Given that E Fe 3- - Fe 2-0-0-36 V and E Fe 2- - Fe 0-0-439 V respectively- Therefore value of E Fe 3- - Fe 2-0 is-A- -3-0-036-2-0-439- VB- -0-036-0-439 VC- -3-0-036-2-0-439- VD- -0-036-0-439 V

The correct option is A [3( 0.036) 2 ( 0.439)] VBy calculation we get [3( 0.036) 2 ( 0.439)] V ...

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Question

Given that $E_{F e^{3 +} / F e^{2 +}}^{0} = - 0.36 V$ and $E_{F e^{2 +} / F e}^{0} = - 0.439 V$ respectively. Therefore value of $E_{F e^{3 +} . F e^{2 +}}^{0}$ is:

(-0.036 - 0.439) V

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(-0.036 + 0.439) V

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[3(- 0.036) + 2(-0.439)] V

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[3(-0.036) - 2 (-0.439)] V

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E_{F e^{3 +} / F e}^{0} = 0.036 V, E_{F e^{2 +} / F e}^{0} = - 0.439 V

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

E_{F e^{3 +} / F e}^{o} = - 0.036 V, E_{F e^{2 +} / F e}^{o} = - 0.439 V

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

E_{F e^{3 +} | F e}^{⊖}

E_{F e^{2 +} | F e}^{⊖}

- 0.36 V

- 0.439 V

E_{F e^{3 +} | F e^{2 +}}^{⊖}

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}

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:

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