Hg22- - 2 e- - 2Hg - E0 - 0-789 V Hg2- - 2 e- - Hg - E0 - 0-854 VCalculate the equilibrium constant for Hg22- Hg - Hg2- -

The correct option is C 6.3 × 10^ 3 Hg_2^2++2e^ → 2Hg ⟹ #160; E^0 = 0.789 V Hg → Hg^2++2e^ ⟹ E^0 = 0.854 V Hg_2^2+→ Hg+Hg^2+⟹ 0.065 V ...

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Byju's Answer

Standard XII

Chemistry

Equilibrium Constant from Nernst Equation

Hg22+ + 2 e- ...

Question

$H g_{2}^{2 +} + 2 e^{-} \to 2 H g ⟹ E^{0} = 0.789 V$
$H g^{2 +} + 2 e^{-} \to H g ⟹ E^{0} = 0.854 V$
Calculate the equilibrium constant for $H g_{2}^{2 +} \to H g + H g^{2 +}$ .

3.1 \times 10^{- 3}

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3.1 \times 10^{- 4}

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6.3 \times 10^{- 3}

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6.3 \times 10^{- 4}

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Solution

The correct option is C $6.3 \times 10^{- 3}$
$H g_{2}^{2 +} + 2 e^{-} \to 2 H g ⟹ E^{0} = 0.789 V$
$H g \to H g^{2 +} + 2 e^{-} ⟹ E^{0} = - 0.854 V$

$H g_{2}^{2 +} \to H g + H g^{2 +} ⟹ - 0.065 V$

$Δ G = - 2 \times (- 0.065) \times 96500 = - 8.314 \times 298 l n K_{e q}$
$K_{e q} = 6.3 \times 10^{- 3}$ .

Hence, option C is correct.

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Similar questions

Q. An excess of

H g

was added to

10^{- 3} M

acidified solution of

F e^{3 +}

ions. Its was found that only

4.6

% of the ions remained as

F e^{3 +}

at equilibrium at

25^{\circ} C

. Calculate

E^{\circ}

for

2 H g / H g_{2}^{2 +}

25^{\circ} C

for

2 H g + 2 F e^{3 +} ⇌ H g_{2}^{2 +} + 2 F e^{2 +}

Q. Calculate the standard potential for the reaction,

H g_{2} C l_{2} + C l_{2} \to 2 H g^{+} + 4 C l^{-}

Given:

H g_{2} C l_{2} + 2 e^{-} \to 2 H g + 2 C l^{-}; E^{\circ} = 0.270 v o l t

H g_{2}^{2 +} \to 2 H g^{2 +} + 2 e^{-}; E^{\circ} = - 0.92 v o l t

2 H g \to H g_{2}^{2 +} + 2 e^{-}; E^{\circ} = - 0.79 v o l t

C l_{2} + 2 e^{-} \to 2 C l^{-}; E^{\circ} = 1.36 v o l t

Q. Calculate

K_{c}

for

H g^{2 +} + H g ⇌ H g_{2}^{2 +}

Give that

E_{2 H g / H g^{2 +}}^{o} = - 0.788 V

and

E_{H g_{2}^{2 +} / 2 H g^{2 +}}^{o} = - 0.920 V

Write answer as nearest integer after dividing by 100.

Q. An excess of liquid Hg was added to

10^{- 3}

M acidified solution of

F e^{3 +}

ions. It was found that only 4.6% of the ions remained as

F e^{3 +}

at equilibrium at

25^{o}

C. Calculate

E^{o}

for

2 H g / H g_{2}^{2 +}

25^{o}

C for,

2 H g + 2 F e^{3 +} ⇌ H g_{2}^{2 +} + 2 F e^{2 +}

and

E_{F e^{2 +} / F e^{3 +}}^{o} = - 0.7712 V

Q. In which direction can the reaction,

2 H g (l) + 2 A g^{+} (a q .) ⇌ 2 A g (s) + H g_{2}^{2 +} (a q .)

proceed spontaneously at the following concentrations of the ions participating in the reactions (i) and (ii)?
(i)

[A g^{+}] = 10^{- 4} m o l L^{- 1}

and

[H g_{2}^{2 +}] = 10^{- 1} m o l L^{- 1}

(ii)

[A g^{+}] = 10^{- 1} m o l L^{- 1}

and

[H g_{2}^{2 +}] = 10^{- 4} m o l L^{- 1}

Given:

E_{H g_{2}^{2 +} / H g}^{\circ} = 0.79 V; E_{A g^{+} / A g}^{\circ} = 0.80 V

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Hg2+2+2e−→2Hg⟹ E0=0.789VHg2++2e−→Hg⟹ E0=0.854VCalculate the equilibrium constant for Hg2+2→Hg+Hg2+ .

The correct option is C 6.3×10−3Hg2+2+2e−→2Hg⟹E0=0.789 VHg→Hg2++2e−⟹E0=−0.854 VHg2+2→Hg+Hg2+⟹−0.065 VΔG=−2×(−0.065)×96500=−8.314×298 lnKeqKeq=6.3×10−3.Hence, option C is correct.

$H g_{2}^{2 +} + 2 e^{-} \to 2 H g ⟹ E^{0} = 0.789 V$
$H g^{2 +} + 2 e^{-} \to H g ⟹ E^{0} = 0.854 V$
Calculate the equilibrium constant for $H g_{2}^{2 +} \to H g + H g^{2 +}$ .