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Chapter 13: Problem 54

\( \mathrm{~A} 1.0 \mathrm{M} \mathrm{KBr}(\mathrm{aq})\) solution was electrolyzed by using inert electrodes. Write (a) the cathode reaction; (b) the anode reaction. (c) With no overpotential or passivity at the electrodes, what is the minimum potential that must be supplied to the cell for the onset of electrolysis?

Short Answer

Expert verified
The cathode reaction is: 2 H₂O(l) + 2 e⁻ → H₂(g) + 2 OH⁻(aq). The anode reaction is: 2 Br⁻(aq) → Br₂(l) + 2 e⁻. The minimum potential required to initiate electrolysis is 1.90 V.

Step by step solution

01

Identifying the Cathode Reaction

At the cathode, reduction occurs. Since KBr is in aqueous solution, water can also undergo reduction. The possible reductions are K⁺ gaining an electron to become K(s), which is unlikely due to its high reduction potential, and H₂O gaining electrons to form H₂(g) and OH⁻(aq). The reaction with a higher likelihood based on standard electrode potentials is H₂O reduction, therefore the cathode reaction is: 2 H₂O(l) + 2 e⁻ → H₂(g) + 2 OH⁻(aq).
02

Determining the Anode Reaction

At the anode, oxidation occurs. The possible reactions are Br⁻ losing an electron to become Br₂ or H₂O losing electrons to form O₂. Since Br⁻ is present in a higher concentration than water and has a lower oxidation potential than water, Br⁻ is more likely to be oxidized. Therefore, the anode reaction is: 2 Br⁻(aq) → Br₂(l) + 2 e⁻.
03

Calculating the Minimum Potential for Electrolysis

To find the minimum potential, we sum the standard reduction potentials of the cathode and anode reactions. For the reduction of water at the cathode, the potential is approximately −0.83 V (E° H₂/H₂O), and the oxidation of bromide at the anode has a potential of about +1.07 V (E° Br₂/Br⁻). The minimum potential is the difference between the anode and cathode potentials: Minimum potential = E°(anode) − E°(cathode) = 1.07 − (-0.83) = 1.90 V. This is the minimum potential without overpotential or passivity.

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Key Concepts

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

Cathode and Anode Reactions in Electrolysis
During electrolysis, chemical reactions occur at the electrodes as a direct current is applied. The cathode is where the reduction takes place, meaning it is the site of electron gain. In our KBr solution example, the reduction of water is preferred over the reduction of potassium ions due to the more favorable standard electrode potential, leading to the production of hydrogen gas and hydroxide ions: 2 H2O(l) + 2 e- → H2(g) + 2 OH-(aq).
The anode, on the other hand, is where oxidation occurs, or the loss of electrons. In our example, bromide ions (Br-) are oxidized to bromine (Br2), which aligns with its lower oxidation potential and higher concentration compared to water: 2 Br-(aq) → Br2(l) + 2 e-.
Standard Electrode Potentials
The standard electrode potential is a measure of the individual potential of a reversible electrode at standard state, which refers to a concentration of 1 molar, at a pressure of 1 atmosphere and a temperature of 298K (25°C). This potential is used to predict the direction of electron flow. In electrolysis, the electrode with the higher reduction potential will usually gain electrons, acting as the cathode, whereas the lower potential electrode will lose electrons and act as the anode.
For example, standard reaction potentials for the cathode reaction involving water is approximately -0.83 V while for the bromide anode reaction it is +1.07 V. The signs indicate whether a substance is likely to be reduced (-) or oxidized (+). The larger the difference in potential, the greater the driving force for the reaction.
Minimum Potential for Electrolysis
To initiate electrolysis, an external potential must be applied that is higher than the combined standard electrode potentials of the cathode and anode reactions. This minimum potential ensures that the electrochemical reactions occur despite resistive forces within the cell. Without considering overpotential or electrode passivity, the minimum potential is simply the algebraic sum of the cathode and anode potentials.
In the KBr example, the calculation of the minimum potential necessary for the onset of electrolysis is the difference between the anode and cathode potentials: Minimum potential = E°(anode) − E°(cathode) = 1.07 V − (-0.83 V) = 1.90 V. This value indicates the least amount of voltage that must be applied across the cell for electrolysis to take place.

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Most popular questions from this chapter

The entropy change of a cell reaction can be determined from the change of the cell potential with temperature. (a) Show that \(\Delta S^{\circ}=n F\left(E_{\text {cell }, 2^{\circ}}-E_{\text {cell }, 1^{\circ}}\right) /\left\langle T_{2}-T_{1}\right) .\) Assume that \(\Delta S^{\circ}\) and \(\Delta H^{\circ}\) are constant over the temperature range considered. (b) Calculate \(\Delta S^{\circ}\) and \(\Delta H^{\circ}\) for the cell reaction \(\mathrm{Hg}_{2} \mathrm{Cl}_{2}(\mathrm{~s})+\) \(\mathrm{H}_{2}(\mathrm{~g}) \rightarrow 2 \mathrm{Hg}(\mathrm{l})+2 \mathrm{H}^{+}(\mathrm{aq})+2 \mathrm{Cl}^{-}(\mathrm{aq})\), given that \(E^{\circ}=+0.2699 \mathrm{~V}\) at \(293 \mathrm{~K}\) and \(+0.2669 \mathrm{~V}\) at \(303 \mathrm{~K}\).

