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If a lead storage battery is charged at too high a voltage, gases are produced at each electrode. (It is possible to recharge a lead-storage battery only because of the high overpotential for gas formation on the electrodes.) (a) What are these gases? (b) Write a cell reaction to describe their formation.

Short Answer

Expert verified
The gases formed during overcharging a lead storage battery are Hydrogen and Oxygen. The correspondive cell reactions are: at the negative electrode \(2H^{+} + 2e^{-} \rightarrow H_{2}\) and at the positive electrode \(O_{2} + 4H^{+} + 4e^{-} \rightarrow 2H_{2}O\).

Step by step solution

01

Identify the gases formed at the electrode

When a lead battery is recharged at high voltage, gases are evolved at both electrodes. The gases that are formed are Hydrogen (H2) at the negative electrode, and Oxygen (O2) at the positive electrode.
02

Write the cell reactions

The reactions at the electrodes representing the formation of gases are:Negative electrode (anode):\[2H^{+} + 2e^{-} \rightarrow H_{2}\]Positive electrode (cathode):\[O_{2} + 4H^{+} + 4e^{-} \rightarrow 2H_{2}O\]

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Key Concepts

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

Gas Formation in Lead Batteries
Understanding why gases form during the charging process of lead storage batteries is crucial for any student studying electrochemistry. In these batteries, when the charging voltage exceeds a certain threshold, it disrupts the balance of the electrochemical reactions within the cells, leading to electrolysis of water, which is a component of the battery's acid electrolyte. This process creates hydrogen gas at the negative electrode, also known as the anode, and oxygen gas at the positive electrode, the cathode.

Hydrogen and oxygen gas formation may not seem concerning at first glance, but it can lead to several issues, such as decreased battery efficiency, potential battery case rupture due to the build-up of pressure, and even safety hazards if the gases ignite. Therefore, understanding gas formation helps in designing safe charging methods for lead-acid batteries and enhances the efficiency and lifespan of the batteries.

Furthermore, overpotential, a necessary condition for the battery's reusability, is what allows for the recharging without excessive gas formation under normal conditions. It's this overpotential that prevents the battery from decomposing too much of the water component into gases during regular charging cycles.
Electrode Reactions
Electrode reactions are at the heart of any electrochemical cell, and lead storage batteries are no exception. When a lead-acid battery is recharged, the electrical energy supplied reverses the discharging reactions. At the negative electrode, lead and sulfate ions recombine to form lead sulfate and electrons are released in the process. Conversely, at the positive electrode, lead dioxide, lead sulfate, and hydrogen ions react to produce water and release electrons.

Understanding the electrode reactions helps students appreciate how a battery stores and releases energy. It's like unwinding a spring to store energy and then letting it release energy as it returns to its original state. The reactions are highly specific and need to happen in coordinated harmony to maintain the battery's function and health. Disruption to these reactions, such as from charging at too high a voltage, leads to the formation of hydrogen and oxygen gases which are discussed in the lead battery gas formation concept.
Overpotential
Overpotential is a term that often perplexes students, but it's a fundamental aspect of electrochemistry, especially when related to batteries. In simple terms, overpotential represents the extra amount of energy needed to drive an electrochemical reaction, beyond what is predicted by the standard electrode potential. In context with lead storage batteries, overpotential is the additional voltage that is applied during charging to overcome the thermodynamic stability of water and other inefficiencies within the battery, such as resistance.

Why is overpotential so important for lead batteries? It's because the overpotential for hydrogen and oxygen gas formation on the electrodes is high, which means that water won't decompose into these gases until a significantly higher voltage is applied. This ensures that we can recharge a lead-acid battery efficiently without expending a lot of the energy into gas formation under normal charging conditions. If this was not the case, batteries would be far more cumbersome to charge and would have limited reusable lifetimes due to rapid water loss and gas buildup.

