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

2Hg + 2Fe^3+→ Hg_2^2+ + 2Fe^2+ At equilibrium, excess 10^ 3× 510010^ 3× 952× 10010^ 3× 95100 At equilibrium, E_cell = 0 0 = E_cell^∘ 0.0591 ...

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Standard XII

Chemistry

EMF

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

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Solution

$2 H g + 2 F e^{3 +} \to H g_{2}^{2 +} + 2 F e^{2 +}$
At equilibrium, excess $\frac{10^{- 3} \times 5}{100} \frac{10^{- 3} \times 95}{2 \times 100} \frac{10^{- 3} \times 95}{100}$
At equilibrium, $E_{c e l l} = 0$
$0 = E_{c e l l}^{\circ} - \frac{0.0591}{2} log \frac{[H g_{2}^{2 +}] [F e^{2 +}]^{2}}{[F e^{3 +}]^{2}}$
$= (E_{H g / H g_{2}^{2 +}}^{\circ} + E_{F e^{3 +} / F e^{2 +}}^{\circ}$
$- \frac{0.0591}{2} log \frac{(\frac{10^{- 3} \times 95}{2 \times 100}) {(\frac{10^{- 3} \times 95}{100})}^{2}}{{(\frac{10^{- 3} \times 5}{100})}^{2}}$
$E_{H g / H g_{2}^{2 +}}^{\circ} = - 0.77 + \frac{0.0591}{2} log \frac{(95)^{3} \times 10^{- 5}}{25 \times 2}$
$= - (0.77 + 0.0226)$
$= - 0.7926 v o l t$
$E_{H g_{2}^{2 +} / H g}^{\circ} = + 0.7926 v o l t$ .

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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 +}

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. 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. 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

and

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

. What will be the concentrations 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}

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

and

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

, what will be the concentration of ions at equilibrium?

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An excess of liquid mercury is added to an acidified solution of 1.0×10−3M Fe3+. It is found that 5% of Fe3+ remains at equilibrium at 25∘C. Calculate E∘Hg2+2/Hg assuming that the only reaction that occurs is2Hg+2Fe3+→Hg2+2+2Fe2+(Given, EFe3+/Fe2+=0.77 volt).