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For the cell Zn(s)|Zn^(2+) (2M)||Cu^(2...

For the cell
`Zn(s)|Zn^(2+) (2M)||Cu^(2+) (0.5 M)|Cu(s)`
(a) Write equation for each half reaction.
(b) Calculate the cell potential at `25^@C`

Text Solution

Verified by Experts

(a) Half reaction oxidation half reaction:
(i) `Zn(s) to Zn^(2+) (aq) +2e^(-)`
Reduction half reaction:
(ii) `Cu^(2+) (aq) +2e^(-) to Cu(s) `
`E_(cell)^@=E_(cathode)-E_(anode)`
`=+0.34-(-0.76)=1.1`
`E_(cell)^@=E_(cell)^@-0.059/2log""(Products)/(Reactants)`
`Zn_((s))+Cu^(2+) (0.5M)to Cu_((s))+Zn^(2+)(2M)`
`E_(cell)^@=1.1-0.059/2 log""2/0.5`
`=1.1-0.0295 times 0.6020`
`=1.1 -0.0177`
=1.082V
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