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Chapter 18: Problem 144

Given the following two standard reduction potentials, $$\mathrm{M}^{3+}+3 \mathrm{e}^{-} \longrightarrow \mathrm{M} \quad \mathscr{E}^{\circ}=-0.10 \mathrm{~V}$$ $$\mathrm{M}^{2+}+2 \mathrm{e}^{-} \longrightarrow \mathrm{M} \quad \mathscr{E}^{\circ}=-0.50 \mathrm{~V}$$ solve for the standard reduction potential of the half- reaction$$\mathrm{M}^{3+}+\mathrm{e}^{-} \longrightarrow \mathrm{M}^{2+}$$

Short Answer

Expert verified
The standard reduction potential of the half-reaction M³⁺ + e⁻ → M²⁺ is \(E^0 = -0.60\) V.

Step by step solution

01

1. Write down the given half-reactions and their standard reduction potentials

The initial half-reactions and their standard reduction potentials are given by: \(1^{st}\): M³⁺ + 3e⁻ → M, E₀ = -0.10 V \(2^{nd}\): M²⁺ + 2e⁻ → M, E₀ = -0.50 V
02

2. Reverse the desired half-reaction

Since the desired half-reaction is: M³⁺ + e⁻ → M²⁺ We need to reverse the second half-reaction in order to relate M³⁺ and M²⁺: \(2^{nd}\): M → M²⁺ + 2e⁻
03

3. Update the standard reduction potential of the reversed half-reaction

When we reverse a half-reaction, we need to change the sign of the standard reduction potential. So, the new standard reduction potential for the reversed second half-reaction is: E₀₍ᵣᵉᵛ₂₎ = +0.50 V
04

4. Combine the two modified half-reactions

Now we have: \(1^{st}\): M³⁺ + 3e⁻ → M, E₀ = -0.10 V \(2^{nd}\): M → M²⁺ + 2e⁻, E₀₍ᵣᵉᵛ₂₎ = +0.50 V To obtain the desired half-reaction, we subtract the reversed second half-reaction from the first half-reaction: M³⁺ + 3e⁻ - (M → M²⁺ + 2e⁻)
05

5. Simplify the combined half-reaction

After combining and simplifying, the desired half-reaction is: M³⁺ + e⁻ → M²⁺
06

6. Determine the standard reduction potential of the desired half-reaction

To find the standard reduction potential of the desired half-reaction, we subtract the standard reduction potential of the reversed second half-reaction from the standard reduction potential of the first half-reaction: E₀₍ᵈᵉˢᵢʳᵉᵈ₎ = E₀₍¹₎ - E₀₍ᵣᵉᵛ₂₎ = -0.10 V - 0.50 V = -0.60 V Thus, the standard reduction potential of the desired half-reaction is: M³⁺ + e⁻ → M²⁺, E₀ = -0.60 V

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Key Concepts

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

Electrochemistry
Electrochemistry lies at the heart of many important scientific processes, from the batteries powering our everyday devices to the synthesis of chemical compounds. It is the branch of chemistry that deals with the relationship between electrical energy and chemical change. In electrochemistry, we study redox (reduction-oxidation) reactions, where electrons are transferred between species, altering their oxidation states.

Standard reduction potentials, represented as E°, are essential in this field as they provide us the quantitative measure of the tendency of a chemical species to gain electrons and thus be reduced. Measured under standard conditions (1 M concentration, 1 atm pressure, and 25°C), these potentials are crucial for predicting the direction of electron flow in an electrochemical cell and understanding the thermodynamics of redox reactions.
Half-Reactions
Half-reactions are the two parts that make up a complete redox reaction: one for the reduction and one for the oxidation. They are often used in the context of galvanic cells, where two different metals are connected by an electrolyte solution that allows the flow of ions.

By breaking down the overall reaction into half-reactions, we gain clarity on how many electrons are involved in each process and the individual standard reduction potentials. For instance, a metal ion like M³⁺ can be reduced to M. The standard reduction potential associated with this process helps in calculating the overall potential of a galvanic cell. Moreover, if we do not have the standard reduction potential for a particular half-reaction, it can often be deduced by manipulating other known half-reactions and their potentials, as was exemplified in the given exercise.
Galvanic Cells
Galvanic cells, or voltaic cells, are the fundamental components of batteries, transforming chemical energy into electrical energy through spontaneous redox reactions. Each galvanic cell consists of two half-cells linked by both an external circuit and a salt bridge or porous disk that allows ions to move while keeping the two solutions separate.

Each half-cell contains an electrode and an electrolyte. The electrode where oxidation occurs (loss of electrons) is called the anode, and the electrode where reduction takes place (gain of electrons) is called the cathode. The potential difference between the anode and cathode, which drives the electric current, is determined by their respective standard reduction potentials. Understanding how to calculate and use standard reduction potentials, therefore, allows students to predict which direction electrons will flow within a galvanic cell and the cell's ability to do electrical work.

