Consider the following cell reaction-2Fes - O2 g -4H-aq - 2Fe2-aq - 2H2 Ol - E- - 1-67 VAt -Fe2-10-3M- PO2-0-1 atm and pH - 3- the cell potential at 25 - C isgiven- 2-303RTF - 0-0591

Nernst equation for the cell reaction E=E^∘ 0.0591nlog[Fe^2+]^2P_O_2.[H^+]^4 Given, E^∘ = 1.67 V nbsp; Concentration of Fe^2+,[Fe^2+] = 10^ 3 M Concentratio ...

You visited us 1 times! Enjoying our articles? Unlock Full Access!

Byju's Answer

Standard XI

Chemistry

Nernst Equation

Consider the ...

Question

Consider the following cell reaction:
$2 F e (s) + O_{2} (g) + 4 H^{+} (a q) \to 2 F e^{2 +} (a q) + 2 H_{2} O (l); E^{\circ} = 1.67 V$
At $[F e^{2 +}] = 10^{- 3} M, P_{(O_{2})} = 0.1 a t m$ and pH = 3, the cell potential at $25^{\circ} C$ is
(given: $\frac{2.303 R T}{F} = 0.0591$ )

Open in App

Solution

Nernst equation for the cell reaction
$E = E^{\circ} - \frac{0.0591}{n} l o g \frac{[F e^{2 +}]^{2}}{P_{O_{2}} . [H^{+}]^{4}}$

Given, $E^{\circ} = 1.67 V$
Concentration of $F e^{2 +}, [F e^{2 +}] = 10^{- 3} M$
Concentration of $H^{+}, [H^{+}] = 10^{- 3} M$
Pressure of oxygen, $P_{(O_{2})} = 0.1 a t m$
number of electrons involved, n = 4
Hence,

$E = 1.67 - \frac{0.0591}{4} l o g \frac{(10^{- 3})^{2}}{0.1 \times (10^{- 3})^{4}} = 1.57 V$

Suggest Corrections

Similar questions

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. 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^{-} = 1.67 V

[F e^{2 +}] = 10^{- 3} M, p (O_{2}) = 0.1

atm and

p H

3

.
The cell potential at

25^{0}

C 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^{o} = 1.67 V

[F e^{2 +}] = 10^{- 3} M, P_{O_{2}} = 0.1

atm and

p H = 3

, the cell potential at 25

^{o} C

is:

Q. What is the cell potential (standard emf,

E^{o}

) for the reaction below ?

[E_{(F e^{2 +} (a q) / F e)}^{o} = 0.44 V

and

E_{(O_{2} (g) / H_{2} O / O H^{-})}^{o} = + 0.4 V]

2 F e (s) + O_{2} (g) + 2 H_{2} O (l) ⇋ 2 F e^{2 +} (a q) + 4 O H^{-} (a q)

Q. 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

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

and

P_{H_{2}} = 0.01 atm

, if the cell potential is 0.43

V

25^{o} C

the

p H

is:

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

Join BYJU'S Learning Program

Consider the following cell reaction: 2Fe(s)+O2(g)+4H+(aq)→2Fe2+(aq)+2H2O(l) ; E∘=1.67 V At [Fe2+]=10−3M,P(O2)=0.1 atm and pH = 3, the cell potential at 25 ∘C is (given: 2.303RTF=0.0591)

Nernst equation for the cell reaction E=E∘−0.0591nlog[Fe2+]2PO2.[H+]4 Given, E∘=1.67 V Concentration of Fe2+,[Fe2+]=10−3 M Concentration of H+,[H+]=10−3 M Pressure of oxygen, P(O2)=0.1 atm number of electrons involved, n = 4 Hence, E=1.67−0.05914log(10−3)20.1×(10−3)4=1.57 V

Consider the following cell reaction:
$2 F e (s) + O_{2} (g) + 4 H^{+} (a q) \to 2 F e^{2 +} (a q) + 2 H_{2} O (l); E^{\circ} = 1.67 V$
At $[F e^{2 +}] = 10^{- 3} M, P_{(O_{2})} = 0.1 a t m$ and pH = 3, the cell potential at $25^{\circ} C$ is
(given: $\frac{2.303 R T}{F} = 0.0591$ )