(a) Why is it impossible to measure the standard reduction potential of a single half-reaction? (b) Describe how the standard reduction potential of a half-reaction can be determined.

Short Answer

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(a) It is impossible to measure the standard reduction potential of a single half-reaction because the reduction half-reaction and the oxidation half-reaction occur simultaneously in a redox reaction, and the measured voltage or potential is a combined potential of both reactions. The standard reduction potential of a half-reaction cannot be measured in isolation, as it always depends on the other half-reaction in the overall redox reaction. (b) To determine the standard reduction potential of a half-reaction, a reference electrode with a known standard reduction potential, such as the Standard Hydrogen Electrode (SHE), is used. An electrochemical cell is set up with the reference electrode on one side and the half-reaction of interest on the other side. The cell potential (E_cell) is measured under standard conditions, and it is equal to the standard reduction potential of the half-reaction of interest when measured against the SHE. By measuring the cell potential, the standard reduction potential of the half-reaction can be determined.

Step by step solution

01

Understanding standard reduction potential

The standard reduction potential is a measure of the tendency of a chemical species to accept electrons and become reduced in a redox reaction (reduction-oxidation reaction). In a redox reaction, there are two half-reactions: a reduction half-reaction and an oxidation half-reaction. The standard reduction potential is expressed in volts (V) and is measured at standard conditions (25°C, 1 atm pressure, and 1 M concentration for all species).
02

Why the standard reduction potential of a single half-reaction cannot be measured

It is impossible to measure the standard reduction potential of a single half-reaction because in a redox reaction, the reduction half-reaction and the oxidation half-reaction occur simultaneously. When one species accepts electrons and undergoes reduction, another species must lose electrons and undergo oxidation. This means that the voltage or potential we measure in an electrochemical cell is a combined potential of both the reduction and oxidation reactions. The standard reduction potential of a half-reaction cannot be measured in isolation since it always depends on the other half-reaction in the overall redox reaction. In other words, there is no absolute value of the standard reduction potential for an individual species, as it is a relative quantity that is measured in reference to another reaction.
03

Determining the standard reduction potential of a half-reaction

To determine the standard reduction potential of a half-reaction, we use a reference electrode with a known standard reduction potential, most commonly the Standard Hydrogen Electrode (SHE). The SHE has an assigned potential of 0 V. To measure the standard reduction potential of a half-reaction, an electrochemical cell is set up with the reference electrode on one side and the half-reaction of interest on the other side. The redox potential of the cell, called the cell potential (E_cell), is measured under standard condition and is equal to the difference between the standard reduction potential of the half-reaction of interest (E) and the standard reduction potential of the reference electrode (E_ref), which is 0 V for SHE: E_cell = E - E_ref Since E_ref is 0 V for the SHE, we get: E_cell = E Thus, the cell potential is equal to the standard reduction potential of the half-reaction of interest when measured against the SHE. By measuring the cell potential, we can determine the standard reduction potential of the half-reaction.

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Most popular questions from this chapter

Given the following reduction half-reactions: $$ \mathrm{Fe}^{3+}(a q)+\mathrm{e}^{-} \longrightarrow \mathrm{Fe}^{2+}(a q) $$ \(E_{\mathrm{red}}^{\circ}=+0.77 \mathrm{~V}\) $$ \mathrm{S}_{2} \mathrm{O}_{6}{ }^{2-}(a q)+4 \mathrm{H}^{+}(a q)+2 \mathrm{e}^{-} \longrightarrow 2 \mathrm{H}_{2} \mathrm{SO}_{3}(a q) $$ \(E_{\mathrm{red}}^{\circ}=+0.60 \mathrm{~V}\) $$ \begin{array}{r} \mathrm{N}_{2} \mathrm{O}(g)+2 \mathrm{H}^{+}(a q)+2 \mathrm{e}^{-} \longrightarrow \mathrm{N}_{2}(g)+\mathrm{H}_{2} \mathrm{O}(l) \\ E_{\mathrm{red}}^{\circ}=-1.77 \mathrm{~V} \\ \mathrm{VO}_{2}^{+}(a q)+2 \mathrm{H}^{+}(a q)+\mathrm{e}^{-} \longrightarrow \mathrm{VO}^{2+}+\mathrm{H}_{2} \mathrm{O}(l) \\ E_{\mathrm{red}}^{\circ}=+1.00 \mathrm{~V} \end{array} $$ (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 by \(\mathrm{VO}_{2}{ }^{+}(a q) .(\mathbf{b})\) 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) What is the difference between a battery and a fuel cell? (b) Can the "fuel" of a fuel cell be a solid? Explain.

Using standard reduction potentials (Appendix E), calculate the standard emf for each of the following reactions: $$ \begin{array}{l} \text { (a) } \mathrm{Cl}_{2}(g)+2 \mathrm{I}^{-}(a q) \longrightarrow 2 \mathrm{Cl}^{-}(a q)+\mathrm{I}_{2}(s) \\ \text { (b) } \mathrm{Ni}(s)+2 \mathrm{Ce}^{4+}(a q) \longrightarrow \mathrm{Ni}^{2+}(a q)+2 \mathrm{Ce}^{3+}(a q) \\ \text { (c) } \mathrm{Fe}(s)+2 \mathrm{Fe}^{3+}(a q) \longrightarrow 3 \mathrm{Fe}^{2+}(a q) \\ \text { (d) } 2 \mathrm{NO}_{3}^{-}(a q)+8 \mathrm{H}^{+}(a q)+3 \mathrm{Cu}(s) \longrightarrow \\ 2 \mathrm{NO}(g)+4 \mathrm{H}_{2} \mathrm{O}(l)+3 \mathrm{Cu}^{2+}(a q) \end{array} $$

This oxidation-reduction reaction in acidic solution is spontaneous: $$ \begin{array}{r} 5 \mathrm{Fe}^{2+}(a q)+\mathrm{MnO}_{4}^{-}(a q)+8 \mathrm{H}^{+}(a q) \longrightarrow \\ 5 \mathrm{Fe}^{3+}(a q)+\mathrm{Mn}^{2+}(a q)+4 \mathrm{H}_{2} \mathrm{O}(l) \end{array} $$ A solution containing \(\mathrm{KMnO}_{4}\) and \(\mathrm{H}_{2} \mathrm{SO}_{4}\) is poured into one beaker, and a solution of \(\mathrm{FeSO}_{4}\) is poured into another. A salt bridge is used to join the beakers. A platinum foil is placed in each solution, and a wire that passes through a voltmeter connects the two solutions. (a) Sketch the cell, indicating the anode and the cathode, the direction of electron movement through the external circuit, and the direction of ion migrations through the solutions. (b) Sketch the process that occurs at the atomic level at the surface of the anode. (c) Calculate the emf of the cell under standard conditions. (d) Calculate the emf of the cell at \(298 \mathrm{~K}\) when the concentrations are the fol- $$ \text { lowing: } \mathrm{pH}=0.0,\left[\mathrm{Fe}^{2+}\right]=0.10 \mathrm{M},\left[\mathrm{MnO}_{4}^{-}\right]=1.50 \mathrm{M} $$ \(\left[\mathrm{Fe}^{3+}\right]=2.5 \times 10^{-4} \mathrm{M},\left[\mathrm{Mn}^{2+}\right]=0.001 \mathrm{M}\)

(a) Magnesium metal is used as a sacrificial anode to protect underground pipes from corrosion. Why is the magnesium referred to as a "sacrificial anode"? (b) Looking in Appendix \(\mathrm{E}\); suggest what metal the underground pipes could be made from in order for magnesium to be successful as a sacrificial anode.

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