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For the cell : Zn"|"Zn^(2+) (a=1)"||"Cu^...

For the cell : `Zn"|"Zn^(2+) (a=1)"||"Cu^(2+) (a=1)"|"Cu`
Given that `E_(Zn//Zn^(2+))=0.761 V, E_(Cu^(2+)//Cu)=0.339V`
(i) Write the cell reaction.
(ii) Calculate the emf and free energy change at 298 K involved in the cell. [Faraday's constant F = 96500 coulomb `eq^(-1)`]

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To solve the problem step by step, we will follow the instructions given in the question. ### Step 1: Determine the Cell Reaction 1. Identify the oxidation and reduction half-reactions: - Zinc (Zn) is oxidized to Zn²⁺. - Copper ions (Cu²⁺) are reduced to copper (Cu). 2. Write the half-reactions: ...
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