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The following electrochemical cell is se...

The following electrochemical cell is set up at 298 K:
`Zn//Zn^(2+) (aq) (1 M) || Cu^(2+) (aq) (1 M)//Cu`
Given: `E^(@)Zn^(2+)//Zn =-0.761 V, E^(@) Cu^(2+)//Cu=+0.339 V`
(1) Write the cell reaction.
(2) Calculate the emf and free energy change at 298 K

Text Solution

Verified by Experts

(1) Cell reactions :
At anode `Zn to Zn^(2+) +2e^(-) E^(@)=0.761V`
At cathode `Cu^(2+) +2e^(-) to Cu E^(@) = +0.339V`
Overall Cell Reaction :
`Zn(s) +Cu^(2+)(aq) to Zn^(2+) (aq) +Cu(s)`
(2) EMF `(E^(@))=E^(@)` cathode `-E^(@)` anode
`=0.339 -(-0.761)`
`E^(@)=1.1V`
Free energy change
`DeltaG^(@)= -nFE^(@)`
`= -2xx96500xx1.1`
`=-212300J`
`DeltaG^(@)= -212.3kJ`
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