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

01

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.

02

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 \(E_{anode}^o\): \(E_{anode}^o = E_{cathode}^o - E_{cell}^o\)

03

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:

04

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: \( \mathrm{HgO}(s) + \mathrm{Zn}(s) \rightarrow \mathrm{Hg}(l) + \mathrm{ZnO}(s) \)

05

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: \(E_{anode}^o = 0.098 V - 1.35 V = -1.252 V\)

06

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.

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