Consider the following cell reaction- 2Fes - O2g - 4H-aq - 2Fe2-aq - 2H2Ol E- - 1-67 VAt -Fe2- - 10-3 M- pO2 - 0-1 atm and pH - 3-The cell potential at 250C is-

The correct option is A 1.57 V 2Fe(s) + O_2(g) + 4H^+(aq) → 2Fe^2+(aq) + 2H_2O(l) E= E ^ 0 0.0591 4 log[ F e ^ 2+ ] #160; ^ 2 [ H ^ + ] ...

340

You visited us 340 times! Enjoying our articles? Unlock Full Access!

Nernst Equation

Consider the ...

Question

Consider the following cell reaction:
$2 F e (s) + O_{2} (g) + 4 H^{+} (a q) \to 2 F e^{2 +} (a q) + 2 H_{2} O (l)$
$E^{-} = 1.67 V$
At $[F e^{2 +}] = 10^{- 3} M, p (O_{2}) = 0.1$ atm and $p H$ = $3$ .
The cell potential at $25^{0}$ C is:

1.47 V

No worries! We‘ve got your back. Try BYJU‘S free classes today!

1.77 V

No worries! We‘ve got your back. Try BYJU‘S free classes today!

1.87 V

No worries! We‘ve got your back. Try BYJU‘S free classes today!

1.57 V

Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses

Open in App

Solution

Suggest Corrections

Similar questions

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

p H = 3

25^{o}

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

25^{\circ} C

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

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

25^{\circ} C

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

E^{o}

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

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)

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)

Join BYJU'S Learning Program