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Most popular questions from this chapter

a. In the electrolysis of an aqueous solution of \(\mathrm{Na}_{2} \mathrm{SO}_{4}\), what reactions occur at the anode and the cathode (assuming standard conditions)? b. When water containing a small amount \((-0.01 M)\) of sodium sulfate is electrolyzed, measurement of the volume of gases generated consistently gives a result that the volume ratio of hydrogen to oxygen is not quite \(2: 1 .\) To what do you attribute this discrepancy? Predict whether the measured ratio is greater than or less than \(2: 1 .\) (Hint: Consider overvoltage.)

Consider the following galvanic cell at \(25^{\circ} \mathrm{C}\) : $$\mathrm{Pt}\left|\mathrm{Cr}^{2+}(0.30 M), \mathrm{Cr}^{3+}(2.0 M)\right|\left|\mathrm{Co}^{2+}(0.20 M)\right| \mathrm{Co}$$ The overall reaction and equilibrium constant value are $$2 \mathrm{Cr}^{2+}(a q)+\mathrm{Co}^{2+}(a q) \longrightarrow{2 \mathrm{Cr}^{3+}(a q)+\mathrm{Co}(s)} \quad K=2.79 \times 10^{7}$$ Calculate the cell potential, \(\mathscr{E}\), for this galvanic cell and \(\Delta G\) for the cell reaction at these conditions.

The following standard reduction potentials have been determined for the aqueous chemistry of indium: $$\begin{array}{ll}\mathrm{In}^{3+}(a q)+2 \mathrm{e}^{-} \longrightarrow \operatorname{In}^{+}(a q) & \mathscr{E}^{\circ}=-0.444 \mathrm{~V} \\ \mathrm{In}^{+}(a q)+\mathrm{e}^{-} \longrightarrow \operatorname{In}(s) & \mathscr{E}^{\circ}=-0.126 \mathrm{~V}\end{array}$$ a. What is the equilibrium constant for the disproportionation reaction, where a species is both oxidized and reduced, shown below? $$3 \operatorname{In}^{+}(a q) \longrightarrow 2 \operatorname{In}(s)+\operatorname{In}^{3+}(a q)$$ b. What is \(\Delta G_{\mathrm{f}}^{\circ}\) for \(\mathrm{In}^{+}(a q)\) if \(\Delta G_{\mathrm{f}}^{\circ}=-97.9 \mathrm{~kJ} / \mathrm{mol}\) for \(\mathrm{In}^{3+}(a q)\) ?

The measurement of pH using a glass electrode obeys the Nernst equation. The typical response of a pH meter at \(25.00^{\circ} \mathrm{C}\) is given by the equation $$\mathscr{C}_{\text {meas }}=\mathscr{E}_{\text {ref }}+0.05916 \mathrm{pH}$$ where \(\mathscr{E}_{\text {ref }}\) contains the potential of the reference electrode and all other potentials that arise in the cell that are not related to the hydrogen ion concentration. Assume that \(\mathscr{E}_{\text {ref }}=0.250 \mathrm{~V}\) and that \(\mathscr{C}_{\text {tme\pi }}=0.480 \mathrm{~V}\) a. What is the uncertainty in the values of \(\mathrm{pH}\) and \(\left[\mathrm{H}^{+}\right]\) if the uncertainty in the measured potential is \(\pm 1 \mathrm{mV}(\pm 0.001 \mathrm{~V})\) ? b. To what precision must the potential be measured for the uncertainty in \(\mathrm{pH}\) to be \(\pm 0.02 \mathrm{pH}\) unit?

The blood alcohol \(\left(\mathrm{C}_{2} \mathrm{H}_{5} \mathrm{OH}\right)\) level can be determined by titrating a sample of blood plasma with an acidic potassium dichromate solution, resulting in the production of \(\mathrm{Cr}^{3+}(a q)\) and carbon dioxide. The reaction can be monitored because the dichromate ion \(\left(\mathrm{Cr}_{2} \mathrm{O}_{7}^{2-}\right)\) is orange in solution, and the \(\mathrm{Cr}^{3+}\) ion is green. The unbalanced redox equation is $$\mathrm{Cr}_{2} \mathrm{O}_{7}^{2-}(a q)+\mathrm{C}_{2} \mathrm{H}_{5} \mathrm{OH}(a q) \longrightarrow \mathrm{Cr}^{3+}(a q)+\mathrm{CO}_{2}(g)$$ If \(31.05 \mathrm{~mL}\) of \(0.0600 M\) potassium dichromate solution is required to titrate \(30.0 \mathrm{~g}\) blood plasma, determine the mass percent of alcohol in the blood.
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