Problem 78 - Chemistry The Central Science

Question

Mercuric oxide dry-cell batteries are often used where a highenergy density is required, such as in watches and cameras. The two half-cell reactions that occur in the battery are $$ \begin{aligned} \mathrm{HgO}(s)+\mathrm{H}_{2} \mathrm{O}(l)+2 \mathrm{e}^{-} & \longrightarrow \mathrm{Hg}(l)+2 \mathrm{OH}^{-}(a q) \\ \mathrm{Zn}(s)+2 \mathrm{OH}^{-}(a q) & \longrightarrow \mathrm{ZnO}(s)+\mathrm{H}_{2} \mathrm{O}(l)+2 \mathrm{e}^{-} \end{aligned} $$ (a) Write the overall cell reaction. (b) The value of \(E_{\text {red }}^{\circ}\) 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) The overall cell reaction is: \( \mathrm{HgO}(s) + \mathrm{Zn}(s) \rightarrow \mathrm{Hg}(l) + \mathrm{ZnO}(s) \) (b) The standard reduction potential for the anode reaction is: \(E_{anode}^o = -1.252 V\) (c) The potential difference of the anode reaction in an acidic medium occurs because acidic environments introduce additional protons (H+), which can change the concentrations and reaction rates of the reactants, affecting the overall standard reduction potential and creating a new equilibrium condition for the overall cell reaction.

Step 1: Find the overall cell reaction

To find the overall cell reaction, we need to add both half-cell reactions together. First, note that we don't need to balance the electrons since they are already balanced in each half-cell reaction. The two half-cell reactions are: \( \mathrm{HgO}(s)+\mathrm{H}_{2} \mathrm{O}(l)+2 \mathrm{e}^{-} \rightarrow \mathrm{Hg}(l)+2 \mathrm{OH}^{-}(a q) \) \( \mathrm{Zn}(s)+2 \mathrm{OH}^{-}(a q) \rightarrow \mathrm{ZnO}(s)+\mathrm{H}_{2} \mathrm{O}(l)+2 \mathrm{e}^{-} \) Add the two half-cell reactions to find the overall cell reaction.

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

We're given the standard reduction potential (\(E_{red}^o\)) for the cathode reaction and the overall cell potential. To find the standard reduction potential for the anode reaction, let's use the formula: \(E_{cell}^o = E_{cathode}^o - E_{anode}^o\) Here, \(E_{cell}^o = 1.35 V\) and \(E_{cathode}^o = 0.098 V\). By rearranging the formula, we can find the value of \(E_{anode}^o\): \(E_{anode}^o = E_{cathode}^o - E_{cell}^o\)

Step 3: Discuss the potential difference of the anode reaction in an acidic medium

Finally, we need to address why the potential of the anode reaction varies when the reaction takes place in an acidic medium. To answer this question, we need to consider how acidic conditions can affect the chemical species involved in the reaction. Solution:

Step 1: Find the overall cell reaction

To find the overall cell reaction, add both half-cell reactions together: \( \mathrm{HgO}(s)+\mathrm{H}_{2} \mathrm{O}(l)+2 \mathrm{e}^{-} + \mathrm{Zn}(s)+2 \mathrm{OH}^{-}(a q) \rightarrow \mathrm{Hg}(l)+2 \mathrm{OH}^{-}(a q) + \mathrm{ZnO}(s)+\mathrm{H}_{2} \mathrm{O}(l)+2 \mathrm{e}^{-} \) This simplified reaction is: \( \mathrm{HgO}(s) + \mathrm{Zn}(s) \rightarrow \mathrm{Hg}(l) + \mathrm{ZnO}(s) \)

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

Using the formula \(E_{anode}^o = E_{cathode}^o - E_{cell}^o\), where \(E_{cell}^o = 1.35 V\) and \(E_{cathode}^o = 0.098 V\), we get: \(E_{anode}^o = 0.098 V - 1.35 V = -1.252 V\)

Step 3: Discuss the potential difference of the anode reaction in an acidic medium

The potential difference of the anode reaction in an acidic medium happens because of the differences in chemical environments. Acidic environments can introduce additional protons (H+) into the reaction, which can change the reactants' concentrations and reaction rates. This affects the overall standard reduction potential and creates a new equilibrium condition for the overall cell reaction.

Key Concepts

Redox Reactions

Redox reactions, or reduction-oxidation reactions, are essential processes that involve the transfer of electrons between chemical species. In a redox reaction, one molecule is reduced by gaining electrons, while another is oxidized by losing electrons. These reactions are fundamental to electrochemistry and form the basis of how batteries work. To illustrate this with the exercise's example, consider the half-reactions of a mercuric oxide dry-cell battery:

• Reduction half-reaction: \( \mathrm{HgO}(s) + \mathrm{H}_{2} \mathrm{O}(l) + 2 \mathrm{e}^{-} \rightarrow \mathrm{Hg}(l) + 2 \mathrm{OH}^{-}(aq) \)
• Oxidation half-reaction: \( \mathrm{Zn}(s) + 2 \mathrm{OH}^{-}(aq) \rightarrow \mathrm{ZnO}(s) + \mathrm{H}_{2} \mathrm{O}(l) + 2 \mathrm{e}^{-} \)

These two half-reactions are combined to yield the overall cell reaction. Through balancing the electrons on both sides, these individual processes make a complete redox reaction, driving the flow of electrical current. Understanding the distinction between oxidation and reduction is crucial for grasping the underlying principles of electrochemical cells.

Standard Cell Potential

Standard cell potential, denoted as \(E_{cell}^o\), measures the potential voltage difference between two electrodes under standard conditions. These conditions typically include a temperature of \(25^{\circ}C\), a pressure of \(1\) atm for gases, and concentrations of \(1\, \text{M}\) for all aqueous solutions. In our exercise, the overall cell potential for the mercuric oxide dry-cell battery is given as \(+1.35 \, \mathrm{V}\). This value is found by using the equation:\[E_{cell}^o = E_{cathode}^o - E_{anode}^o\] This formula derives the overall cell potential from the reduction potentials of the individual half-cell reactions. Here:

• \(E_{cathode}^o = 0.098 \, \mathrm{V}\)
• \(E_{anode}^o\) is found by rearranging the formula, \(E_{anode}^o = E_{cathode}^o - E_{cell}^o\), resulting in \(-1.252 \, \mathrm{V}\).

The positive value of \(E_{cell}^o\) indicates a spontaneous reaction, key for understanding why batteries can release energy without external input.

Half-Cell Reactions

Half-cell reactions are the individual components of redox reactions that occur at the electrodes in electrochemical cells. Each half-cell reaction entails either the gain or loss of electrons:

• **The cathode reaction** is where reduction occurs, meaning electrons are gained. In our example, \( \mathrm{HgO}(s) + \mathrm{H}_{2} \mathrm{O}(l) + 2 \mathrm{e}^{-} \rightarrow \mathrm{Hg}(l) + 2 \mathrm{OH}^{-}(aq) \), mercury oxide is reduced to mercury.
• **The anode reaction** involves oxidation, meaning electrons are lost. For the exercise, \( \mathrm{Zn}(s) + 2 \mathrm{OH}^{-} (aq) \rightarrow \mathrm{ZnO}(s) + \mathrm{H}_{2} \mathrm{O}(l) + 2 \mathrm{e}^{-} \), zinc is oxidized to zinc oxide.

Half-cell reactions are crucial for determining the direction and magnitude of the electron flow. By understanding these reactions, we can predict the behavior of the overall electrochemical cell in different conditions, such as in various pH environments, which can influence the reaction dynamics as seen with the shift in potential in acidic conditions.