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Nernst Equation

An excess of ...

Question

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 $2 H g / H g_{2}^{2 +}$ at $25^{o}$ C for,
$2 H g + 2 F e^{3 +} ⇌ H g_{2}^{2 +} + 2 F e^{2 +}$ and $E_{F e^{2 +} / F e^{3 +}}^{o} = - 0.7712 V$ .

A

-0.7912 V

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B

-0.7721 V

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C

-0.9922 V

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D

None of these

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Solution

The correct option is A -0.7912 V
The given reaction is
$2 H g + 2 F e^{3 +} ⇌ H g_{2}^{2 +} + 2 F e^{2 +}$
Before reaction Excess $10^{- 3}$ 0 0
After reaction Excess $\frac{4.6 \times 10^{- 3}}{100}$ $\frac{95.4 \times 10^{- 3}}{100 \times 2}$ $\frac{95.4 \times 10^{- 3}}{100}$
For cell at equilibrium;
$E_{c e l l} = 0 = E_{O P_{H g / H g_{2}^{2 +}}} + E_{R P_{F e^{3 +} / F e^{2 +}}}$
$0 = E_{O P_{H g / H g_{2}^{2 +}}}^{o} - \frac{0.059}{2} l o g [H g_{2}^{2 +}] + E_{R P_{F e^{3 +} / F e^{2 +}}}^{o} + \frac{0.059}{2} l o g \frac{[F e^{3 +}]^{2}}{[F e^{2 +}]^{2}}$
or $0 = E_{O P_{H g / H g_{2}^{2 +}}}^{o} + 0.771 + \frac{0.059}{2} l o g \frac{[F e^{3 +}]^{2}}{[F e^{2 +}]^{2} [H g_{2}^{2 +}]}$
$(∵ E_{O P_{F e^{2 +} / F e^{3 +}}}^{o}) = - 0.77 V ∴ E_{R P_{F e^{3 +} / F e^{2 +}}}^{o} = + 0.771 V)$
$E_{O P_{H g / H g^{2 +}}}^{p} = - 0.771 - \frac{0.059}{2} l o g \frac{[4.6 \times 10^{- 3} / 100]^{2}}{{[\frac{95.4 \times 10^{- 3}}{100}]}^{2} [\frac{95.4 \times 10^{- 3}}{2 \times 100}]}$
$= - 0.7912$ V

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Similar questions

Q. An excess of

H g

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

E^{\circ}

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

25^{\circ} C

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

.

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)
Given : $E_{F e^{3 +} / F e^{2 +}}^{⊖}$ = 0.77 V.

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

{H g}_{2}^{2 +} = 0.5 M

{F e}^{2 +} = 0.03 M

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

, what will be the concentration of ions at equilibrium?

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

. What will be the concentrations of ions at equilibrium?

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