Given that-E-Fe2-Fe-0-44 V and E-Fe3-Fe2-0-77 VE-Fe3-Fe is-

Fe^2++2e^ → Fe, E^∘= 0.44 V Fe^3++e^ → Fe^2+, E^∘=0.77 V Fe^3++3e^ → Fe, E^∘=? We know, Δ G^∘= nFE^∘ ∴ Δ G_3^∘=Δ G_1^∘+Δ G^∘_2 3FE^∘= 2F( 0.44)+( 1)F(0.77) ...

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Byju's Answer

Standard XII

Chemistry

Gibb's Energy and Nernst Equation

Given that:E∘...

Question

Given that:
$E_{F e^{2 +} / F e}^{\circ} = - 0.44 V and E_{F e^{3 +} / F e^{2 +}}^{\circ} = 0.77 V$
$E_{F e^{3 +} / F e}^{\circ}$ is:

1.21 V

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0.33 V

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- 0.036 V

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0.036 V

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Solution

The correct option is C $- 0.036 V$
$F e^{2 +} + 2 e^{-} \to F e, E^{\circ} = - 0.44 V$
$F e^{3 +} + e^{-} \to F e^{2 +}, E^{\circ} = 0.77 V$
$F e^{3 +} + 3 e^{-} \to F e, E^{\circ} = ?$

We know, $Δ G^{\circ} = - n F E^{\circ}$
$∴ Δ G_{3}^{\circ} = Δ G_{1}^{\circ} + Δ G_{2}^{\circ}$
$- 3 F E^{\circ} = - 2 F (- 0.44) + (- 1) F (0.77)$
$\Rightarrow - 3 F E^{\circ} = - F (- 0.88 + 0.77)$
$\Rightarrow - 3 F E^{\circ} = 0.11 F$
$∴ E^{\circ} = - \frac{0.11 F}{3 F} = - 0.036 V$
Hence,
$E_{F e^{3 +} / F e}^{\circ} = - 0.036 V$

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Similar questions

Q. Given

E_{F e^{2 +} | F e}^{⊖}

and

E_{F e^{3 +} | F e^{2 +}}^{⊖}

are -0.44 and 0.77 V respectively.
If

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

, and

F e

blocks are kept together, then:

Given standard $E^{⊖}$ :

$F e^{3 +} + 3 e^{-} \to F e$ ; $E^{⊖} = - 0.036 V$

$F e^{2 +} + 2 e^{-} \to F e$ ; $E^{⊖} = - 0.440 V$

The $E^{⊖}$ of $F e^{3 +} + e^{-} \to F e^{2 +}$ is:

Q. For the reaction,

F e^{3 +} + 3 e ⇌ F e, E^{o}

is -0.036 V and the standard electrode potential(SEP) for

F e^{3 +} + e ⇌ F e^{2 +}

is 0.77 V.
Hence,

E^{o}

for

F e^{2 +} + 2 e ⇌ F e

is:

Q. Which of the following statements are correct for the given electrochemical cells?
(I)

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

(II)

F e | F e^{2 +} | | F e^{3 +}, F e^{2 +} | P t

(III)

F e | F e^{3 +} | | F e^{3 +}, F e^{2 +} | P t

Q. What is the standard reduction potential

(E^{o})

for

F e^{3 +} \to F e

?
Given that:

F e^{2 +} + 2 e^{-} \to F e; E_{F e^{2 +} / F e}^{o} = - 0.47 V

F e^{3 +} + e^{-} \to F e^{2 +}; E_{F e^{3} / F e^{2 +}}^{o} = + 0.77 V

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Given that: E∘Fe2+/Fe=−0.44 V and E∘Fe3+/Fe2+=0.77 V E∘Fe3+/Fe is:

Given that:
$E_{F e^{2 +} / F e}^{\circ} = - 0.44 V and E_{F e^{3 +} / F e^{2 +}}^{\circ} = 0.77 V$
$E_{F e^{3 +} / F e}^{\circ}$ is: