Chhattisgarh PMT Chhattisgarh PMT Solved Paper-2007

  • question_answer
    34 kcal heat is liberated when \[1\text{ }g\text{ }{{H}_{1}}(ECE=1.044\times {{10}^{-8}})\] is converted into water. Minimum voltage required for it is

    A)  0.75 V                                  

    B)  1.5 V

    C)  3.0V                                     

    D)  4.5V

    Correct Answer: B

    Solution :

    Given,   \[m=1g={{10}^{-3}}g\] Electrochemical equivalent \[Z=1.044\times {{10}^{-8}}kg/C\] Heat  \[Q=34kcal=34\times 1000cal\]                 \[=34\times 1000\times 4.2J\] Mass liberated on electrode during electrolysis,                 \[m=Zq\] \[\therefore \]  \[q=\frac{m}{Z}\] \[\therefore \] Required potential, \[V=\frac{H}{q}=\frac{34\times 1000\times 4.2\times 1.044\times {{10}^{-8}}}{{{10}^{-3}}}\] \[=1.5volt\]


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