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Electrochemistry Test - 72

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Electrochemistry Test - 72
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  • Question 1
    1 / -0
    In the galvanic cell, $$Cu|{ Cu }^{ 2+ }(1M)\parallel { Ag }^{ + }(1M)|Ag$$, the electrons will travel in the external circuit:
    Solution

  • Question 2
    1 / -0
    $${ E }_{ cell }^{ o }$$ for the reaction $$Co(s)+{ Ni }^{ 2+ }\quad \longrightarrow { Co }^{ 2+ }+Ni(s)$$ is $$+0.03$$ volt. If cobalt metal is added to an aqueous solution  having $$\left[ { Ni }^{ 2+ } \right] =1M$$
    Solution

  • Question 3
    1 / -0

    Directions For Questions

    The process of corrosion of iron is known as rusting. Rust appears to be a hydrated ferric oxide, $$Fe_{2}O_{3}\cdot xH_{2}O.$$ Both $$O_{2}$$ and $$H_{2}O$$ are required of rusting. Iron can be protected from rusting by use of the following methods:
    (i) Applying paints, lacquers, and enamels on the surface of the iron.
    (ii) By forming a firm and coherent protective coating of ferrosoferric oxide. This is done by passing steam over a hot iron.
    (iii) By coating a thin film of zinc, tin, nickel, chromium, aluminum, etc., on the surface of the iron.
    If plated with zinc, the reaction products of zinc with $$O_{2}$$ and $$H_{2}O$$ from a coating which sticks to the surface and thereby protects iron.

    ...view full instructions

    Corrosion is an electrochemical process. It involves:
    Solution

  • Question 4
    1 / -0
    The standard reduction potential data at $${25}^{o}C$$ is given below:

    (i) $${ E }^{ o }\left( { Fe }^{ 3+ },{ Fe }^{ 2+ } \right) =-0.77V$$
    (ii) $${ E }^{ o }\left( { Fe }^{ 2+ },{ Fe }^{  } \right) =-0.44V$$
    (iii) $${ E }^{ o }\left( { Cu }^{ 2+ },{ Cu }^{  } \right) =+0.34V\quad $$
    (iv) $${ E }^{ o }\left( { Cu }^{ + },{ Cu }^{  } \right) =+0.52V$$
    (v) $${ E }^{ o }({ O }_{ 2\ (g)}+{ H }^{ + }+4e\rightarrow 4O{ H }^{ - })=+1.23V$$
    (vi) $${ E }^{ o }({ O }_{ 2\ (g)}+2{ H }_{ 2 }O+4e\rightarrow 4{ OH }^{ - })=+0.40V$$
    (vii) $${ E }^{ o }\left( { Cr }^{ 3+ },{ Cr }^{  } \right) =-0.74V$$
    (viii) $${ E }^{ o }\left( { Cr }^{ 2+ },{ Cr }^{  } \right) =+0.91V\quad $$

    Match $${E}^{o}$$ of the redox pair in List-I with the values given in List-II and select the correct answer using the code given below the lists

    List IList II
    (P) $${E}^{o}({ Fe }^{ 3+ },{ Fe })$$(1) $$-0.18V$$
    (Q) $${E}^{o}(4{ H }_{ 2 }O\rightleftharpoons 4{ H }^{ + }+4{ OH }^{ - })$$(2) $$-0.4V$$
    (R) $${E}^{o}({ Cu }^{ + },{ Cu }\rightarrow 2{Cu}^{+})$$(3) $$-0.04V$$
    (S) $${E}^{o}({Cr}^{3+},{Cr}^{2+})$$(4) $$-0.83V$$
    Solution

  • Question 5
    1 / -0
    $$\cfrac { 1 }{ 2 } { H }_{ 2 }(g)+AgCl(s)={ H }^{ + }(aq)+{ Cl }^{ - }(aq)+Ag(s)$$ occurs in the galvanic cell:
    Solution
    $$\cfrac { 1 }{ 2 } { H }_{ 2 }(g)+AgCl(s)={ H }^{ + }(aq)+{ Cl }^{ - }(aq)+Ag(s)$$

    $$Pt/{ H }_{ 2 }(g),HCl(sol)\parallel AgCl(s)/{ Ag }^{ 2+ }$$

    Hence, option C is correct.
  • Question 6
    1 / -0
    One coulomb is the charge of
    Solution

  • Question 7
    1 / -0
    Unit of Faraday is
    Solution
    In the International System of Units ( SI ), the coulomb (C) is the preferred unit of electric charge quantity. It is equivalent to one ampere-second (1 A.s) and represents approximately $$6.24 * 10^{ 18}$$ electric charge carriers. To convert from coulombs to faradays, multiply by $$1.04 *10^{ -5}$$ .
  • Question 8
    1 / -0
    Which of the following facts about the chemical cell and concentration cell is correct?
    Solution

  • Question 9
    1 / -0
    When one of ampere current flows for $$1\ sec$$ through a conductor, this quantity of electricity is known as
    Solution
    $$\text{When one of ampere current flows for 1 sec through a conductor, this quantity of electricity is known as E.M.F.}$$.
    therefore option (C) is a correct answer.
  • Question 10
    1 / -0
    Hydrogen-oxygen fuel cells are used in space-craft to supply:
    Solution
    Fuel-cells are used to provide power and drinking water to astronauts in space programme.

    Option D is correct.
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