Redox Reactions And Electrochemistry Ques 718
Question: EMF of which of the following cells at 298 K is highest-
Given, $ E_{( M{g^{2+}}/mg )}^{0}=-2.37V; $ $ E_{( Cu^{2}/Cu )}^{0}=+2.34V; $ $ E_{( F{e^{2+}}/Fe )}^{0}=-0.44V; $ $ E_{( S{n^{2+}}/Sn )}^{0}=-0.14V; $ $ E_{( \frac{1}{2}Br_2/B{r^{-}} )}^{0}=+1.08V; $
Options:
A) $ Mg(s)|M{g^{2+}}(0.001M)| $ $ |C{u^{2+}}(0.0001M)|Cu(s) $
B) $ Fe(s)|F{e^{2+}}(0.001M)| $ $ |{H^{+}}(1M)|H_2(g)(1bar)|Pt(s) $
C) $ Sn(s)S{n^{2+}}(0.050M)| $ $ |{H^{+}}(0.020M)|H_2(g)(1bar)|Pt(s) $
D) $ Pt(s)|Br_2(l)|B{r^{+}}(0.010M)| $ $ |{H^{+}}(0.030M)|H_2(g)(1bar)|Pt(s) $
Show Answer
Answer:
Correct Answer: A
Solution:
[a] (i) Cell equation : $ Mg(s)+C{u^{2+}}(aq)\to M{g^{2+}}(aq)+Cu(s)(n=2) $
Nernst equation: $ E_{cell}=E_{cell}^{0}-\frac{0.0591}{2}\log \frac{[ M{g^{2+}} ]}{[ C{u^{2+}} ]} $
EMF of the cell, $ E_{cell}=[ 0.34-(-2.37) ]-\frac{0.0591}{2}\log \frac{[ {10^{-3}} ]}{[ {10^{-4}} ]} $ $ =2.71-0.02955=2.68V $
(ii) Cell equation: $ Fe(s)+2{H^{+}}(aq)\to F{e^{2+}}(aq)+H_2(g)(n=2) $
Nernst equation: $ E_{cell}=E_{cell}^{0}-\frac{0.0591}{2}\log \frac{[ F{e^{2+}} ]}{{{[ {H^{+}} ]}^{2}}} $
EMF of the cell, $ E_{cell}=[ 0-(0.44) ]-\frac{0.0591}{2}\log \frac{[ {10^{-3}} ]}{{{[1]}^{2}}} $ $ =0.44-\frac{0.591}{2}\times (-3) $ $ =0.44+0.0887 $ $ =0.5287V=0.53V $ $ EMF=0.53V $
(iii) Cell equation: $ Sn(s)+2{H^{+}}(aq)\to S{n^{2+}}(aq)+H_2(g) $ Nernst equation:
$ E_{cell}=E_{cell}^{0}-\frac{0.0591}{2}\log \frac{[ S{n^{2+}} ]}{{{[ {H^{+}} ]}^{2}}} $
EMF of the cell,
$ E_{cell}=[0-(-0.14)]-\frac{0.0591}{2}\log \frac{[0.05]}{{{[0.02]}^{2}}} $ $ =0.14-\frac{0.0591}{2}\times (2.097) $ $ =0.14-0.0620=0.08V $ EMF = 0.08V
(iv) Cell equation: $ 2B_r^{-}(l)+2{H^{+}}(aq)\to Br_2(l)+H_2(g)(n=2) $
Nernst equation: $ E_{Cell}=E_{cell}^{0}=\frac{0.0591}{2}\log \frac{1}{{{[ B{r^{-}} ]}^{2}}{{[ {H^{+}} ]}^{2}}} $
EMF of the cell, $ E_{cell}=[ 0-1.08 ]-\frac{0.0591}{2}\log \frac{1}{{{(0.01)}^{2}}\times {{(0.03)}^{2}}} $ $ =-1.08-\frac{0.0591}{2}\log \frac{1}{{{(0.01)}^{2}}\times {{(0.03)}^{2}}} $
$ =-1.08-\frac{0.0591}{2}\log ( 1.11\times 10^{7} ) $
$ =-1.08-\frac{0.0591}{2}(7.0457) $
$ =-1.08-0.208=-1.288V $
$ EMF=-1.288V $
$ (\therefore $ Oxidation will occur at hydrogen electrode and reduction at $ Br_2 $ electrode) Thus, emf is highest for $ Mg-Cu $ cell.
