QJEEM21Ch5830: JEE Main 2021 Chemistry - Class 11 JEEM21Ch5830 Solution: Consider the cell at 25 C Zn

Question

Verified Solution • 2025–26 Session moderate

Consider the cell at 25

^\circ

C

Zn | Zn²⁺ (aq), (1M) || Fe³⁺ (aq), Fe²⁺ (aq) | Pt(s)

The fraction of total iron present as Fe³⁺ ion at the cell potential of 1.500 V is x

\times

10^{$-$ 2}. The value of x is ______________. (Nearest integer)

(Given :

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

,

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

)

Accepted Answer

Zn\mathrel{\mathop{\kern0pt\longrightarrow}\limits_{}} Z{n^{2 + }} + 2{e^ - }

2F{e^{3 + }}\mathrel{\mathop{\kern0pt\longrightarrow}\limits_{}} 2{e^ - } + 2{e^{2 + }}

Z n + 2 F e^{3 +} \overset{}{⟶} Z n^{2 +} + 2 F e^{2 +}

E_{cell}^0 = 0.77 - (0.76)

= 1.53

V

1.50 = 1.53 - {{0.06} \over 2}\log {\left( {{{F{e^{2 + }}} \over {F{e^{3 + }}}}} \right)^2}

\log \left( {{{F{e^{2 + }}} \over {F{e^{3 + }}}}} \right) = {{0.03} \over {0.06}} = {1 \over 2}

{{[F{e^{2 + }}]} \over {[F{e^{3 + }}]}} = {10^{1/2}} = \sqrt {10}

{{[F{e^{3 + }}]} \over {[F{e^{2 + }}]}} = {1 \over {\sqrt {10} }}

{{[F{e^{3 + }}]} \over {[F{e^{2 + }}] + [F{e^{3 + }}]}} = {1 \over {1 + \sqrt {10} }} = {1 \over {4.16}}

= 0.2402

= 24

\times

10^-2

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