Write the half-reactions and devise a galvanic cell (write a cell diagram) to study each of the following reactions: (a) \(\mathrm{AgBr}(\mathrm{s})=\mathrm{Ag}^{+}(\mathrm{aq})+\mathrm{Br}^{-}(\mathrm{aq})\), a solubility equilibrium (b) \(\mathrm{H}^{+}(\mathrm{aq})+\mathrm{OH}^{-}\)(aq) \(\longrightarrow \mathrm{H}_{2} \mathrm{O}(\mathrm{l})\), the Bronsted neutralization reaction (c) \(\mathrm{Cd}(\mathrm{s})+2 \mathrm{Ni}(\mathrm{OH})_{3}(\mathrm{~s}) \rightarrow \mathrm{Cd}(\mathrm{OH})_{2}(\mathrm{~s})+2 \mathrm{Ni}(\mathrm{OH})_{2}(\mathrm{~s})\), the reaction in the nickel-cadmium cell

(a) What is the approximate chemical formula of rust? (b) What is the oxidizing agent in the formation of rust? (c) How does the presence of salt accelerate the rusting process?

Write the half-reactions and the balanced equation for the cell reaction for each of the following galvanic cells: (a) \(\mathrm{Ni}(\mathrm{s})\left|\mathrm{Ni}^{2+}(\mathrm{aq})\right|\left|\mathrm{Ag}^{+}(\mathrm{aq})\right| \mathrm{Ag}(\mathrm{s})\) (b) \(\mathrm{C}(\mathrm{gr})\left|\mathrm{H}_{2}(\mathrm{~g}) / \mathrm{H}^{+}(\mathrm{aq}) \| \mathrm{Cl}^{-}(\mathrm{aq})\right| \mathrm{Cl}_{2}(\mathrm{~g}) \mid \mathrm{Pt}(\mathrm{s})\) (c) \(\mathrm{Cu}(\mathrm{s})\left|\mathrm{Cu}^{2+}(\mathrm{aq})\right|\left|\mathrm{Ce}^{4+}(\mathrm{aq}), \mathrm{Ce}^{3+}(\mathrm{aq})\right| \operatorname{Pt}(\mathrm{s})\) (d) \(\operatorname{Pt}(\mathrm{s})\left|\mathrm{O}_{2}(\mathrm{~g})\right| \mathrm{H}^{+}(\mathrm{aq}) \| \mathrm{OH}^{-}(\mathrm{aq})\left|\mathrm{O}_{2}(\mathrm{~g})\right| \mathrm{Pt}(\mathrm{s})\) (c) \(\operatorname{Pt}(\mathrm{s})\left|\mathrm{Sn}^{4+}(\mathrm{aq}), \mathrm{Sn}^{2+}(\mathrm{aq}) \| \mathrm{Cl}^{-}(\mathrm{aq})\right| \mathrm{Hg}_{2} \mathrm{Cl}{ }_{2}\) (s) \(\mid \mathrm{Hg}(\mathrm{l})\)

Balance each of the following skeletal equations by using oxidation and reduction half-reactions. All the reactions take place in basic solution. Identify the oxidizing agent and reducing agent in cach reaction. (a) Production of chlorite ions from dichlorine heptoxide: \(\mathrm{Cl}_{2} \mathrm{O}_{7}(\mathrm{~g})+\mathrm{H}_{2} \mathrm{O}_{2}(\mathrm{aq}) \rightarrow \mathrm{ClO}_{2}{ }^{-}(\mathrm{aq})+\mathrm{O}_{2}(\mathrm{~g})\) (b) Action of permanganate ions on sulfide ions: \(\mathrm{MnO}_{4}^{-}(\mathrm{aq})+\mathrm{S}^{2-}(\mathrm{aq}) \longrightarrow \mathrm{S}(\mathrm{s})+\mathrm{MnO}_{2}(\mathrm{~s})\) (c) Reaction of hydrazine with chlorate ions: \(\mathrm{N}_{2} \mathrm{H}_{4}(\mathrm{~g})+\mathrm{ClO}_{3}{ }^{-}(\mathrm{aq}) \rightarrow \mathrm{NO}(\mathrm{g})+\mathrm{Cl}^{-}\)(aq) (d) Reaction of plumbate ions and hypochlorite ions: \(\mathrm{Pb}(\mathrm{OH})_{4}{ }^{2-}(\mathrm{aq})+\mathrm{ClO}^{-}(\mathrm{aq}) \longrightarrow \mathrm{PbO}_{2}(\mathrm{~s})+\mathrm{Cl}^{-}(\mathrm{aq})\)
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