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The EMF of the cell, Cr|Cr^(+3) (0.1M)...

The `EMF` of the cell,
`Cr|Cr^(+3) (0.1M) ||Fe^(+2) (0.01M)|Fe`
(Given: `E^(0) Cr^(+3)|Cr =- 0.75 V, E^(0) Fe^(+2)|Fe =- 0.45 V)`

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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