2. Solved Papers
  3. J & K CET Engineering

J & K CET Engineering J and K - CET Engineering Solved Paper-2011

  • question_answer
    A cell is constituted by coupling the two electrodes \[\text{Sn/S}{{\text{n}}^{\text{2+}}}\]and \[Cu/C{{u}^{2+}}.\] If \[{{E}^{o}}(S{{n}^{2+}},Sn),{{E}^{o}}(C{{u}^{2+}},Cu)\]and \[{{E}^{o}}\](cell) are \[-0.14\,V,0.34\,V\]and 0.48 V respectively, the correct representation of the cell is

    A) \[Sn(s)|S{{n}^{2+}}(0.1\,M)||C{{u}^{2+}}(1.0\,M)|Cu(s)\]

    B)  \[Sn(s)|S{{n}^{2+}}(1.0\,M)||C{{u}^{+}}(1.0\,M)|Cu(s)\]

    C)  \[Sn(s)|S{{n}^{2+}}(1.0\,M)||C{{u}^{2+}}(1.0\,M)|Cu(s)\]

    D)  \[Cu(s)|C{{u}^{2+}}(1.0\,M)||S{{n}^{2+}}(1.0\,M)|Sn(s)\]

    Correct Answer: C

    Solution :

     Lower value of standard reduction potential represents anode while higher value of standard   reduction potential represents cathode. Hence, copper (Cu) electrode acts as cathode and tin (Sn) electrode acts as anode. Half-cell reaction: Cathode (reduction): \[C{{u}^{2+}}+2{{e}^{-}}\xrightarrow{{}}Cu\] Anode (oxidation): \[Sn\xrightarrow{{}}S{{n}^{2+}}+2{{e}^{-}}\] In the representation of the cell, anode keeps on the left and cathode on the right. Hence, the correct representation of the cells \[Sn(s)|S{{n}^{2+}}(1.0M)||C{{u}^{2+}}(1.0\,M)|Cu(s)\]


You need to login to perform this action.
You will be redirected in 3 sec spinner