Problem 76 - Chemistry The Central Science In Si Units

Question

Mercuric oxide dry-cell batteries are often used where a flat discharge voltage and long life are required, such as in watches and cameras. The two half-cell reactions that occur in the battery are $$ \begin{array}{l} \mathrm{HgO}(s)+\mathrm{H}_{2} \mathrm{O}(l)+2 \mathrm{e}^{-} \longrightarrow \mathrm{Hg}(l)+2 \mathrm{OH}^{-}(a q) \\ \mathrm{Zn}(s)+2 \mathrm{OH}^{-}(a q) \longrightarrow \mathrm{ZnO}(s)+\mathrm{H}_{2} \mathrm{O}(l)+2 \mathrm{e}^{-} \end{array} $$ (a) Write the overall cell reaction. (b) The value of \(E_{\mathrm{red}}^{\circ}\) for the cathode reaction is \(+0.098 \mathrm{~V}\). The overall cell potential is \(+1.35 \mathrm{~V}\). Assuming that both half- cells operate under standard conditions, what is the standard reduction potential for the anode reaction? (c) Why is the potential of the anode reaction different than would be expected if the reaction occurred in an acidic medium?

Step-by-Step Solution

Answer

(a) \( \mathrm{HgO}(s) + \mathrm{Zn}(s) \rightarrow \mathrm{Hg}(l) + \mathrm{ZnO}(s) \); (b) \( -1.252 \, \text{V} \); (c) Alkaline medium affects the reaction differently than acidic medium.

Step 1: Write the overall cell reaction

To find the overall cell reaction, combine the two half-cell reactions provided:1. Cathode reaction: \( \mathrm{HgO}(s)+\mathrm{H}_{2} \mathrm{O}(l)+2 \mathrm{e}^{-} \rightarrow \mathrm{Hg}(l)+2 \mathrm{OH}^{-}(aq) \)2. Anode reaction: \( \mathrm{Zn}(s)+2 \mathrm{OH}^{-}(aq) \rightarrow \mathrm{ZnO}(s)+\mathrm{H}_{2} \mathrm{O}(l)+2 \mathrm{e}^{-} \)By adding these reactions, the electrons cancel out, leading to:\[ \mathrm{HgO}(s) + \mathrm{Zn}(s) \rightarrow \mathrm{Hg}(l) + \mathrm{ZnO}(s) \]

Step 2: Find the standard reduction potential for the anode reaction

Given:- \( E_{\text{cell}}^{\circ} = +1.35 \, \text{V} \)- \( E_{\text{red}}^{\circ} \text{ (cathode)} = +0.098 \, \text{V} \)The overall cell potential is related to the cathode and anode potentials by:\[ E_{\text{cell}}^{\circ} = E_{\text{cathode}}^{\circ} - E_{\text{anode}}^{\circ} \]Rearranging gives:\[ E_{\text{anode}}^{\circ} = E_{\text{cathode}}^{\circ} - E_{\text{cell}}^{\circ} \]Plugging in the given values:\[ E_{\text{anode}}^{\circ} = 0.098 \, \text{V} - 1.35 \, \text{V} = -1.252 \, \text{V} \]

Step 3: Explain the effect of medium on the anode reaction potential

In acidic medium, the availability of protons can stabilize the positive charge better, potentially lowering the energy required for oxidation reactions. However, in an alkaline medium like this, with hydroxide ions, the reaction dynamics are different. The different reaction environment affects the electrochemical behavior making the potentials differ from those if the reaction occurred in acidic conditions.

Key Concepts

Redox Reactions

Redox reactions, short for reduction-oxidation reactions, are fundamental to understanding electrochemistry. They involve the transfer of electrons between substances. One substance will lose electrons (oxidation), and the other will gain them (reduction).
For instance, in the mercuric oxide dry-cell battery exercise, we observe two half-reactions:

• The cathode half-reaction: \( \mathrm{HgO}(s) + \mathrm{H}_{2} \mathrm{O}(l) + 2 \mathrm{e}^{-} \rightarrow \mathrm{Hg}(l) + 2 \mathrm{OH}^{-}(aq) \) involves reduction as mercury gains electrons.
• The anode half-reaction: \( \mathrm{Zn}(s) + 2 \mathrm{OH}^{-}(aq) \rightarrow \mathrm{ZnO}(s) + \mathrm{H}_{2} \mathrm{O}(l) + 2 \mathrm{e}^{-} \) involves oxidation as zinc loses electrons.

The overall reaction combines these to show the complete electron transfer:
\[ \mathrm{HgO}(s) + \mathrm{Zn}(s) \rightarrow \mathrm{Hg}(l) + \mathrm{ZnO}(s) \]
Mastering the balance and purpose of redox reactions is essential for exploring advanced electrochemical systems.

Standard Reduction Potentials

Standard reduction potentials \( (E^\circ) \) are key to predicting how easily a species can gain electrons. It's a measure of the tendency for a chemical species to be reduced, and is measured under standard conditions, typically 25°C, 1 M concentration for solutes, or 1 atm pressure for gases.
Each half-reaction in electrochemistry has its own standard reduction potential, which helps to determine the overall cell potential. The formula below is used in the battery problem:
\[ E_{\text{cell}}^{\circ} = E_{\text{cathode}}^{\circ} - E_{\text{anode}}^{\circ} \]
The given cathode potential was \( +0.098 \, \text{V} \), and we calculated the anode potential as \( -1.252 \, \text{V} \) using the The key takeaway is that comparing standard reduction potentials lets you understand which half-reaction will occur at the cathode or anode. In simple terms, a higher \( E^\circ \) means a stronger tendency to be reduced.

Battery Chemistry

Battery chemistry is all about converting chemical energy into electrical energy. It involves electrochemical cells where redox reactions take place. Understanding how batteries work enhances our grasp of both everyday technology and future innovations.
In the example of the mercuric oxide battery:

• Two electrodes (cathode and anode) are submerged in an electrolyte.
• Oxidation happens at the anode, producing electrons.
• Reduction occurs at the cathode, consuming electrons.
• The movement of electrons from the anode to the cathode creates electric current.

The electrochemical reaction is influenced by factors like the medium (acidic or alkaline). In our exercise, hydroxide ions are involved because it’s an alkaline medium, impacting how the standard potentials develop.
Batteries can be optimized for specific uses, such as providing a stable voltage for watches or cameras, by understanding these chemical processes.