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NEET AIPMT SOLVED PAPER SCREENING 2011

  • question_answer
    Standard electrode potential for \[\text{S}{{\text{n}}^{\text{4+}}}\text{/S}{{\text{n}}^{\text{2+}}}\] couple is +0.15 V and that for the \[C{{r}^{3+}}/Cr\] couple is -0.74. These two couples in their standard state are connected to make a cell. The cell potential will be

    A)  + 0.89 V       

    B)        + 0.18 V

    C)  + 1.83 V        

    D)         + 1.199 V

    Correct Answer: A

    Solution :

    \[E{{_{Sn}^{o}}_{^{4+}/S{{n}^{2+}}}}=+0.15\,\text{V}\] \[E{{_{Cr}^{o}}_{^{3+}/Cr}}=-\,0.74\,\text{V}\] \[E_{cell}^{o}={{E}^{o}}_{cathode(RP)}-{{E}^{o}}_{anode(RP)}\]                 \[=0.15-(-0.74)\]                 \[=+\,0.89\,V.\]


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