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Kohlrausch Law

If EFe+2/ F...

Question

If $E_{F e^{+ 2} / F e}^{0} = - 0.441 V$ and $E_{F e^{+ 3} / F e^{+ 2}}^{0}$ and $E_{F e^{+ 3} / F e^{+ 2}}^{0} = 0.771 V$ the standard EMF of the reaction $F e + 2 F e^{+ 3} \to 3 F e^{+ 2}$ will be:

0.330 V

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1.653 V

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1.212 V

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0.111 V

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Solution

The correct option is D $1.212 V$
Given: $E_{F e^{2 +} / F e}^{o} = - 0.441 V$
$F e ⟶ F e^{2 +} + 2 e^{-}, E^{o} = + 0.441 V ⟶ (i)$ & $E_{F e^{3 +} / F e^{2^{+}}}^{o} = 0.771 V$
So, $F e^{3 +} + e^{-} ⟶ F e^{2 +}, E^{o} = 0.771 V ⟶ (i i)$
Cell reaction, (i) + (ii)
$F e + 2 F e^{3 +} ⟶ 3 F e^{2 +}$
At anode, $F e ⟶ F e^{2 +} + 2 e^{-}$
Cathode $2 F e^{3 +} + 2 e^{-} ⟶ 2 F e^{2 +}$
So, $E_{C e l l}^{o} = E_{C a t h o d e}^{o} - E_{A n o d e}^{o}$
$= E_{F e^{3 +} / F e^{2 +}}^{o} - E_{F e^{2 +} / F e}^{o}$
$= (0.771) - (- 0.441)$
$= + 1.212 V$ .

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Similar questions

E_{F e^{2 +} / F e}^{o} = - 0.441 V

E_{F e^{3 +} / F e^{2 +}}^{o} = 0.771 V

E^{o}

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

E_{F e^{2 +} / F e}^{0} = - 0.441 V

E_{F e^{3 +} / F e^{2 +}}^{0} = 0.771 V

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

E_{F e^{2 +} / F e}^{0}

-

E_{F e^{3 +} / F e^{2 +}}^{0}

E M F

F E + 2 F E^{3 +} \to 3 F E^{2 +}

E_{F e (s) / F e^{2 +} (a q)}^{0} = 0.44 V

E_{F e^{2 +} (a q) / F e^{3 +} (a q)}^{0} = - 0.77 V

F e (s) + 2 F e^{3 +} (a q) \to 3 F e^{2 +} (a q)

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

I^{-}

35^{o} C

E_{{F e}^{+ 3} / {F e}^{+ 2}}^{o} = + 0.77 V; E_{I_{2} / I^{-}}^{o} = 0.536 V

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