For an electrochemical cell
Sn(s) | Sn2+ (aq,1M)||Pb2+ (aq,1M)|Pb(s)
the ratio ${{\left[ {S{n^{2 + }}} \right]} \over {\left[ {P{b^{2 + }}} \right]}}$ when this cell attains
equilibrium is _________.
(Given $E_{S{n^{2 + }}|Sn}^0 = - 0.14V$,
$E_{P{b^{2 + }}|Pb}^0 = - 0.13V$, ${{2.303RT} \over F} = 0.06$)
Answer (integer)
2
Solution
Cell reaction is :
<br><br>Sn(s) + Pb<sup>+2</sup>(aq) $\to$ Sn<sup>+2</sup>(aq) + Pb(s)
<br><br>Apply Nernst equation :
<br><br>E<sub>cell</sub> = $E_{cell}^0$ - $${{0.06} \over 2}\log {{\left[ {S{n^{ + 2}}} \right]} \over {\left[ {P{b^{ + 2}}} \right]}}$$ ....(1)
<br><br>$\Rightarrow$ $E_{cell}^0$ = -0.13 + 0.14 = 0.01 V
<br><br>At equilibrium : E<sub>cell</sub> = 0
<br><br>Substituting in (1), we get
<br><br>0 = 0.01 - $${{0.06} \over 2}\log {{\left[ {S{n^{ + 2}}} \right]} \over {\left[ {P{b^{ + 2}}} \right]}}$$
<br><br>$\Rightarrow$ $\log {{\left[ {S{n^{ + 2}}} \right]} \over {\left[ {P{b^{ + 2}}} \right]}}$ = ${1 \over 3}$
<br><br>$\Rightarrow$ ${{\left[ {S{n^{2 + }}} \right]} \over {\left[ {P{b^{2 + }}} \right]}}$ = 2.15
About this question
Subject: Chemistry · Chapter: Electrochemistry · Topic: Electrochemical Cells
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