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Calculate the equilibrium constant for the reaction, `2Fe^(3+)+3I^(-)hArr2Fe^(2+)+I_(3)^(-)`, the standard reduction potentials in acidic conditions are 0.77V and 0.54V respectively for `Fe^(3+)//Fe^(2+)` and `I_(3)^(-)//I^(-)` couples.

Text Solution

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The correct Answer is:
`6.2605 xx 10^(7)`
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Reaction : 2Fe^(3+)+3I^(-) hArr 2Fe^(2+) + I_3^(-) The standard reduction potentials in acidic conditions are 0.77 V and 0.54 V respectively for cathodic and anodic reactions. The equilibrium constant for the reaction is approximately . (Given 10^(7.79) = 6.26 xx10^7)