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Electrochemistry Test - 66

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Electrochemistry Test - 66
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  • Question 1
    1 / -0
    A Voltaic cell consists of an electrode of solid silver immersed in 0.10M $${ AgNO }_{ 3 }$$ Solution and an electrode of unknown metal X immersed in a 0.10M solution $${ X(NO }_{ 3 })_{ 2 }$$. A porous barrier separates the two half of the cell. Also given:

    $$E\left( { Ag^{ + } }/{ Ag } \right)$$ = 0.80V and $${ E }_{ cell }$$ = 1.05V and at 25 degree Celsius

    $${ X_{ (s) } }/{ { X }^{ 2+ } }(1.0M)\parallel { Ag }^{ + }(0.1M)/{ Ag }_{ (s) }$$

    Which of the following statements regarding the cell and X is correct?
    Solution

  • Question 2
    1 / -0
    For a cell reaction represented by the equation $$Cu(s)+2Ag^{+}(aq)\rightarrow Cu^{2+}(aq)+2Ag(s)$$ the value of $$\Delta G^{\circ}$$ could be obtained by 
  • Question 3
    1 / -0
    Which one of the following metals could not be obtained on electrolysis of aqueous solution of its salts ?
    Solution
    The reduction potential of $$Mg$$ is less than that of water $$(E^o=-0.83\ V)$$. Hence their ions in the aqueous solution cannot be reduced instead water will be reduced
    $$2H_2O +2e^- \to H_2+2OH^-$$.
  • Question 4
    1 / -0
     

    Solution

  • Question 5
    1 / -0
    Number of electrons lost during electrolysis of 0.355 g of $$Cl^-$$ is:
  • Question 6
    1 / -0
    Without losing its concentration $$ZnCl_2$$ solution cannot be kept in contact with :
  • Question 7
    1 / -0
    $${{\text{E}}^{\text{o}}}_{{\text{C}}{{\text{u}}^{{\text{2 + }}}}{\text{|Cu = }}}\,{\text{ + 0}}{\text{.337}}\,{\text{V}}\,$$, $${{\text{E}}^{\text{o}}}_{{\text{Z}}{{\text{n}}^{{\text{2 + }}}}{\text{|Zn = }}}\,{\text{-0.762}}\, {\text{V}}\,$$ The EMF of the cell, $${\text{Zn|Z}}{{\text{n}}^{{\text{2 + }}}}\left( {{\text{0}}{\text{.1M}}} \right)||\,{\text{C}}{{\text{u}}^{{\text{2 + }}}}\left( {{\text{0}}{\text{.01M}}} \right){\text{|Cu}}$$ is :
    Solution

  • Question 8
    1 / -0
    On the basis of the following $$E^0$$ the strongest oxidising agent is:
    $$\begin{array}{l}{\left[ {Fe{{\left( {CN} \right)}_6}} \right]^{4 - }} \to {\left[ {Fe{{\left( {CN} \right)}_6}} \right]^{3 - }} + {e^ - };{E^0} =  - 0.35V\\F{e^{2 + }} \to F{e^{3 + }} + {e^ - };{E^0} =  - 0.77V\end{array}$$ 
  • Question 9
    1 / -0
    An amino acid migrates towards the cathode when the electric field is applied. This implies that 
    Solution

  • Question 10
    1 / -0
    $$\frac{1}{2}F_{2}+e^{-}\rightarrow F^{-};\ E^{\circ}=+3.02V$$ 
    The electrode potential for given reaction: $$F_{2}+2e^{-}\rightarrow 2F^{-}$$ 
    Solution

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