For the reaction-H2g 1 atm - 2AgCls- 2Ags - 2H- aq 0-1 M - 2Cl-aq 0-1 M 1 atm- G-42927 joules at 25-C-What is the emf of the cell in which of the reaction takes place- -

2AgCl + H_2(1 atm) nbsp; → 2Ag + 2H^+ (aq) (0.1 M) + 2Cl^ (aq.) (0.1 M) (1 atm) Given nbsp;Δ G^∘= 42927 J E^∘= Δ G^∘nF= ( 42927)2 × 96500=0.22 V E_ce ...

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Standard XII

Chemistry

Spontaneity

For the react...

Question

For the reaction,
$H_{2} (g) (1 atm) + 2 A g C l (s) \to 2 A g (s) + 2 H^{+} (aq) (0.1 M) + 2 C l^{-} (aq) (0.1 M) (1 atm)$
$Δ G^{\circ} = - 42927 joules$ at $25^{\circ} C$ .
What is the emf of the cell in which of the reaction takes place.

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Solution

$2 A g C l + H_{2} (1 atm) \to 2 A g + 2 H^{+} (aq) (0.1 M) + 2 C l^{-} (aq.) (0.1 M) (1 atm)$

Given $Δ G^{\circ} = - 42927 J$

$E^{\circ} = - \frac{Δ G^{\circ}}{n F} = \frac{- (- 42927)}{2 \times 96500} = 0.22 V$

$E_{cell} = E^{\circ} - \frac{0.0591}{n} log \frac{[H^{+}]^{2} [C l^{-}]^{2}}{[H_{2}]}$

$\Rightarrow E_{cell} = 0.22 - \frac{0.0591}{2} log \frac{(0.1)^{2} (0.1)^{2}}{1}$

$\Rightarrow E_{cell} = 0.22 - \frac{0.0591}{2} log 10^{- 4}$

$\Rightarrow E_{cell} = 0.22 + 0.118$

$E_{cell} = 0.338 V$

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Similar questions

Q. For the reaction,

Δ G^{0} = - 42927

joules at

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)

Calculate the emf of the cell of given reaction.

Q. The standard free energy change for the given reaction is -

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)

A cell using the above reaction is operated at

25^{\circ} C

under conditions,

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

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

Calculate the emf of the cell.

Q. 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 .)

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

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.

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]

Δ G^{\circ}

of the cell reaction,

A g C l (s) + \frac{1}{2} H_{2} (g) \to A g (s) + H^{+} + C l^{-} is - 21.52 K J . Δ G^{\circ} of 2 A g C l (s) + H_{2} (g) \to 2 A g (s) + 2 H^{+} 2 C l^{-}

is:

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For the reaction, H2(g)(1 atm)+2AgCl(s)→2Ag(s)+2H+(aq)(0.1 M)+2Cl−(aq)(0.1 M)(1 atm) ΔG∘=−42927 joules at 25∘C. What is the emf of the cell in which of the reaction takes place. ​​​​​​​

2AgCl+H2(1 atm)→2Ag+2H+(aq)(0.1 M)+2Cl−(aq.)(0.1 M)(1 atm) Given ΔG∘=−42927 J E∘=−ΔG∘nF=−(−42927)2×96500=0.22 V Ecell=E∘−0.0591nlog[H+]2[Cl−]2[H2] ⇒Ecell=0.22−0.05912log(0.1)2(0.1)21 ⇒Ecell=0.22−0.05912log10−4 ⇒Ecell=0.22+0.118 Ecell=0.338 V

For the reaction,
$H_{2} (g) (1 atm) + 2 A g C l (s) \to 2 A g (s) + 2 H^{+} (aq) (0.1 M) + 2 C l^{-} (aq) (0.1 M) (1 atm)$
$Δ G^{\circ} = - 42927 joules$ at $25^{\circ} C$ .
What is the emf of the cell in which of the reaction takes place.