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Mercuric oxide dry-cell batteries are often used where a high-energy density is

ID: 854750 • Letter: M

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
HgO(s)+H2O(l)+2e??Hg(l)+2OH?(aq)Zn(s)+2OH?(aq)?ZnO(s)+H2O(l)+2e?.

Part A

Write the overall cell reaction.

Express your answer as a chemical equation. Identify all of the phases in your answer.

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Part B

The value of E?red for the cathode reaction is +0.098 V. The overall cell potential is +1.35 V. Assuming that both half-cells operate under standard conditions, what is the standard reduction potential for the anode reaction?

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
HgO(s)+H2O(l)+2e??Hg(l)+2OH?(aq)Zn(s)+2OH?(aq)?ZnO(s)+H2O(l)+2e?.

Part A

Write the overall cell reaction.

Express your answer as a chemical equation. Identify all of the phases in your answer.

SubmitMy AnswersGive Up

Part B

The value of E?red for the cathode reaction is +0.098 V. The overall cell potential is +1.35 V. Assuming that both half-cells operate under standard conditions, what is the standard reduction potential for the anode reaction?

Explanation / Answer

Cell reactions

HgO(S) +H2O + 2e--------------------------> Hg(l)+ 2OH-(aq)      at Cathode

Zn(s) + 2OH-(aq) ----------------------------> ZnO(s) + H2O + 2e-    at Anode

-----------------------------------------------------------------------------------------

Overall cell reaction HgO(s) +Zn(s) --------------------> Hg(l) + ZnO (s)

E0cell = Reduction potential of cathode(E0cathode ) - reduction potential of anode(E0anode)

Therefore,

Reduction potential of the Anode (E0anode)= Reduction potential of cathode(E0cathode ) - Ecell

                                                              = 0.098V-1.35

                                                              = -1.252V