A For the reaction2AgCIs - H2g 1atm - 2Ag s - 2H- 0-1 M - 2CI- 0-1 M- G- - - 43600 J at 25- C-Calculate the e-m-f- of the cell-log 10-n - -n-b Define fuel cell and write its two advantages-

(a) nbsp; Δ G^∘ = nFE^∘_cell ⇒ 43600 = 2 × 96500 × E^∘_cell E^∘_cell = 0.226 V E_cell = E^∘_cell 0.0591/n log [H^+]^2 [CI^ ]^2/[H_2] = 0.226 0.0591 ...

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Byju's Answer

Standard XII

Chemistry

Third Law of Thermodynamics

a For the rea...

Question

(a) For the reaction

$2 A g C I (s) + H_{2} (g) (1 a t m) ⟶ 2 A g (s) + 2 H^{+} (0.1 M) + 2 C I^{-} (0.1 M),$

$Δ G^{\circ}$ = - 43600 J at $25^{\circ} C .$

Calculate the e.m.f. of the cell

$[l o g 10^{- n} = - n]$

(b) Define fuel cell and write its two advantages.

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Solution

(a) $Δ G^{\circ} = - n F E_{c e l l}^{\circ}$

$\Rightarrow - 43600 = - 2 \times 96500 \times E_{c e l l}^{\circ}$

$E_{c e l l}^{\circ}$ = 0.226 V

$E_{c e l l} = E_{c e l l}^{\circ} - \frac{0.0591}{n} l o g \frac{[H^{+}]^{2} [C I^{-}]^{2}}{[H_{2}]}$

= $0.226 - \frac{0.0591}{2} l o g \frac{(0.1)^{2} (0.1)^{2}}{(1)}$

= $0.226 - \frac{0.0591}{2} l o g 10^{- 4}$

= $0.226 - \frac{0.0591}{2} \times - 4$

= 0.226 + 0.118 = 0.344 V

(b) Cell which converts the chemical energy, produced by the combustion of fuels into electrical energy is called fuel cell.

Ex : $H_{2} - O_{2}$ fuel cell.

Advantages :

(i) High efficiency

(ii) Non-polluting

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Similar questions

Q. (a) For the reaction

2 A g C I (s) + H_{2} (g) (1 a t m) ⟶ 2 A g (s) + 2 H^{+} (0.1 M) + 2 C I^{-} (0.1 M),

Δ G^{\circ}

= - 43600 J at

25^{\circ} C .

Calculate the e.m.f. of the cell

[l o g 10^{- n} = - n]

(b) Define fuel cell and write its two advantages.

Q. For the reaction.

2 A g C l (s) + H_{2} (g) (1 a t m) \to 2 A g (s) + 2 H^{+} (0.1 M) + 2 C l^{-} (0.1 M),

Δ G^{o} = - 43600

J at

25^{o}

C.
Calculate the e.m.f of the cell.

[l o g 10^{- n} = - n]

Q. Define fuel cell and write its two advantages.

Q. The e.m.f. of a cell corresponding to the reaction

Z n + 2 H^{+} (a q) \to Z n^{2 +} (0.1 M) + H_{2} (g) (1 a t m)

is 0.26 volt at

25^{\circ}

C. The pH of the solution at the hydrogen electrode is:

(E_{Z n^{2 +} / Z n}^{\circ} = - 0.76

V and

E_{H^{+} / H_{2}}^{\circ} = 0)

Q. The emf of a cell corresponding to the reaction,

Z n + 2 H^{+} (a q .) \to Z n^{2 +} (0.1 M) + H_{2} (g) 1 a t m

0.28

volt at

25^{\circ} C

. Write the half-cell reactions and calculate the

p H

of the solution at the hydrogen electrode.

E_{Z n^{2 +} / Z n}^{\circ} = - 0.76 v o l t

and

E_{H^{+} / H_{2}}^{\circ} = 0

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(a) For the reaction 2AgCI(s)+H2(g)(1atm)⟶2Ag(s)+2H+(0.1M)+2CI−(0.1M), ΔG∘ = - 43600 J at 25∘C. Calculate the e.m.f. of the cell [log 10−n=−n] (b) Define fuel cell and write its two advantages.

(a) For the reaction

$2 A g C I (s) + H_{2} (g) (1 a t m) ⟶ 2 A g (s) + 2 H^{+} (0.1 M) + 2 C I^{-} (0.1 M),$

$Δ G^{\circ}$ = - 43600 J at $25^{\circ} C .$

Calculate the e.m.f. of the cell

$[l o g 10^{- n} = - n]$

(b) Define fuel cell and write its two advantages.