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Redox Reactions and Couple

Whats the pot...

Question

Whats the potential of the reaction below given the half-reaction potentials:
$2 F e_{2} + + C l_{2} \to 2 F e^{3 +} + 2 C l^{-}$
$F e^{3 +} + e^{-} \to F e^{2 +}; E = 0.77 V$
$C l_{2} + 2 e^{-} \to 2 C l^{-}; E = 1.36 V$

0.18 V

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0.59 V

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1.05 V

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2.13 V

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2.90 V

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Solution

The correct option is B $0.59 V$
$2 F e^{2 +} + C l_{2} \to 2 F e^{3 +} + 2 C l^{-}$
$E_{r e a c t i o n}^{0} = E_{c a t h o d e}^{0} - E_{a n o d e}^{0}$
$= 1.36 - 0.77$
$= 0.59 V$

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Similar questions

2 F e^{2 +} + C l_{2} \to 2 F e^{3 +} + 2 C l

F e^{3 +} + e^{-} \to F e^{2 +}

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C l_{2} + 2 e^{-} \to 2 C l^{-}

E_{r e d}^{o}

{C l}_{2} + {2 e}^{-} \to {2 C l}^{-}

{P t}^{2 +} + {2 C l}^{-} \to P t + {C l}_{2}, E_{c e l l} = - 0.15 V

{P t}^{2 +} + {2 e}^{-} \to P t, E = 1.20 V

H g_{2} C l_{2} + C l_{2} \to 2 H g^{+} + 4 C l^{-}

H g_{2} C l_{2} + 2 e^{-} \to 2 H g + 2 C l^{-}; E^{\circ} = 0.270 v o l t

H g_{2}^{2 +} \to 2 H g^{2 +} + 2 e^{-}; E^{\circ} = - 0.92 v o l t

2 H g \to H g_{2}^{2 +} + 2 e^{-}; E^{\circ} = - 0.79 v o l t

C l_{2} + 2 e^{-} \to 2 C l^{-}; E^{\circ} = 1.36 v o l t

{C l}_{2} + {2 e}^{-} ⟶ 2 {C l}^{-}

{P t}^{2} + 2 C l ⟶ P t + {C l}_{2}, E_{c e l l}^{0} = - 0.15 V

{P t}^{2} + {2 e}^{-} ⟶ P t, E^{0} = 1.20 V

F e^{2 +} ⇌ F e; E^{\circ} = - 0.44 V

F e^{3 +} ⇌ F e^{2 +}; E^{\circ} = + 0.77 V

F e^{2 +}, F e^{3 +}

F e

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