E- for F2 - 2e- 2F- is 2-8 V E- for 0-5 F2 - e- F- is-

The correct option is D 2.8 V F_2+2e^ ⟶ 2F^ #160; #160; #160; F^o=2.8V 12F_2+e^ ⟶ F^ #160; #160; #160; E^o=2.8V ...

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E∘ for F2 +...

Question

$E^{\circ}$ for $F_{2} + 2 e^{-} \to 2 F^{-}$ is $2.8 V$
$E^{\circ}$ for $0.5 F_{2} + e^{-} \to F^{-}$ is:

2.8 V

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1.4 V

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- 2.8 V

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- 1.4 V

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Solution

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E° for F2 + 2e- -> 2F- is 2.8V, E° for 1÷2 F2 + e -> F- is

Ans : 2.8V

Please give detailed explanation as I won't be able to understand concise answers .

Given :

Oxidation :

$H_{2} O_{2} \to O_{2} + 2 H^{\oplus} + 2 e^{-}$

$E^{⊖} = - 0.69 V$ ;

$2 F^{⊖} \to F_{2} + 2 e^{-}$

$E^{⊖} = - 2.87 V$ ;

Reduction :

$H_{2} O_{2} + 2 H^{\oplus} + 2 e^{-} \to 2 H_{2} O$

$E^{⊖} = - 1.77 V$ ;

$2 I^{⊖} \to I_{2} + 2 e^{-}$ $E^{⊖} = - 0.54 V$ ;

Which of the following statements is/are correct ?

E_{O_{2} / H_{2} O}^{o} = + 1.23 V

E_{S_{2} O_{8}^{2 -} / S O_{4}^{2 -}}^{o} = + 2.05 V

E_{B r_{2} / B r^{-}}^{o} = + 1.09 V

E_{A u^{3 +} / A u}^{o} = + 1.4 V

F_{2} + 2 e^{-} \to 2 F^{-} (a q); E = + 2.87 V

C a^{+} + 2 e^{-} \to C a (s); E = - 2.76 V

F_{2} (g) + 2 e^{-} \to 2 F^{-} (a q); E^{0} = 2.87 V C l_{2} (g) + 2 e^{-} \to C l^{-} (a q); E^{0} = 1.36 V F e^{3 +} (a q) + e^{-} \to F e^{2 +} (a q); E^{0} = 0.77 V

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