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Calculate the equilibrium constant for the reaction 2Fe3+(aq) + 2I– (aq) → 2Fe2+(aq) + I2(s) E0cell = 0.236V

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Cell reaction is,
`2Fe^(3+)+2I^(-) to 2Fe^(2+)+I_(2)`
Half cell reaction and overall reaction are,
Oxidation: `2I^(-) to I_(2)+2e^(-)]xx3`
Reduction: `Fe^(3+)+3l^(-) to Fe]xx2`
Overall `2Fe^(3+)+6I^(-) to 2Fe+3I_(2)`
Here, `n=6 and E_(cell)^(@)=0.235V`
`logK_(C)=(E_(cell)^(@)xxn)/(0.0591V)`
`=(0.235Vxx6)/(0.0591V)=23.8579`
`K_(C)`=antilog `23.8579=7.209xx10^(23)`
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