Consider the following cell reaction- 2Fes -O2g-4H-aq- 2Fe2-aq-2H2Ol- Eo-1-67VAt -Fe2-10-3M- PO2-0-1 atm and pH-3- the cell potential at 25 oC is-

The correct option is D 1.57 V pH=3 ⇒ [H^+]=10^ 3M Now, E_cell=E^o 0.0594log [Fe^2+]^2[H^+]^4 P_ O_2 #160; #160; #160; #160; #160; ...

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Nernst Equation

Consider the ...

Question

Consider the following cell reaction:

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

1.47

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1.77

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1.87

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1.57

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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

p H

3

25^{0}

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)

25^{o} C

P t (s) | F e^{2 +} (a q, 0.1 M), F e^{3 +} (a q, 0.01 M); E_{F e^{3 +} / F e^{2 +}}^{0} = + 0.77 V

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