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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

Verified by Experts

The correct Answer is:
`K_C=6.26xx10^7`
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Explore conceptually related problems

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

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

Knowledge Check

  • 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)

    A
    `6.26xx10^(-7)`
    B
    `5.33xx10^(-4)`
    C
    `6.26xx10^(7)`
    D
    `5.33xx10^(4)`
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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.

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