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Most popular questions from this chapter

A voltaic cell, with \(E_{\text {cell }}=0.180 \mathrm{V},\) is constructed as follows: $$\mathrm{Ag}(\mathrm{s})\left|\mathrm{Ag}^{+}\left(\operatorname{satd} \mathrm{Ag}_{3} \mathrm{PO}_{4}\right) \| \mathrm{Ag}^{+}(0.140 \mathrm{M})\right| \mathrm{Ag}(\mathrm{s})$$ What is the \(K_{\mathrm{sp}}\) of \(\mathrm{Ag}_{3} \mathrm{PO}_{4} ?\)

Ultimately, \(\Delta G_{\mathrm{f}}^{\mathrm{Q}}\) values must be based on experimental results; in many cases, these experimental results are themselves obtained from \(E^{\circ}\) values. Early in the twentieth century, G. N. Lewis conceived of an experimental approach for obtaining standard potentials of the alkali metals. This approach involved using a solvent with which the alkali metals do not react. Ethylamine was the solvent chosen. In the following cell diagram, \(\mathrm{Na}(\text { amalg, } 0.206 \%)\) represents a solution of \(0.206 \%\) Na in liquid mercury. 1\. \(\mathrm{Na}(\mathrm{s}) | \mathrm{Na}^{+}(\text {in ethylamine }) | \mathrm{Na}(\text { amalg }, 0.206 \%)\) \(E_{\text {cell }}=0.8453 \mathrm{V}\) Although Na(s) reacts violently with water to produce \(\mathrm{H}_{2}(\mathrm{g}),\) at least for a short time, a sodium amalgam electrode does not react with water. This makes it possible to determine \(E_{\text {cell }}\) for the following voltaic cell. 2\. \(\mathrm{Na}(\text { amalg }, 0.206 \%)\left|\mathrm{Na}^{+}(1 \mathrm{M}) \| \mathrm{H}^{+}(1 \mathrm{M})\right|\) $$\mathrm{H}_{2}(\mathrm{g}, 1 \mathrm{atm}) \quad E_{\mathrm{cell}}=1.8673 \mathrm{V}$$ (a) Write equations for the cell reactions that occur in the voltaic cells (1) and (2) (b) Use equation (20.14) to establish \(\Delta G\) for the cell reactions written in part (a). (c) Write the overall equation obtained by combining the equations of part (a), and establish \(\Delta G^{\circ}\) for this overall reaction. (d) Use the \(\Delta G^{\circ}\) value from part (c) to obtain \(E_{\text {cell }}^{\circ}\) for the overall reaction. From this result, obtain \(E_{\mathrm{Na}^{+}}^{\circ} / \mathrm{Na}\) Compare your result with the value listed in Appendix D.

Of the following statements concerning electrochemical cells, the correct ones are: (a) The cathode is the negative electrode in both voltaic and electrolytic cells. (b) The function of a salt bridge is to permit the migration of electrons between the half-cell compartments of an electrochemical cell. (c) The anode is the negative electrode in a voltaic cell. (d) Electrons leave the cell from either the cathode or the anode, depending on what electrodes are used. (e) Reduction occurs at the cathode in both voltaic and electrolytic cells. (f) If electric current is drawn from a voltaic cell long enough, the cell becomes an electrolytic cell. (g) The cell reaction is an oxidationreduction reaction.

Calculate the quantity indicated for each of the following electrolyses. (a) the mass of \(\mathrm{Zn}\) deposited at the cathode in 42.5 min when 1.87 A of current is passed through an aqueous solution of \(\mathrm{Zn}^{2+}\) (b) the time required to produce \(2.79 \mathrm{g} \mathrm{I}_{2}\) at the anode if a current of \(1.75 \mathrm{A}\) is passed through \(\mathrm{KI}(\mathrm{aq})\)

Derive a balanced equation for the reaction occurring in the cell: $$\mathrm{Fe}(\mathrm{s})\left|\mathrm{Fe}^{2+}(\mathrm{aq}) \| \mathrm{Fe}^{3+}(\mathrm{aq}), \mathrm{Fe}^{2+}(\mathrm{aq})\right| \mathrm{Pt}(\mathrm{s})$$ (a) If \(E_{\text {cell }}^{\circ}=1.21 \mathrm{V},\) calculate \(\Delta G^{\circ}\) and the equilibrium constant for the reaction. (b) Use the Nernst equation to determine the potential for the cell: $$\begin{array}{r} \mathrm{Fe}(\mathrm{s}) | \mathrm{Fe}^{2+}\left(\mathrm{aq}, 1.0 \times 10^{-3} \mathrm{M}\right) \| \mathrm{Fe}^{3+}\left(\mathrm{aq}, 1.0 \times 10^{-3} \mathrm{M}\right) \\ \mathrm{Fe}^{2+}(\mathrm{aq}, 0.10 \mathrm{M}) | \mathrm{Pt}(\mathrm{s}) \end{array}$$ (c) In light of (a) and (b), what is the likelihood of being able to observe the disproportionation of \(\mathrm{Fe}^{2+}\) into \(\mathrm{Fe}^{3+}\) and Fe under standard conditions?

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