The standard free energy change for the reaction- H2g - 2AgCls- 2Ags - 2H-aq- - 2Cl-aq-is -10-26 kcal mol-1 at 25-C- A cell using the above reaction is operated at 25-C under PH2 - 1 atm- -H- and -Cl- - 0-1- Calculate the emf of the cell-

The correct option is C 0.340 volt Δ G^∘= 10.26 K cal mol^ 1 Δ G^∘= nFE^∘ E^∘= 10.26× 10^3× 4.159/ 2× 96500 E^∘=0.22109 E = E ...

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EMF

The standard ...

Question

The standard free energy change for the reaction,
$H_{2} (g) + 2 A g C l (s) \to 2 A g (s) + 2 H^{+} (a q .) + 2 C l^{-} (a q .)$
is $- 10.26 k c a l m o l^{- 1}$ at $25^{\circ} C$ . A cell using the above reaction is operated at $25^{\circ} C$ under $P_{H_{2}} = 1 a t m . [H^{+}]$ and $[C l^{-}] = 0.1$ . Calculate the emf of the cell.

0.140 v o l t

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0.240 v o l t

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0.340 v o l t

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0.440 v o l t

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10.26 k c a l m o l^{- 1}

25^{\circ} C .

H_{2} (g) + 2 A g C l (s) \to 2 A g (s) + 2 H^{+} (a q) + 2 C l^{-} (a q)

25^{\circ} C

P_{H_{2}} = 1 a t m

[H^{+}] and [C l^{-}] = 0.1 M

10.26 k c a l m o l^{- 1}

25^{\circ} C .

H_{2} (g) + 2 A g C l (s) \to 2 A g (s) + 2 H^{+} (a q) + 2 C l^{-} (a q)

25^{\circ} C

P_{H_{2}} = 1 a t m

[H^{+}] and [C l^{-}] = 0.1 M

\frac{2.303 R T}{F} = 0.059, R = 8.314 J K^{- 1}, F = 96500 C

2 A g C l (s) + H_{2} (g) \to 2 A g (s) + 2 H^{+} (a q) + 2 C l^{-} (a q); E^{o} = 0.19 V

[C l^{-}] = 10^{- 3} M

P_{H_{2}} = 0.01 atm

V

25^{o} C

p H

(Take \frac{2.303 R \times 298}{F} = 0.06)

Δ G^{0} = - 42927

25^{0} C

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

C d (s) | C d C l_{2} (a q, 0.1 M) | | A g C l (s) | A g (s)

C d (s) + 2 A g C l (s) \to 2 A g (s) + C d^{2 +} (a q) + 2 C l^{-} (a q)

0.6915 V

0.6753 V

Δ S

Δ H

F = 96485 C m o l^{- 1}

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