For the cell E Fe 2- -Fe o -0-44V E Fe 3- - Fe 2- o -0-77V Fe-2 Fe 3- - 3 Fe 2- Calculate E o -

The correct option is C 1.21V Fe+2 Fe ^ 3+ → 3 Fe ^ 2+ Anode : #160; Fe→ Fe ^ 2+ Cathode : #160; 2 Fe ^ 3+ → 2Fe ^ 2+ ∴ E _ cell = ...

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Carrier Concentration

For the cell ...

Question

For the cell $E_{{F e}^{2 +} / F e}^{o} = - 0.44 V E_{{F e}^{3 +} / {F e}^{2 +}}^{o} = 0.77 V$

$F e + 2 {F e}^{3 +} \to 3 {F e}^{2 +}$ , Calculate $E^{o} :$

1.21 V

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0.11 V

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0.33 V

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1.65 V

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Solution

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Similar questions

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}

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^{o}

E_{{F e}^{3 +} / {F e}^{2 +}} = + 0.76 V

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

E_{{S n}^{2 +} / {S n}} = - 0.14 V

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

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_{C r^{3 +} |}^{o}

_{Cr}

= - 0.72 V, E_{F e^{2 +} | F e}^{o} = - 0.42

| C r^{3 +} (0.1 M) ∥ F e^{2 +} (0.01 M) |

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