JEE MAIN - Chemistry (2021 - 25th July Morning Shift - No. 20)
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 : $$E_{F{e^{3 + }}/F{e^{2 + }}}^0 = 0.77V$$, $$E_{Z{n^{2 + }}/Zn}^0 = - 0.76V$$)
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 : $$E_{F{e^{3 + }}/F{e^{2 + }}}^0 = 0.77V$$, $$E_{Z{n^{2 + }}/Zn}^0 = - 0.76V$$)
Answer
24
Explanation
$$Zn\mathrel{\mathop{\kern0pt\longrightarrow}
\limits_{}} Z{n^{2 + }} + 2{e^ - }$$
$$2F{e^{3 + }}\mathrel{\mathop{\kern0pt\longrightarrow} \limits_{}} 2{e^ - } + 2{e^{2 + }}$$
$$Zn + 2F{e^{3 + }}\buildrel {} \over \longrightarrow Z{n^{2 + }} + 2F{e^{2 + }}$$
$$E_{cell}^0 = 0.77 - (0.76)$$
$$ = 1.53$$ 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
$$2F{e^{3 + }}\mathrel{\mathop{\kern0pt\longrightarrow} \limits_{}} 2{e^ - } + 2{e^{2 + }}$$
$$Zn + 2F{e^{3 + }}\buildrel {} \over \longrightarrow Z{n^{2 + }} + 2F{e^{2 + }}$$
$$E_{cell}^0 = 0.77 - (0.76)$$
$$ = 1.53$$ 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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