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Calculate the emf of the cell Cr∣Cr^(3+)...

Calculate the emf of the cell Cr∣`Cr^(3+)(0.1 M)∣∣Fe^(2+)`(0.01M)∣Fe
(Given: `E^(@)Cr^(3+)`/Cr =−0.75 volt;`E^(@)Fe^(2+)` /Fe =−0.45 volt).

Text Solution

Verified by Experts

Half cell reactions are:
`{:("At anode",:[Crrarr Cr^(+3) +3e^(-)]xx2),("At anode":,underline([Fe^(+2)+2e^(-)rarrFe]xx3)),("Over all reaction":,2Cr+3F^(+2)rarr2Cr^(+3)+3Fe):}`
`E_(cell)^(0) =` Oxidation pot.`+` Reduction pot.
`=0.75 +(-0.45) = 0.30`
`E_(cell) = E^(0) - (0.0591)/(n) log .(["Product"])/(["Reactant"])`
`= 0.30 - (0.0591)/(6) log .([Cr^(+3)]^(2))/([Fe^(+2)]^(3))`
`= 0.30 - (0.0591)/(6) log .[0.1]^(2)/([0.01]^(3))`
`= 0.30 - (0.24)/(6) =0.26` volt.
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