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Electrochemistr...

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  • Question 1
    1 / -0

    Find the potential of a half-cell having reaction, Ag2S+2e2Ag+S2Ag_{2}S+2e\rightarrow2Ag+S^{2-} in a solution buffered
    pH=3pH=3 and which is also saturated with 0.1 M H2S0.1\ M\ H_{2}S. For H2S: K1=108H_{2}S:\ K_{1}=10^{-8} and K2=2×1013, Ksp(Ag2S)=2×1048, EAg,ago=0.8K_{2}=2\times 10^{-13},\ K_{sp}(Ag_{2}S)=2\times 10^{-48},\ E_{Ag,ag}^{o}=0.8

  • Question 2
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    Which of the following represent first law of Faraday?

  • Question 3
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    In ZnZn+2Ag+AgZn|Zn^{+2}|| Ag^+|Ag, how will cell potential be affected if KI is added to Ag+Ag^{+} half cell?

  • Question 4
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    A : Electrochemical cell is based on redox reaction.
    R : Electrochemical cell converts electrical energy into chemical energy.

  • Question 5
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    Which of the following substance has the lowest electric resistivity at room temperature?

  • Question 6
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    The no. of electrons involved in the electro deposition of 63.5 g. of Cu from aq. CuSO4 CuSO_4 is :

  • Question 7
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    In the electrolysis of aqueous NaClNaCl, what volume of Cl2(g)Cl_2(g) is produced in the time that it takes to liberate 5.05.0 liter of H2(g)?H_2(g) ? Assume that both gases are measured at STP.

  • Question 8
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    How many faradays of charge is transferred to produce 11.2 L of H2H_2 at STP in the reaction, in the reaction, NaH+H2ONaOH+H2 NaH +H_2O \rightarrow NaOH +H_ 2 \uparrow ?

  • Question 9
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    2 ml ethanoic acid was taken in test tubes - A, B and C. 2 ml, 4 ml and 8 ml of water was added to the test tubes A, B and C respectively. Which test tube will show clear solution?

  • Question 10
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    Deduce from the following E{E^ \circ } value of half cells, What combination of half cells would result in cell with the largest potential? 
    i) A3A2+e;E=1.5V{A^{3 - }} \to {A^2} + {e^ - };{\rm{ }}{{\rm{E}}^ \circ } = 1.5V
    ii) B2++eB+;E= 2.1V{B^{2 + }} + {e^ - } \to {B^ + };{\rm{ }}{{\rm{E}}^ \circ } =  - 2.1V
    iii) C2++e +C+;E= +0.5V{C^{2 + }} + {e^ - } \to  + {C^ + };{\rm{ }}{{\rm{E}}^ \circ } =  + 0.5V
    iv) DD2++2e;E= 1.5VD \to {D^{2 + }} + 2{e^ - };{\rm{ }}{{\rm{E}}^ \circ } =  - 1.5V

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