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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.

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The correct Answer is:
`K_C=6.26xx10^7`
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Calculate the euilibrium constant for the reaction, 2Fe^(3+) + 3I^(-) hArr 2Fe^(2+) + I_(3)^(-) . The standard reduction potential in acidic conditions are 0.77 V and 0.54 V respectivelu for Fe^(3+)//Fe^(2+) and I_(3)^(-)//I^(-) couples.

Calculate the equilibrium constant for the reaction : 2Fe^(3+)+3I^(Θ)hArr2Fe^(2+)+I_(3)^(Θ) The standard reduction potential in acidic conditions is 0.78V and 0.54V , respectively, for Fe^(3+)|Fe^(2+) and I_(3)^(c-)|I^(c-) couples

Write the half reaction for the reaction 2Fe^(+3) + 2I^(-) to 2Fe^(+2) + I_(2)

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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)

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