Given that EoFe2- - Fe - -0-44 V- EoFe2- - Fe3- - -0-77 VIf Fe2- Fe3- and Fe solid are kept together- then-

The correct option is B Fe^3+ decreasesCell reaction : 2Fe^3+ + Fe → 3Fe^2+ E^⊖_cell = 0.77 ( 0.44) = 1.21 V #160;= +ve #160 ...

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Nernst Equation

Given that ...

Question

Given that $E_{F e 2 + | F e}^{o} = - 0.44 V; E_{F e^{2 +} | F e^{3 +}}^{o} = - 0.77 V$
If $F e^{2 +}, F e^{3 +}$ and $F e$ solid are kept together, then:

F e^{3 +}

increases

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F e^{3 +}

decreases

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

remain unchanged

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F e^{2 +}

decreases

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Solution

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Similar questions

E_{F e^{2 +} / F e}^{o} = - 0.44 V, E_{F e^{2 +} / F e^{3 +}}^{o} = - 0.77 V

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

F e

E_{F e^{2 +} | F e}^{⊖}

E_{F e^{3 +} | F e^{2 +}}^{⊖}

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

F e

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}

(E^{o})

F e^{3 +} \to F e

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

F e^{2 +} / F e (E^{o} = - 0.44 V), F e^{3 +} / F e^{2 +} (E^{o} = 0.77 V)

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

F e

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