is 0.80 V. This value serves as a reference point against which we can measure changes in condition, such as changes in concentration using the Nernst equation. Think of 贰掳 as the baseline potential recorded in the 'chemical scorebook' that tells us the inherent strength of each half-reaction in the electron tug-of-war.
Reaction Quotient
Lastly, let's unwrap the concept of the reaction quotient, or Q. You might know about the equilibrium constant K, which describes the ratio of concentrations of products to reactants at equilibrium. The reaction quotient is its movie trailer鈥攊t gives us a sneak peek of how close or far the reaction is from reaching its climax, the equilibrium state.

When we talk about Q in the context of the Nernst equation, we're focusing on the ratio of the concentrations of the ions involved in the half-reaction, whether or not the reaction is at equilibrium. If Q is equal to the equilibrium constant K, the system is at equilibrium, and the cell's voltage will be zero.

However, if Q is not equal to K, it tells us if the reaction will proceed forward or in reverse to reach equilibrium. This affects the cell's voltage and hence the reduction potential as explained with the Nernst equation in the exercise. By understanding Q, students can predict the direction of the reaction and calculate the current voltage of the cell.

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Most popular questions from this chapter

Given the following reduction half-reactions: \(\mathrm{Fe}^{3+}(a q)+\mathrm{e}^{-} \longrightarrow \mathrm{Fe}^{2+}(a q)\) \(E_{\mathrm{red}}^{\circ}=+0.77 \mathrm{~V}\) \(\mathrm{~S}_{2} \mathrm{O}_{6}^{2-}(a q)+4 \mathrm{H}^{+}(a q)+2 \mathrm{e}^{-} \longrightarrow 2 \mathrm{H}_{2} \mathrm{SO}_{3}(a q)\) \(E_{\mathrm{red}}^{\circ}=+0.60 \mathrm{~V}\) \(\mathrm{~N}_{2} \mathrm{O}(g)+2 \mathrm{H}^{+}(a q)+2 \mathrm{e}^{-} \longrightarrow \mathrm{N}_{2}(g)+\mathrm{H}_{2} \mathrm{O}(l)\) \(E_{\mathrm{red}}^{\circ}=-1.77 \mathrm{~V}\) \(\mathrm{VO}_{2}^{+}(a q)+2 \mathrm{H}^{+}(a q)+\mathrm{e}^{-} \longrightarrow \mathrm{VO}^{2+}(a q)+\mathrm{H}_{2} \mathrm{O}(l)\) \(E_{\mathrm{red}}^{\circ}=+1.00 \mathrm{~V}\) (a) Write balanced chemical equations for the oxidation of \(\mathrm{Fe}^{2+}(a q)\) by \(\mathrm{S}_{2} \mathrm{O}_{6}{ }^{2-}(a q)\), by \(\mathrm{N}_{2} \mathrm{O}(a q)\), and \(\mathrm{by} \mathrm{VO}_{2}{ }^{+}(a q)\). (b) Calculate \(\Delta G^{\circ}\) for each reaction at \(298 \mathrm{~K}\). (c) Calculate the equilibrium constant \(K\) for each reaction at \(298 \mathrm{~K}\).

(a) Write the anode and cathode reactions that cause the corrosion of iron metal to aqueous iron(II). (b) Write the balanced half-reactions involved in the air oxidation of \(\mathrm{Fe}^{2+}(a q)\) to \(\mathrm{Fe}_{2} \mathrm{O}_{3} \cdot 3 \mathrm{H}_{2} \mathrm{O}\)

If you were going to apply a small potential to a steel ship resting in the water as a means of inhibiting corrosion, would you apply a negative or a positive charge? Explain.

If the equilibrium constant for a two-electron redox reaction at \(298 \mathrm{~K}\) is \(1.5 \times 10^{-4}\), calculate the corresponding \(\Delta G^{\circ}\) and \(E_{\text {cell }}^{\circ}\).

(a) What is meant by the term reduction? (b) On which side of a reduction half-reaction do the electrons appear? (c) What is meant by the term reductant? (d) What is meant by the term reducing agent?
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