Calculate the value of equilibrium constant for the reaction : `2Fe^(3+)+2I^(-) to 2Fe^(2+)+I_(2)` Given that `E_(cell)^(@)=0.235" V "`
Calculate the value of equilibrium constant for the reaction :
`2Fe^(3+)+2I^(-) to 2Fe^(2+)+I_(2)`
Given that `E_(cell)^(@)=0.235" V "`
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Correct Answer - `8.97xx10^(7)`
`logK_(C)=(nE_(cell)^(@))/(0.0591)=(2xx0.235)/(0.0591)=7.953,K_(C)="Antilog "7.953=8.97xx10^(7)`.
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