An excess of liquid mercury is added to an acidified solution of 1-0 - 10-3 M Fe3- It is found that 5

2Hg + 2Fe^3+ #160;→ Hg^2+_2 + 2Fe^2+ Fe^3+ ion concentration #160;originally = #160; 10^ 3 M = 5100 × 10^ 3 = 5 × 10^ 5 M Fe^3+ ...

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Standard XII

Chemistry

Nernst Equation

An excess of ...

Question

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.

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Solution

$2 H g + 2 F e^{3 +} \to H g_{2}^{2 +} + 2 F e^{2 +}$
$F e^{3 +}$ ion concentration originally = $10^{- 3}$ M
$= \frac{5}{100} \times 10^{- 3} = 5 \times 10^{- 5} M$
$F e^{3 +}$ converted into $F e^{2 +}$ ion
= $(1.0 \times 10^{- 3}) - (5 \times 10^{- 5}) = 0.95 \times 10^{- 3}$
$H g_{2}^{2 +}$ ions at eq = $\frac{0.95 \times 10^{- 3}}{2}$
$E_{c e l l} = E_{c e l l}^{⊖} - \frac{0.059}{n} l o g \frac{[H g_{2}^{2 +}] [F e^{2} +]^{2}]}{[F e^{3 +}]^{2}}$
At equivalent, $E_{c e l l} = 0$
$∴ 0 = E^{⊖} - \frac{0.0591}{2} l o g \frac{\frac{0.95 \times 10^{- 3}}{2} [0.95 \times 10^{- 3}]^{2}}{[5 \times 10^{- 5}]^{2}}$
$E_{c e l l}^{⊖} = \frac{0.0591}{2} l o g [95]^{3} 10^{- 5} 50 = - 0.0276$
$E_{c e l l}^{⊖} = E_{F e^{3 +} I F e^{2 +}}^{⊖} - E_{H g_{2}^{2 +} I H g}^{⊖}$
$- 0.0276 = 0.77 - E_{H g_{2}^{2 +} I H g}^{⊖}$
$E_{H g^{2 +} I H g}^{⊖} = 0.77 + 0.0276$
= 0.7976 V

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Similar questions

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

% of

F e^{3 +}

remains at equilibrium at

25^{\circ} C

. Calculate

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 +}

(Given,

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

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. 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

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for

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

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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. The standard electrode potential (reduction) of

P t / F e^{2 -}, F e^{3 +}

and

P t / S n^{4 +}, S n^{2 +}

are 0.77 V and 0.15 v respectively at

25^{o}

C The standard EMF of the reaction

S n^{4 +} + 2 F e^{2 +} \to S n^{2 +} + 2 F e^{3 +}

is:

K_{a}

for the ionisation of

F e^{3 +}

F e (O H)^{2 +}

and

H^{\oplus}

6.5 \times 10^{- 3}

what is the maximum pH value which could be used so that at least 95% of the total

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(write your answer to nearest integer)?

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An excess of liquid mercury is added to an acidified solution of 1.0×10−3 M Fe3+. It is found that 5 % of Fe3+ remains that the only reaction that occurs is:2Hg+2Fe2+→Hg2+2+2Fe2+. (write your answer to nearest integer)Given : E⊖Fe3+/Fe2+ = 0.77 V.