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 : $E_{F{e^{3 + }}/F{e^{2 + }}}^0 = 0.77V$, $E_{Z{n^{2 + }}/Zn}^0 = - 0.76V$)
Answer (integer)
24
Solution
$Zn\mathrel{\mathop{\kern0pt\longrightarrow}
\limits_{}} Z{n^{2 + }} + 2{e^ - }$<br><br>$$2F{e^{3 + }}\mathrel{\mathop{\kern0pt\longrightarrow}
\limits_{}} 2{e^ - } + 2{e^{2 + }}$$<br><br>$Zn + 2F{e^{3 + }}\buildrel {} \over
\longrightarrow Z{n^{2 + }} + 2F{e^{2 + }}$<br><br>$E_{cell}^0 = 0.77 - (0.76)$<br><br>$= 1.53$ V<br><br>$$1.50 = 1.53 - {{0.06} \over 2}\log {\left( {{{F{e^{2 + }}} \over {F{e^{3 + }}}}} \right)^2}$$<br><br>$$\log \left( {{{F{e^{2 + }}} \over {F{e^{3 + }}}}} \right) = {{0.03} \over {0.06}} = {1 \over 2}$$<br><br>${{[F{e^{2 + }}]} \over {[F{e^{3 + }}]}} = {10^{1/2}} = \sqrt {10}$<br><br>${{[F{e^{3 + }}]} \over {[F{e^{2 + }}]}} = {1 \over {\sqrt {10} }}$<br><br>$${{[F{e^{3 + }}]} \over {[F{e^{2 + }}] + [F{e^{3 + }}]}} = {1 \over {1 + \sqrt {10} }} = {1 \over {4.16}}$$<br><br>= 0.2402<br><br>= 24 $\times$ 10<sup>-2</sup>
About this question
Subject: Chemistry · Chapter: Electrochemistry · Topic: Electrochemical Cells
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