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If the cell potential is proportional to work and the standard reduction potential for the hydrogen ion is zero, does this mean that the reduction of the hydrogen ion requires no work?

Short Answer

Expert verified
Although the standard reduction potential of the hydrogen ion is zero, it does not mean that the reduction of the hydrogen ion requires no work. The zero value is used as a reference point in electrochemistry. The work associated with its reduction depends on the particular redox reaction it participates in and the cell's overall potential. When paired with another half-reaction in an electrochemical cell, the overall cell potential may be positive or negative, leading to work being done by or on the cell.

Step by step solution

01

Cell potential, also called the electromotive force (EMF) or electrode potential, is a measure of the driving force of a chemical reaction in an electrochemical cell. In other words, it is the potential difference between the two electrodes participating in the redox reaction. Cell potential is proportional to the overall work done in a cell for converting chemical energy into electrical energy or vice versa. The relationship between cell potential and work can be expressed as: \[ W = nFE \] where, W is the work done (Joule), n is the number of moles of electrons transferred, F is the Faraday constant (96,485 C/mol), and E is the cell potential (Voltage). #Step 2: Reduction and standard reduction potentials#

Reduction is the process in which an element gains electrons, usually from another element or ion. The reduction potential is the ability of a species, such as an ion or a molecule, to undergo reduction, or accept electrons. The standard reduction potential (E°) represents the tendency of a chemical species to accept electrons under standard conditions (1 M concentration of reactants and products, 1 atm gas pressure, and 25°C temperature). The more positive the standard reduction potential, the more likely a species is to be reduced. #Step 3: Standard reduction potential of hydrogen ion#
02

By definition, the standard reduction potential of the hydrogen ion (H⁺) is set to be zero. The reduction half-reaction for hydrogen ions can be written as: \[ 2H⁺ + 2e⁻ \rightarrow H_2\] The zero value of the standard reduction potential is used as a reference point in electrochemistry. #Step 4: Does the reduction of hydrogen ion require no work?#

Although the standard reduction potential of the hydrogen ion is zero, it does not necessarily mean that the reduction of the hydrogen ion requires no work. Remember that the cell potential is proportional to work, but only when considering the overall redox reaction in a cell. The hydrogen ion's standard reduction potential of zero indicates that, on its own, the reduction of the hydrogen ion is neither particularly spontaneous nor non-spontaneous. However, when paired with another half-reaction (which would be the oxidation half-reaction occurring at the other electrode) in an electrochemical cell, the overall cell potential may be positive or negative, leading to work being done by or on the cell. So, although the hydrogen ion has a standard reduction potential of zero, the work associated with its reduction depends on the particular redox reaction it participates in and the cell's overall potential.

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Most popular questions from this chapter

Assign oxidation numbers to all the atoms in each of the following: a. \(\mathrm{HNO}_{3}\) b. \(\mathrm{CuCl}_{2}\) c. \(\mathrm{O}_{2}\) d. \(\mathrm{H}_{2} \mathrm{O}_{2}\) e. \(C_{6} H_{12} O_{6}\) f .\(\mathrm {Ag} \) g. \(\mathrm{PbSO}_{4}\) h. \(\mathrm{PbO}_{2}\) i. \(\quad \mathrm{Na}_{2} \mathrm{C}_{2} \mathrm{O}_{4}\) j . \(\mathrm{CO}_{2}\) k. \(\left(\mathrm{NH}_{4}\right)_{2} \mathrm{Ce}\left(\mathrm{SO}_{4}\right)_{3}\) 1\. \(\mathrm{Cr}_{2} \mathrm{O}_{3}\)

An electrochemical cell consists of a nickel metal electrode immersed in a solution with \(\left[\mathrm{Ni}^{2+}\right]=1.0 \mathrm{M}\) separated by a porous disk from an aluminum metal electrode. a. What is the potential of this cell at \(25^{\circ} \mathrm{C}\) if the aluminum electrode is placed in a solution in which \(\left[\mathrm{Al}^{3+}\right]=7.2 \times\) \(10^{-3} M ?\) b. When the aluminum electrode is placed in a certain solution in which \(\left[\mathrm{Al}^{3+}\right]\) is unknown, the measured cell potential at \(25^{\circ} \mathrm{C}\) is \(1.62 \mathrm{V}\). Calculate \(\left[\mathrm{Al}^{3+}\right]\) in the unknown solution. (Assume Al is oxidized.)

