Question

Mercuric oxide dry-cell batteries are often used where a high energy density is required, such as in watches and cameras. The two half-cell reactions that occur in the battery are \(\mathrm{HgO}(s)+\mathrm{H}_{2} \mathrm{O}(l)+2 \mathrm{e}^{-}--\rightarrow \mathrm{Hg}(l)+2 \mathrm{OH}^{-}(a q)\) \(\mathrm{Zn}(s)+2 \mathrm{OH}^{-}(a q) \longrightarrow \mathrm{ZnO}(s)+\mathrm{H}_{2} \mathrm{O}(l)+2 \mathrm{e}^{-}\) (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?

Short answer

The overall cell reaction for the Mercuric oxide dry-cell battery is: \( \mathrm{HgO}(s) + \mathrm{Zn}(s) \rightarrow \mathrm{Hg}(l) + \mathrm{ZnO}(s) \). The standard reduction potential for the anode reaction is -1.252 V. The potential of the anode reaction is different in an acidic medium because the species involved and the reactions taking place are different, which affects the redox potentials.

Step-by-step solution

1. Find the overall cell reaction

To find the overall cell reaction, we should add the two half-cell reactions together. They are given in the exercise as follows: \( \mathrm{HgO}(s) + \mathrm{H}_{2} \mathrm{O}(l) + 2 \mathrm{e}^{-} \rightarrow \mathrm{Hg}(l) + 2 \mathrm{OH}^{-}(a q) \) \( \mathrm{Zn}(s) + 2 \mathrm{OH}^{-}(a q) \rightarrow \mathrm{ZnO}(s) + \mathrm{H}_{2} \mathrm{O}(l) + 2 \mathrm{e}^{-} \) When we add them together, we can cancel out the 2 electrons in both reactions and the extra water molecule, which gives us the overall cell reaction: \( \mathrm{HgO}(s) + \mathrm{H}_{2} \mathrm{O}(l) + \mathrm{Zn}(s) \rightarrow \mathrm{Hg}(l)+2 \mathrm{OH}^{-}(a q) + \mathrm{ZnO}(s) + \mathrm{H}_{2} \mathrm{O}(l) \) Simplifying the reaction by removing the extra water molecule: \( \mathrm{HgO}(s) + \mathrm{Zn}(s) \rightarrow \mathrm{Hg}(l) + \mathrm{ZnO}(s) \)

2. Calculate the standard reduction potential for the anode reaction

We are given the standard reduction potential for the cathode reaction (\(E_{cathode}^{\circ}\)) and the overall cell potential (\(E_{cell}^{\circ}\)): \( E_{cathode}^{\circ} = +0.098V \) \( E_{cell}^{\circ} = +1.35V \) The relationship between the overall cell potential, the standard reduction potentials for the anode and cathode reactions, is given by the following equation: \( E_{cell}^{\circ} = E_{cathode}^{\circ} - E_{anode}^{\circ} \) We can solve for the standard reduction potential for the anode reaction (\(E_{anode}^{\circ}\)): \( E_{anode}^{\circ} = E_{cathode}^{\circ} - E_{cell}^{\circ} \) \( E_{anode}^{\circ} = +0.098V - (+1.35V) = -1.252V \)

3. Explain the difference in the anode reaction potential in an acidic medium

The potential of the anode reaction in the mercuric oxide dry-cell battery is different than when it occurs in an acidic medium because the species involved and the reactions taking place are different. In the given battery, the reactions involve zinc, mercuric oxide and hydroxide ions, while in an acidic medium, the reactions would involve protons (H+) instead of hydroxide ions (OH-). The change in the chemical species involved in the reactions affects the redox potentials, and hence can result in a different standard reduction potential for the anode reaction in an acidic medium.

Key concepts

Standard Reduction Potential

Understanding standard reduction potential, often symbolized as E°, is fundamental to grasping electrochemistry. This value signifies the tendency of a chemical species to gain electrons and thereby be reduced, under standard state conditions. Standard conditions refer to a set solvent (usually water), a pressure of 1 atmosphere, and a concentration of 1 molar at 25°C.

Each half-reaction in an electrochemical cell has an associated reduction potential. In a table of standard reduction potentials, the higher the value on the scale, the greater the substance’s ability to gain electrons. It is this property that allows us to predict the direction of electron flow in an electrochemical cell. For instance, in the exercise, the cathode half-reaction involving mercuric oxide was given a standard reduction potential of +0.098 V.

What's particularly interesting is that these potentials are relative measures. All standard reduction potentials are determined with respect to the standard hydrogen electrode (SHE), which is arbitrarily assigned a reduction potential of 0 V. Different conditions such as temperature, or if the medium is acidic or basic, can alter these potentials, as we may notice in the problem's context involving the anode reaction.

Half-Cell Reactions

In electrochemistry, a half-cell corresponds to one part of a two-part setup, where redox reactions occur. A half-cell is essentially an electrode immersed in an electrolyte where either oxidation or reduction takes place. When we talk about half-cell reactions, we refer to the separate oxidation or reduction reactions that occur in their respective half-cells before combining to form the overall cell reaction.

The power of separating these reactions enables us to analyze them, measuring the potentials they contribute to the cell, and also experimenting with different combinations of half-cells to create batteries with desired properties. In the provided solution, we observe that by combining the half-cell reactions, we find the overall cell reaction for the battery: the reduction at mercury oxide’s side, and the oxidation at zinc’s layer, coupling to form the full electrochemical cell.

Galvanic Cells

A Galvanic cell, also known as a voltaic cell, is an electrochemical cell that derives electrical energy from spontaneous redox reactions taking place within it. It consists of two half-cells connected by a salt bridge, along with the electrodes. Electrons flow from the anode (the oxidation half-cell) to the cathode (the reduction half-cell), creating a current that can power electronic devices.

The mercuric oxide dry-cell battery mentioned in the exercise is a type of galvanic cell. Here, the Galvanic cell operates under standard conditions and the half-cell reactions drive the current through the external circuit. One critical aspect of galvanic cells is that they function due to the difference in reduction potentials between the two electrodes, which provides the electromotive force (EMF) necessary for the flow of electrons.

Electrochemical Cell Potential

The electrochemical cell potential, often denoted as E°_cell, is a quantitative measure of the ability of the redox reaction occurring in the cell to drive an electric current. It’s the overall difference in potential between the two half-cell reactions, effectively the voltage that the cell can produce.

It is calculated as the difference between the cathode (reduction) and anode (oxidation) potentials, as seen in the equation used in the given exercise: E°_cell = E°_cathode – E°_anode. In the example provided, the overall cell potential is found by subtracting the standard reduction potential of the anode from that of the cathode, resulting in +1.35 V. This illustrates the driving force of the battery and helps quantify how much electrical energy can be harnessed from the electrochemical reactions.