Consider the following cell reaction-2 AgCl s - H 2 g - 2 Ag s -2 H - aq -2 Cl - aq - E 0-0-19 VAt - Cl -10-3 M and P H 2-0-01 atm- if the cell potential is 0-43 V at 25- C the pH is- Take -2-303 R - 298- F -0-06

Using Nernst Equation, E_cell = E^∘_cell 0.062 log[H^+]^2 [Cl^ ]^2P_H_2 ⇒ 0.43=0.19 0.062 log[H^+]^2× (10^ 3)^210^ 2 ⇒0.43=0.19 0.062 log[H^+]^2× (10^ 2)^2 ⇒ 0. ...

You visited us 6 times! Enjoying our articles? Unlock Full Access!

Byju's Answer

Standard XII

Chemistry

Nernst Equation

Consider the ...

Question

Consider the following cell reaction:
$2 A g C l (s) + H_{2} (g) \to 2 A g (s) + 2 H^{+} (a q) + 2 C l^{-} (a q); E^{o} = 0.19 V$
At $[C l^{-}] = 10^{- 3} M$ and $P_{H_{2}} = 0.01 atm$ , if the cell potential is 0.43 $V$ at $25^{o} C$ the $p H$ is:
$(Take \frac{2.303 R \times 298}{F} = 0.06)$ .

Open in App

Solution

Using Nernst Equation,
$E_{c e l l} = E_{c e l l}^{\circ} - \frac{0.06}{2} log \frac{[H^{+}]^{2} [C l^{-}]^{2}}{P_{H_{2}}}$
$\Rightarrow 0.43 = 0.19 - \frac{0.06}{2} log \frac{[H^{+}]^{2} \times (10^{- 3})^{2}}{10^{- 2}} \Rightarrow 0.43 = 0.19 - \frac{0.06}{2} log [H^{+}]^{2} \times (10^{- 2})^{2} \Rightarrow 0.43 = 0.19 - 0.06 log [H^{+}] + 0.12 \Rightarrow 0.12 = 0.06 p H \Rightarrow p H = 2$

Suggest Corrections

Similar questions

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. 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.

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:

Q. Consider the following cell reaction:

2 F e (s) + O_{2} (g) + 4 H^{+} (a q) \to 2 F e^{2 +} (a q) + 2 H_{2} O (l); E^{\circ} = 1.67 V

[F e^{2 +}] = 10^{- 3} M, P_{(O_{2})} = 0.1 a t m

and pH = 3, the cell potential at

25^{\circ} C

is
(given:

\frac{2.303 R T}{F} = 0.0591

)

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.

Join BYJU'S Learning Program

Consider the following cell reaction: 2AgCl(s)+H2(g)→2Ag(s)+2H+(aq)+2Cl−(aq);Eo=0.19 V At [Cl−]=10−3 M and PH2=0.01 atm, if the cell potential is 0.43 V at 25 oC the pH is: (Take 2.303R×298F=0.06).

Using Nernst Equation, Ecell=E∘cell−0.062 log[H+]2[Cl−]2PH2 ⇒0.43=0.19−0.062 log[H+]2×(10−3)210−2⇒0.43=0.19−0.062 log[H+]2×(10−2)2⇒0.43=0.19−0.06 log[H+]+0.12⇒0.12=0.06 pH⇒pH=2

Consider the following cell reaction:
$2 A g C l (s) + H_{2} (g) \to 2 A g (s) + 2 H^{+} (a q) + 2 C l^{-} (a q); E^{o} = 0.19 V$
At $[C l^{-}] = 10^{- 3} M$ and $P_{H_{2}} = 0.01 atm$ , if the cell potential is 0.43 $V$ at $25^{o} C$ the $p H$ is:
$(Take \frac{2.303 R \times 298}{F} = 0.06)$ .