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For the galvanic cell: Ag|AgCI(s)|KCI(0....

For the galvanic cell: `Ag|AgCI(s)|KCI(0.2M)||KBr(0.001M)|AgBr(s)|Ag`, calculate the EMF generated and assign correct polarity to each electrode for a spontaneous process after taking into account the cell reaction at `25^(@)C`.
`[K_(sp(AgCI)) = 2.8 xx 10^(-10), K_(sp(AgBr)) = 3.3 xx 10^(-13)]`

Text Solution

Verified by Experts

`[Ag^(+)]` in left hand electrode chamber `=(2.8xx10^(-10))/(0.2)=1.4xx10^(-9) M`
`[Ag^(+)]` in right hand electrode chamber `=(3.3xx10^(-13))/(0.001)`
emf `=-0.0.0592 log. ([Ag^(+)]_(anode))/([Ag^(+)]_(cathode))=0.00592 log. (1.4xx10^(-9))/(3.3xx10^(-10))=-0.037 V`
Therefore, the cell as written is non-spontaneous and its reverse will be spontaneous with emf = 0.037 V.
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