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Standard XII

Chemistry

Equilibrium Constant from Nernst Equation

Calculate equ...

Question

Calculate equilibrium constant for
$I_{2} + I^{-} ⇌ I_{3}^{-}$
at $298 K$ from the following information:
$I_{2} (a q .) + 2 e^{-} \to 2 I^{-}; E^{\circ} = 0.6197 v o l t$
$I_{3}^{-} + 2 e^{-} \to 3 I^{-}; E^{\circ} = 0.5355 v o l t$ .

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Solution

$I_{2} + 2 e^{-} \to 2 I^{-}; E^{\circ} = 0.6197 v o l t$
$3 I^{-} \to I_{3}^{-} + 2 e^{-} - ---------------- -; E^{\circ} = - 0.5355 v o l t$
$I_{2} + I^{-} ⇌ I_{3}^{-}; E^{\circ} = 0.6197 - 0.5355$
$= 0.0842 v o l t$
$K = a n t i l o g [\frac{n E^{\circ}}{0.0591}] = a n t i l o g [\frac{2 \times 0.0842}{0.0591}] = 706.9$ .

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Similar questions

Let $I_{1} = {(\frac{π}{4})}^{2} + \sqrt{2}, I_{2} = {(t a n^{- 1} (\frac{1}{e}))}^{2} + \frac{2 e}{\sqrt{e^{2} + 1}}, I_{3} = (t a n^{- 1} e)^{2} + \frac{2}{\sqrt{e^{2} + 1}}$ , then which of the following is true?

Given :

H_{2} O_{2} \to O_{2} + 2 H^{\oplus} + 2 e^{-}

e^{\oplus} = - 0.69 V

H_{2} O_{2} + 2 H^{\oplus} + 2 e^{-} \to 2 H_{2} O

E^{\oplus} = 1.77 V

2 I^{⊖} \to I_{2} + 2 e^{-}

E^{⊖} = - 0.535 V

Assertion:

H_{2} O_{2}

acts as an oxidizing agent for

I^{⊖}

Reason:

E^{⊖}

is positive.

Given :

Oxidation :

$H_{2} O_{2} \to O_{2} + 2 H^{\oplus} + 2 e^{-}$

$E^{⊖} = - 0.69 V$ ;

$2 F^{⊖} \to F_{2} + 2 e^{-}$

$E^{⊖} = - 2.87 V$ ;

Reduction :

$H_{2} O_{2} + 2 H^{\oplus} + 2 e^{-} \to 2 H_{2} O$

$E^{⊖} = - 1.77 V$ ;

$2 I^{⊖} \to I_{2} + 2 e^{-}$ $E^{⊖} = - 0.54 V$ ;

Which of the following statements is/are correct ?

Q. Calculate the half-cell potential at

298 K

for the reaction,

Z n^{2 +} + 2 e^{-} \to Z n

[Z n^{2 +}] = 0.1 M

and

E^{\circ} = - 0.76 v o l t

Q. A saturated solution of iodine in water contains

0.33

g iodine/L. More than this can dissolve in a

K I

solution because of the following equilibrium:

I_{2} (a q) + I^{-} \to I_{3} - (a q)

. A

0.1 M

K I

solution

(0.1 M

I^{-})

actually dissolves

12.5

g of Iodine/L, most of which is converted to

I_{3}^{-} .

Assuming that the concentration of

I_{2}

in all saturated solutions is the same, calculate the equilibrium constant for the above reaction: (Mol. wt. of

I

258 g m o l^{- 1}

)

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Calculate equilibrium constant forI2+I−⇌I−3at 298 K from the following information:I2(aq.)+2e−→2I−;E∘=0.6197 voltI−3+2e−→3I−;E∘=0.5355 volt.

I2+2e−→2I−;E∘=0.6197 volt3I−→I−3+2e−––––––––––––––––––;E∘=−0.5355 voltI2+I−⇌I−3;E∘=0.6197−0.5355=0.0842 voltK=antilog[nE∘0.0591]=antilog[2×0.08420.0591]=706.9.

Calculate equilibrium constant for
$I_{2} + I^{-} ⇌ I_{3}^{-}$
at $298 K$ from the following information:
$I_{2} (a q .) + 2 e^{-} \to 2 I^{-}; E^{\circ} = 0.6197 v o l t$
$I_{3}^{-} + 2 e^{-} \to 3 I^{-}; E^{\circ} = 0.5355 v o l t$ .