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Electrochemistry question

2021 · 25 Jul · Shift 1 · Q20

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JEE Main
Chemistry
Electrochemistry
2021 · 25 Jul · Shift 1 · Q20

Electrochemistry question

2021 · 25 Jul · Shift 1 · Q20

JEE MainChemistryElectrochemistryNumerical+4 / −1

Consider the cell at 25

^\circ

C Zn |

Zn^{2+}

(aq), (1M) ||

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

\times

10

-

2. The value of x is

\underline{\hspace{2cm}}

. (Nearest integer) (Given :

E_{F{e^{3 + }}/F{e^{2 + }}}^0 = 0.77V

,

E_{Z{n^{2 + }}/Zn}^0 = - 0.76V

)

Numerical answer

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