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JEE Main & Advanced Chemistry Electrochemistry / विद्युत् रसायन Question Bank Topic Test - Electrochemistry

  • question_answer
    EMF of which of the following cells at 298 K is highest?
    Given,   \[E_{\left( M{{g}^{2+}}/mg \right)}^{0}=-2.37V;\]
    \[E_{\left( C{{u}^{2}}/Cu \right)}^{0}=+2.34V;\]
    \[E_{\left( F{{e}^{2+}}/Fe \right)}^{0}=-0.44V;\]
    \[E_{\left( S{{n}^{2+}}/Sn \right)}^{0}=-0.14V;\]
    \[E_{\left( \frac{1}{2}B{{r}_{2}}/B{{r}^{-}} \right)}^{0}=+1.08V;\]

    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) \[Pt(s)|B{{r}_{2}}(l)|B{{r}^{+}}(0.010M)|\]\[|{{H}^{+}}(0.030M)|{{H}_{2}}(g)(1bar)|Pt(s)\]

    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:
    \[{{E}_{cell}}=E_{Cell}^{0}-\frac{0.0591}{2}\log \frac{\left[ M{{g}^{2+}} \right]}{\left[ C{{u}^{2+}} \right]}\]
    EMF of the cell,
    \[{{E}_{cell}}=\left[ 0.34-(-2.37) \right]-\frac{0.0591}{2}\log \frac{\left[ {{10}^{-3}} \right]}{\left[ {{10}^{-4}} \right]}\]
    \[=2.71-0.02955=2.68V\]
    (ii) Cell equation:
    \[Fe(s)+2{{H}^{+}}(aq)\to F{{e}^{2+}}(aq)+{{H}_{2}}(g)(n=2)\]
    Nernst equation:
    \[{{E}_{cell}}=E_{cell}^{0}-\frac{0.0591}{2}\log \frac{\left[ F{{e}^{2+}} \right]}{{{\left[ {{H}^{+}} \right]}^{2}}}\]
    EMF of the cell,
    \[{{E}_{cell}}=\left[ 0-(0.44) \right]-\frac{0.0591}{2}\log \frac{\left[ {{10}^{-3}} \right]}{{{[1]}^{2}}}\]
    \[=0.44-\frac{0.591}{2}\times (-3)\]
    \[=0.44+0.0887\]
    \[=0.5287V=0.53V\]
    \[EMF=0.53V\]
    (iii) Cell equation: 
    \[Sn(s)+2{{H}^{+}}(aq)\to S{{n}^{2+}}(aq)+{{H}_{2}}(g)\]
    Nernst equation:
    \[{{E}_{cell}}=E_{cell}^{0}-\frac{0.0591}{2}\log \frac{\left[ S{{n}^{2+}} \right]}{{{\left[ {{H}^{+}} \right]}^{2}}}\]
    EMF of the cell,
    \[{{E}_{cell}}=[0-(-0.14)]-\frac{0.0591}{2}\log \frac{[0.05]}{{{[0.02]}^{2}}}\]
    \[=0.14-\frac{0.0591}{2}\times (2.097)\]
    \[=0.14-0.0620=0.0\text{8V}\]
    EMF = 0.08V
    (iv) Cell equation:
    \[2B_{r}^{-}(l)+2{{H}^{+}}(aq)\to B{{r}_{2}}(l)+{{H}_{2}}(g)(n=2)\]
    Nernst equation:
    \[{{E}_{Cell}}=E_{Cell}^{0}=\frac{0.0591}{2}\log \frac{1}{{{\left[ B{{r}^{-}} \right]}^{2}}{{\left[ {{H}^{+}} \right]}^{2}}}\]
    EMF of the cell,
    \[{{E}_{cell}}=\left[ 0-1.08 \right]-\frac{0.0591}{2}\log \frac{1}{{{(0.01)}^{2}}\times {{(0.03)}^{2}}}\]\[=-1.08-\frac{0.0591}{2}\log \frac{1}{{{(0.01)}^{2}}\times {{(0.03)}^{2}}}\]
    \[=-1.08-\frac{0.0591}{2}\log \left( 1.11\times {{10}^{7}} \right)\]
    \[=-1.08-\frac{0.0591}{2}(7.0457)\]
    \[=-1.08-0.208=-1.288\text{ }V\]
    \[EMF=-1.288V\]
    (\[\therefore \]Oxidation will occur at hydrogen electrode and reduction at \[B{{r}_{2}}\] electrode)
    Thus, emf is highest for \[Mg-Cu\] cell.


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