In what direction, can the reaction::
$2 N a C l + F e_{2} (S O_{4})_{3} ⇌ 2 F e S O_{4} + C l_{2} + N a_{2} S O_{4}$ proceed spontaneously?
(Given: $E_{F e^{3 +} / F e^{2 +}}^{\circ} = 0.77 V; E_{C l^{-} / C l_{2}}^{\circ} = 1.36 V$ ).

forward

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backward

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first backward and then forward

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Solution

The correct option is A forward
$E_{c e l l}^{0}$ = reduction potential of cathode - reduction potential of anode
or we have
$E_{c e l l}^{0}$ = reduction potential of cathode + oxidation potential of anode
here in this reaction, iron is getting reduced and Cl is getting oxidized
$E_{c e l l}^{0}$ = 0.77 + 1.36
$E_{c e l l}^{0}$ = 2.13 V
$E_{c e l l}^{0}$ is positive;
hence, the reaction moves forward.

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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 +}; E = 0.77 V

C l_{2} + 2 e^{-} \to 2 C l^{-}; E = 1.36 V

E_{F e^{2 +} | F e}^{\circ} = - 0.04 V

E_{F e^{3 +} | F e^{2 +}}^{\circ} = 0.77 V

E_{F e^{3 +} | F e^{+}}^{\circ}

E_{{F e}^{2 +} | F e}^{o} = - 0.44 V

E_{{F e}^{3 +} | {F e}^{2 +}}^{o} = - 0.77 V

E_{{{F e}^{3 +} | F e}^{o}}^{o}

2 N a O H cold and dilute + C l_{2} \to

2 F e S O_{4} + H_{2} S O_{4} + C l_{2} \to

C l_{2} excess + 3 F_{2} 573 K - -- \to

2 N a + {C l}_{2} \to 2 N a C l

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