Calculate equilibrium constant for `I_(2)+I^(-) hArr I_(3)^(-)` at 298 K from the following information : `{:(I_(2) (aq.)+2e^(-) rarr 2I^(-),,E^(@)=0.6197" volt"),(I_(3)^(-)+2e^(-) rarr 3I^(-),,E^(@)=0.5355" volt"):}`

Calculate equilibrium constant for `I_(2)+I^(-) hArr I_(3)^(-)` at 298

1.	Calculate equilibrium constant for `I_(2)+I^(-) hArr I_(3)^(-)` at 298 K from the following information : `{:(I_(2) (aq.)+2e^(-) rarr 2I^(-),,E^(@)=0.6197" volt"),(I_(3)^(-)+2e^(-) rarr 3I^(-),,E^(@)=0.5355" volt"):}`
Answer» Correct Answer - 706.9 `{:(I_(2)+2e^(-) rarr 2I^(-)," "E^(@)=0.6197" volt"),(3I^(-) rarr I_(3)^(-)+2e^(-)," "E^(@)=-0.5355" volt"),(bar(I_(2)+I^(-) hArr I_(3)^(-),)" "E^(@)=0.6197-0.5355),(" "=0.0842" volt"):}` `K="antilog" [(nE^(@))/0.0591]="antilog"[(2xx0.0842)/0.0591]=706.9`

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Calculate equilibrium constant for `I_(2)+I^(-) hArr I_(3)^(-)` at 298 K from the following information : `{:(I_(2) (aq.)+2e^(-) rarr 2I^(-),,E^(@)=0.6197" volt"),(I_(3)^(-)+2e^(-) rarr 3I^(-),,E^(@)=0.5355" volt"):}`

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