A galvanic cell is based on the following half-reactions: $$\begin{array}{ll} \mathrm{Cu}^{2+}(a q)+2 \mathrm{e}^{-} \longrightarrow \mathrm{Cu}(s) & \mathscr{E}^{\circ}=0.34 \mathrm{V} \\ \mathrm{V}^{2+}(a q)+2 \mathrm{e}^{-} \longrightarrow \mathrm{V}(s) & \mathscr{E}^{\circ}=-1.20 \mathrm{V} \end{array}$$ In this cell, the copper compartment contains a copper electrode and \(\left[\mathrm{Cu}^{2+}\right]=1.00 \mathrm{M},\) and the vanadium compartment contains a vanadium electrode and \(V^{2+}\) at an unknown concentration. The compartment containing the vanadium \((1.00 \mathrm{L}\) of solution) was titrated with \(0.0800 M \space \mathrm{H}_{2} \mathrm{EDTA}^{2-},\) resulting in the reaction $$\mathrm{H}_{2} \mathrm{EDTA}^{2-}(a q)+\mathrm{V}^{2+}(a q) \rightleftharpoons \mathrm{VEDTA}^{2-}(a q)+2 \mathrm{H}^{+}(a q) \space \mathrm{K=?}$$ The potential of the cell was monitored to determine the stoichiometric point for the process, which occurred at a volume of \(500.0 \mathrm{mL} \space \mathrm{H}_{2} \mathrm{EDTA}^{2-}\) solution added. At the stoichiometric point, \(\mathscr{E}_{\text {cell }}\) was observed to be \(1.98 \mathrm{V}\). The solution was buffered at a pH of \(10.00 .\) a. Calculate \(\mathscr{E}_{\text {cell }}\) before the titration was carried out. b. Calculate the value of the equilibrium constant, \(K,\) for the titration reaction. c. Calculate \(\mathscr{E}_{\text {cell }}\) at the halfway point in the titration.

A disproportionation reaction involves a substance that acts as both an oxidizing and a reducing agent, producing higher and lower oxidation states of the same element in the products. Which of the following disproportionation reactions are spontaneous under standard conditions? Calculate \(\Delta G^{\circ}\) and \(K\) at \(25^{\circ} \mathrm{C}\) for those reactions that are spontaneous under standard conditions. a. \(2 \mathrm{Cu}^{+}(a q) \rightarrow \mathrm{Cu}^{2+}(a q)+\mathrm{Cu}(s)\) b. \(3 \mathrm{Fe}^{2+}(a q) \rightarrow 2 \mathrm{Fe}^{3+}(a q)+\mathrm{Fe}(s)\) c. \(\mathrm{HClO}_{2}(a q) \rightarrow \mathrm{ClO}_{3}^{-}(a q)+\mathrm{HClO}(a q) \quad\) (unbalanced) Use the half-reactions: \(\mathrm{ClO}_{3}^{-}+3 \mathrm{H}^{+}+2 \mathrm{e}^{-} \longrightarrow \mathrm{HClO}_{2}+\mathrm{H}_{2} \mathrm{O} \quad \mathscr{E}^{\circ}=1.21 \mathrm{V}\) \(\mathrm{HClO}_{2}+2 \mathrm{H}^{+}+2 \mathrm{e}^{-} \longrightarrow \mathrm{HClO}+\mathrm{H}_{2} \mathrm{O} \quad \mathscr{E}^{\circ}=1.65 \mathrm{V}\)

Consider the galvanic cell based on the following halfreactions: $$ \begin{array}{ll} \mathrm{Zn}^{2+}+2 \mathrm{e}^{-} \longrightarrow \mathrm{Zn} & \mathscr{E}^{\circ}=-0.76 \mathrm{V} \\ \mathrm{Fe}^{2+}+2 \mathrm{e}^{-} \longrightarrow \mathrm{Fe} & \mathscr{E}^{\circ}=-0.44 \mathrm{V} \end{array} $$ a. Determine the overall cell reaction and calculate \(\mathscr{E}_{\text {cell. }}\) b. Calculate \(\Delta G^{\circ}\) and \(K\) for the cell reaction at \(25^{\circ} \mathrm{C}\). c. Calculate \(\mathscr{C}_{\text {cell }}\) at \(25^{\circ} \mathrm{C}\) when \(\left[\mathrm{Zn}^{2+}\right]=0.10 \mathrm{M}\) and \(\left[\mathrm{Fe}^{2+}\right]=1.0 \times 10^{-5} \mathrm{M}\)

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