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The following half-reaction takes place in a mercury battery used in hearing aids: $$ \mathrm{HgO}(s)+\mathrm{H}_{2} \mathrm{O}(l)+2 e^{-} \longrightarrow \mathrm{Hg}(l)+2 \mathrm{OH}^{-}(a q) $$ a. Is the half-reaction an oxidation or a reduction? b. What substance is oxidized or reduced? c. At which electrode would this half-reaction occur?

Short Answer

Expert verified
a. Reduction, b. Mercury (Hg), c. Cathode

Step by step solution

01

Identify Oxidation States

First, determine the oxidation states of mercury and oxygen in the reactants and products. For HgO, mercury (Hg) has an oxidation state of +2 and oxygen (O) has an oxidation state of -2. In the products, Hg (liquid mercury) has an oxidation state of 0.
02

Determine the Type of Reaction

Next, determine if the half-reaction involves a gain or loss of electrons. The oxidation state of mercury changes from +2 to 0, indicating a gain of electrons. Since reduction involves gaining electrons, this half-reaction is a reduction.
03

Identify the Substance Being Reduced

Because the oxidation state of mercury (Hg) decreases from +2 in HgO to 0 in liquid mercury (Hg), mercury is being reduced in this reaction.
04

Determine the Electrode

Reduction reactions occur at the cathode in electrochemical cells. Therefore, this half-reaction takes place at the cathode.

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Key Concepts

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

oxidation state
The oxidation state is a key concept in understanding redox reactions in electrochemistry. It indicates the degree of oxidation of an atom in a compound. For instance, mercury in mercury oxide (HgO) has an oxidation state of +2, meaning it has lost two electrons compared to its elemental form. This is in contrast to the oxygen in HgO, which has an oxidation state of -2. When examining reactions, changes in oxidation states can help us determine if a substance is being oxidized (losing electrons) or reduced (gaining electrons). Looking at the half-reaction provided, the mercury’s oxidation state changes from +2 in HgO to 0 in liquid Hg, suggesting that mercury is gaining electrons.
reduction reaction
A reduction reaction is one where a substance gains electrons. This can be remembered by the acronym 'OIL RIG' - Oxidation Is Losing, Reduction Is Gaining (electrons). In the given half-reaction for the mercury battery: \[ \text{HgO}(s) + \text{H}_2\text{O}(l) + 2 e^{-} \longrightarrow \text{Hg}(l) + 2 \text{OH}^{-}(aq) \] we see that mercury (Hg) gains electrons as its oxidation state changes from +2 to 0. This gain of electrons is what qualifies this process as a reduction reaction. Recognizing reduction reactions is crucial in battery chemistry, as they occur at the cathode, where the positive electrode attracts electrons.
cathode
In electrochemical cells, the cathode is the electrode where reduction takes place. It's characterized by the gain of electrons and is essential for the overall chemistry of batteries and other electrochemical processes. For the mercury battery used in hearing aids, the half-reaction shows a gain of electrons by mercury, confirming it as a reduction reaction. This reaction occurs at the cathode, making it the site of the reduction. The cathode plays a critical role in facilitating the flow of electrons into the electrochemical cell, which is crucial for the battery’s functionality. Understanding the role of the cathode can help in comprehending how batteries work and how to optimize their performance.

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Most popular questions from this chapter

What we call "tin cans" are really iron cans coated with a thin layer of tin. The anode is a bar of tin and the cathode is the iron can. An electrical current is used to oxidize the Sn to \(\mathrm{Sn}^{2+}\) in solution, which is reduced to produce a thin coating of \(\mathrm{Sn}\) on the can. a. What half-reaction takes place to tin plate an iron can? b. Why is the iron can the cathode? c. Why is the tin bar the anode?

The following half-reaction takes place in a nickel-cadmium battery used in a cordless drill: $$ \mathrm{Cd}(s)+2 \mathrm{OH}^{-}(a q) \longrightarrow \mathrm{Cd}(\mathrm{OH})_{2}(s)+2 e^{-} $$ a. Is the half-reaction an oxidation or a reduction? b. What substance is oxidized or reduced? c. At which electrode would this half-reaction occur?

Assign oxidation numbers to all the elements in each of the following: a. \(\mathrm{C}_{2} \mathrm{H}_{3} \mathrm{O}_{2}^{-}\) b. \(\mathrm{AlCl}_{3}\) c. \(\mathrm{NH}_{4}^{+}\) d. \(\mathrm{HBrO}_{4}\)

Balance each of the following half-reactions in acidic solution: \((15.2)\) a. \(\mathrm{I}_{2}(s) \longrightarrow \mathrm{I}^{-}(a q)\) b. \(\mathrm{MnO}_{4}^{-}(a q) \longrightarrow \mathrm{Mn}^{2+}(a q)\) c. \(\mathrm{Br}_{2}(l) \longrightarrow \mathrm{BrO}_{3}^{-}(a q)\) d. \(\mathrm{ClO}_{3}^{-}(a q) \longrightarrow \mathrm{ClO}_{4}^{-}(a q)\)

Describe the voltaic cell and half-cell components and write the shorthand notation for the following oxidation-reduction reactions: a. \(\mathrm{Mn}(s)+\mathrm{Sn}^{2+}(a q) \longrightarrow \mathrm{Mn}^{2+}(a q)+\operatorname{Sn}(s)\) b. \(\mathrm{Ni}(s)+2 \mathrm{Ag}^{+}(a q) \longrightarrow \mathrm{Ni}^{2+}(a q)+2 \mathrm{Ag}(s)\)

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