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Electrochemistry Test - 68

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Electrochemistry Test - 68
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Weekly Quiz Competition
  • Question 1
    1 / -0
    For the fuel cell reaction : 2H2(g)+O2(g)2H2O(l);ΔfH298(H2O,l)= 285.5 kJ/mol2{H_2}\left( g \right) + {O_2}\left( g \right) \to 2{H_2}O\left( l \right);\quad {\Delta _f}H_{298}^ \circ \left( {{H_2}O,l} \right) =  - 285.5\ kJ/mol what is ΔS2980\Delta S_{298}^0 for the given fuel cell reaction?

    Given: O2(g)+4H+(aq)+4e2H2O(l);E=1.23V{O_2}\left( g \right) + 4{H^ + }\left( {aq} \right) + 4{e^ - } \to 2{H_2}O\left( l \right)\,; \quad {E^ \circ } = 1.23V
    Solution

  • Question 2
    1 / -0
    The electrolysis of a solution resulted in the formation of H2(g)H_{2}(g) at the cathode and O2(g)O_{2}(g) at the anode. The solution is :
    Solution

  • Question 3
    1 / -0
    An electric current is passed through a copper voltameter and a water voltameter connected in series. If the copper of the copper voltameter now weights 16 mg less, hydrogen liberated at the cathode of the water voltameter measures at STP about:
    Solution

  • Question 4
    1 / -0
    The charge in coulombs of 1 mole of N3 N^{3-} is (The charge on an electron is: 1.602×1019C) 1.602 \times 10^{-19} C )
    Solution

  • Question 5
    1 / -0
    Coulomb is equal to ........
    Solution

    The coulomb (symbolized C) is the standard unit of electric charge in the International System of Units (SI)In terms of SI base units, the coulomb is the equivalent unit of one ampere-second.

    So option A is correct.

  • Question 6
    1 / -0
    The current strength of 3.863 amp3.863\ amp was passed through molten calcium oxide for 4141 minutes and 4040 seconds. The mass of calcium in grams deposited at the cathode is: (Atomic mass of Ca is 40 g/mol,1 f=96500 C)40\ g/mol, 1\ f=96500\ C)
    Solution

  • Question 7
    1 / -0
    Consider the half-cell reaction(s)
    Cu2++eCu+;Eo=0.15V{Cu}^{2+}+{e}^{-}\rightarrow {Cu}^{+};{E}^{o}=0.15V
    Cu2++2eCu;Eo=0.33V{Cu}^{2+}+2{e}^{-}\rightarrow {Cu};{E}^{o}=0.33V
    Eo{E}^{o} for the half-cell reaction: Cu++eCu{Cu}^{+}+{e}^{-}\rightarrow Cu will be:
    Solution

  • Question 8
    1 / -0
    The charge required for the reduction of 1 mol of  MnO4 - {\text{MnO}}_{\text{4}}^{\text{ - }} to MnO2{\text{Mn}}{{\text{O}}_2} is:
    Solution

  • Question 9
    1 / -0
    Consider the following   Eo  \;{{\text{E}}^{\text{o}}}\; values,
    EoFe3 + /Fe2 + = + 0.77V;{{\text{E}}^{\text{o}}}_{{\text{F}}{{\text{e}}^{{\text{3 + }}}}{\text{/F}}{{\text{e}}^{{\text{2 + }}}}} = {\text{ + 0}}{\text{.77V}};EoSn2 + /Sn= 0.14V        {{\text{E}}^{\text{o}}}_{{\text{S}}{{\text{n}}^{{\text{2 + }}}}{\text{/Sn}}} =  - {\text{0}}{\text{.14V}}\;\;\;\;

    Under standard conditions the potential for the reaction is:
    Sn(s) + 2Fe3 + (aq)2Fe2 + (aq) + Sn2 + (aq){\text{Sn}}\left( {\text{s}} \right){\text{ + 2F}}{{\text{e}}^{{\text{3 + }}}}\left( {{\text{aq}}} \right) \to {\text{2F}}{{\text{e}}^{{\text{2 + }}}}{\text{(aq) + S}}{{\text{n}}^{{\text{2 + }}}}\left( {{\text{aq}}} \right)
    Solution

  • Question 10
    1 / -0
    Three faradays of electricity is passed through molten solutions of AgNO3,NiSO4Ag{NO}_{3}, Ni{SO}_{4} and CrCl3Cr{Cl}_{3} kept in three vessels using inert electrodes. The ratio in mol in which the metals Ag,NiAg, Ni and CrCr will be deposited is:
    Solution

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