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Consider the cell : Zn|Zn^(2+)(aq)(1.0...

Consider the cell `:`
`Zn|Zn^(2+)(aq)(1.0M)||Cu^(2+)(aq)(1.0M)||Cu`
Thee standard reduction potentials are `0.350V` for
`Cu^(2+)(aq)+2e^(-)rarrCu ` and `-0.763V` for
`Zn^(2+)(aq)+2e^(-) rarr Zn`
`a.` Write the cell reaction.
`b.` Calculate the `EMF` of the cell.
`c.` Is the reaction spontaneous or not ?

Text Solution

Verified by Experts

The correct Answer is:
`b. 1.113V,` ,`Spontaneous`
`i.` In the cell
`Zn|Zn^(2+)(aq)(0.1M)||Cu^(2+)(aq)(1.0M)|Cu`,
the oxidation half reaction is
`(` at anode `)Znrarr Zn^(2+)+2e^(-) ……(i) `
the reduction half reaction is
`(` at cathode `) Cu^(2+)+2e^(-) rarr Cu …..(ii) `
On adding equations `(i)` and `(ii)`, we get
`Zn+Cu^(2+) rarr Zn^(2+) + Cu`
This is the cell `=E^(c-)._(cathode)-E^(c-)._(anode)`
`=E^(c-)._(Cu^(2+)|Cu)-E^(c-)._(Zn^(2+)|Zn)`
`=0.350-(-0.7630=1.113V`
`iii. " " As the `EMF` of the cell is positive, the cell reaction is spontaneous.
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