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Electrochemistry Test - 32

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Electrochemistry Test - 32
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Weekly Quiz Competition
  • Question 1
    1 / -0
    A metal is released in the electrolysis of a salt. It gets deposited on the :
    Solution
    The metal which is released in the electrolysis is of simple salt, so it would generally be deposited on the cathode.

     For example, suppose we take sodium chloride (NaCl) then on breaking this into simpler form we get,

    $$NaCl\rightarrow Na^++Cl^-$$

    This can be said that as Na acquires positive charge, it shall move to negative electrode (or negative terminal) i.e. cathode.

    Hence , option C is correct .
  • Question 2
    1 / -0
    For reducing one mole of $$Cr_{2} O_{7}^{2-}$$ to $$Cr^{3+}$$, the charge required is:
    Solution
    $$Cr_{2}O_{7}^{2-} + 14H^{+} + 6e^{-} \rightarrow 2Cr^{3+} + 7 H_{2} O$$
    For reducing one mole of $$Cr_{2}O_{7}^{2-}$$, charge required $$= 6\times 96500\ coulomb$$.
  • Question 3
    1 / -0
    The internal resistance of a lead acid battery can be reduced by
  • Question 4
    1 / -0
    The charge required for the reduction of $$1\ mol$$ of $${MnO_{4}}^{-}$$ to $$MnO_{2}$$ is:
    Solution
    $$MnO_{4}^{-} + 4H^{+} + 3e^- \rightarrow MnO_{2} + 2H_{2} O$$

    Oxidation number of Mn in $${MnO_4}^-$$ = +7
    Oxidation number of Mn in $$MnO_2$$ = +4

    So, $$3\ F$$ of charge will be required to reduce $$1\ mole$$ of $$MnO_{4}^{-}$$.

    Hence, option B is correct.
  • Question 5
    1 / -0
    The process of lead action in a simple voltaic cell
    Solution
    The process of lead action in a simple voltaic cell is a chemical action that occurs while the current is flowing and causes hydrogen bubbles to form on the surface of anode. The action is called polarization.
    Some hydrogen bubbles rise to the surface of the electrolyte and escape into the air and some remain on the surface of the anode. If enough bubbles remain around the anode, the bubbles form a barrier that increases internal resistance.
    When the internal resistance of the cell increases, the output current is decreased and the voltage of the cell also decreases.
    Hence it decreases the efficiency of the cell.
  • Question 6
    1 / -0
    The $$ZnCl$$ used in a dry cell is which helps to maintain the moistness of paste contained in the cell between anode and cathode is
    Solution
    The Zinc used in dry cell is which helps to maintain the moistness of paste contained in the cell between anode and cathode is highly hygroscopic
  • Question 7
    1 / -0
    The standard reduction potential for $$Mg^{2+}/Mg$$ is $$-2.37V$$ and for $$Cu^{2+}/Cu$$ is $$0.337$$. The $$E^o_{cell}$$ for the following reaction is $$Mg+Cu^{2+}\rightarrow Mg^{2+}+Cu$$
    Solution
    $$E^o_{cell}=E^o_{cathode}-E^o_{anode}$$
    $$=E^o_{(Cu^{2+}/Cu)}-E^o_{(Mg^{2+}/Mg)}=0.337-(-2.37)V$$
    $$=2.7V$$
  • Question 8
    1 / -0
    Electrolytes which allow large current to pass through them are known as :
    Solution
    Electrolytes which allow large current to pass through them are known as strong electrolytes.
  • Question 9
    1 / -0
    _____________ ionizes in a fused or in aqueous solution state and furnish positive ions other than $$H^+$$ and negative ions other than $$OH^-$$.
    Solution
    Ionic salts ionizes in a fused or in aqueous solution state and furnish positive ions other than $$H^+$$ and negative ions other than $$OH^−$$.    
    For example, aq NaCl ionises to form $$\displaystyle Na^+$$ ions and chloride $$\displaystyle Cl^-$$ ions.
    $$\displaystyle NaCl \rightleftharpoons Na^+ +  Cl^- $$
  • Question 10
    1 / -0
    ______ ionizes in fused or in aqueous solution state and furnish ions in solution.
    Solution
    Bases ionizes in fused or in aqueous solution state and furnish ions in solution.
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