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)`

The `EMF` of the cell, `Cr|Cr^(+3) (0.1M) ||Fe^(+2) (0.01M)|Fe` (Given

1.	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)`
Answer» 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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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)`

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