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An excess of ...

Question

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 $2 H g / H g_{2}^{2 +}$ at $25^{\circ} C$ for
$2 H g + 2 F e^{3 +} ⇌ H g_{2}^{2 +} + 2 F e^{2 +}$ .

A

- 0.9121 v o l t

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B

0.9121 v o l t

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C

- 0.7912 v o l t

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D

0.7912 v o l t

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Solution

The correct option is D $- 0.7912 v o l t$
$2 H g + 2 F e^{+ 3} \to H g_{2}^{2 +} + 2 F e^{2 +}$
$i n i t i a l c o n c e n t r a t i o n o f F e^{3 +} = 10^{- 3} M$
$e q u i l i b r i u m c o n c e n t r a t i o n o f F e^{3 +} = \frac{4.6}{100} 10^{- 3} M = 4.6 \times 10^{- 5}$
$e q u i l i b r i u m c o n c e n t r a t i o n o f F e^{2 +} = 95.4 % o f 10^{- 3} M = 0.954 \times 10^{- 3}$
$e q u i l i b r i u m c o n c e n t r a t i o n o f H g_{2}^{2 +} = H a l f o f t h e e q i l i b r i u m c o n c e n t r a t i o n o f F e^{+ 2}$
$= 0.477 \times 10^{- 3} M$
$E_{c e l l}^{\circ} = \frac{0.0591}{1} log \frac{(0.954)^{2} \times 10^{- 6} \times 0.477 \times 10^{- 3}}{(4.6)^{2} \times 10^{- 10}}$
$E_{c e l l}^{\circ} = 0.0406$
$E_{\frac{H g}{H g^{2 +}}}^{\circ} = E_{c e l l}^{\circ} - E_{r e d}^{\circ}$
$= - 0.0402 - 0.775$
$= - 0.81$
$≃ - 0.7912 v o l t$

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Similar questions

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}

E^{o}

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

25^{o}

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

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

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Q. An excess of liquid mercury is added to an acidified solution of

1.0 \times 10^{- 3} M F e^{3 +}

. It is found that

5

F e^{3 +}

remains at equilibrium at

25^{\circ} C

E_{H g_{2}^{2 +} / H g}^{\circ}

assuming that the only reaction that occurs is

2 H g + 2 F e^{3 +} \to H g_{2}^{2 +} + 2 F e^{2 +}

E_{F e^{3 +} / F e^{2 +}} = 0.77 v o l t)

An excess of liquid mercury is added to an acidified solution of $1.0 \times 10^{- 3}$ M $F e^{3 +}$ . It is found that 5 % of $F e^{3 +}$ remains that the only reaction that occurs is:

$2 H g + 2 F e^{2 +} \to H g_{2}^{2 +} + 2 F e^{2 +}$ . (write your answer to nearest integer)
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Q. For the reaction,

2 {F e}^{3 +} (a q) + {({H g}_{2})}^{2 +} (a q) ⇌ 2 {F e}^{2 +} (a q) + 2 {H g}^{2 +} (a q)

K_{c} = 9.14 \times 10^{- 6}

25

. If the initial concentration of the ions are

[{F e}^{3 +}] = 0.5 M, [{H g}_{2}^{2 +}] = 0.5 M, [{F e}^{2 +}] = 0.03 M

[{H g}^{2 +}] = 0.03 M

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Q. For the reaction:

2 {F e}^{3 +} (a q) + {{(H g}_{2})}^{2 +} (a q) ⇌ 2 {F e}^{2 +} (a q) + {2 H g}^{2 +} (a q)

K_{C} = 9.14 \times 10^{- 6}

at 25C. If the initial concentration of the ions are

{F e}^{3 +} = 0.5 M

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