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Consider the cell at 25 ° C Zn Zn 2 + aq , 1 M ‖ Fe 3 + ( aq ) , Fe 2 + aq Pt s The fraction of total iron present as Fe 3 + ion at the cell potential of 1 . 500 V is x × 10 - 2 . The value of x is ______. (Nearest integer) Given :   E ∘ Fe 3 + | Fe 2 + = 0 . 77 V ,   E ∘ Zn 2 + | Zn = - 0 . 76 V

Correct answer

0

Step-by-step solution

E cell 0 = 0 . 77 - 0 . 76 = 1 . 53   V 1 . 50 = 1 . 53 - 0 . 06 2 log Fe 2 + Fe 3 + 2 log Fe 2 + Fe 3 + = 0 . 03 0 . 06 = 1 2 Fe 2 + Fe 3 + = 10 1 / 2 = 10 Fe 3 + Fe 2 + = 1 10 Fe 3 + Fe 2 + + Fe 3 + = 1 1 + 10 = 1 4 . 16 = 0 . 2402 = 24 × 10 - 2

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