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Electrochemistry Test - 44

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Electrochemistry Test - 44
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  • Question 1
    1 / -0
    1000 mL 1 M CuSO4(aq)CuSO_4(aq) is electrolysed by 9.65 amp current for 100 sec using PtPt-electrode. Which is incorrect statement?
    Solution
    CuSO4Cu2++SO42{ CuSO }_{ 4 }\rightleftharpoons { Cu }^{ 2+ }+{ SO }_{ 4 }^{ 2- } is formed from strong acid H2SO4{ H }_{ 2 }{ SO }_{ 4 } and weak base Cu(OH) 2{ Cu\left( OH \right)  }_{ 2 }. Hence, the CuSO4{ CuSO }_{ 4 } is acidic in nature and it's pHpH is thus less than 77 always.
    \therefore  The incorrect statement is pHpH of solution after electrolysis is 88.
  • Question 2
    1 / -0
    Iron nails rust fast when used for fixing plates or strips of aluminium on a building. Similarly, a water pipe made of iron corrodes faster when connected to a pipe of copper. This happens because of the?
    Solution
    This is due to higher rusting power of iron.
    When iron pipe is connected to copper corrosion is caused by self induced current created. The current is generated due to difference in electrical potential. The water act as a weak electrolyte.
  • Question 3
    1 / -0
    The standard reduction potential for the half-cell reaction, Cl2+2e2ClCl_2+2e^-\rightarrow 2Cl^- will be:
    Pt2++2ClPt+Cl2,Ecello=0.15Pt^{2+}+2Cl^-\rightarrow Pt+Cl_2, E^o_{cell}=-0.15V;

    Pt2++2ePt,Eo=1.20Pt^{2+}+2e^-\rightarrow Pt, E^o=1.20V
    Solution
    A galvanic cell or simple battery is made of two electrodes. Each of the electrodes of a galvanic cell is known as a half cell. In a battery, the two half cells form an oxidizing-reducing couple. When two half cells are connected via an electric conductor and salt bridge, an electrochemical reaction is started.

    The given reaction is, Pt2++2ClPt+Cl2,Ecell0=0.15VPt^{2+}+2Cl^{-}\rightarrow Pt+Cl_2,E^{0}_{cell}=-0.15V

    On reversing the above reaction, Pt+Cl2Pt2++2Cl,Ecell0=0.15VPt+Cl_2\rightarrow Pt^{2+}+2Cl^{-},E^{0}_{cell}=0.15V

    Also given,Pt2++2ePt,Ecell0=1.20VPt^{2+}+2e^{-}\rightarrow Pt,E^{0}_{cell}=1.20V

    As we know the relation,

    Ecell0=Ecathode0Eanode0E^{0}_{cell}=E^{0}_{cathode}-E^{0}_{anode}

    So we get, 

    ECl2/2Cl0=0.15(1.20)=1.35VE^{0}_{Cl_2/2Cl^{-}}=0.15-(-1.20)=1.35V

    Hence, the correct option is B\text{B}
  • Question 4
    1 / -0
    Match the column I with column II and mark the appropriate choice.
    Column I
    Column II
    (A)Electrochemical equivalent(i)Potential difference x Quantity of charge
    (B)Faraday(ii)Mass of substance deposited by one coulomb of charge
    (C)Ampere(iii)Charge carried by one mole of electrons
    (D)Electrical energy(iv)One coulomb of electric charge passed through one second
    Solution
    • Electrochemical equivalent: The weight of a substance deposited or evolved during electrolysis by the passage of a specified quantity of electricity and usually expressed in grams per coulomb
    • Faraday: A measure of the electric charge carried by one mole of electrons, used in electrolysis as the quantity of charge required to deposit or liberate one gram equivalent weight of a substance.
    • Ampere: By definition, if 1 Coulomb of Charge flows through a Conductor in 1 second then the magnitude of the current is said to be 1 A.
    • Electrical Energy: Electrical energy is the work done by electric charge. If current ii ampere flows through a conductor or through any other conductive element of potential difference V volts across it, for time tt second. We know Q=I×tQ=I\times t and hence electrical energy becomes, Potential difference×\timesquantity of charge
  • Question 5
    1 / -0
    How much electricity in terms of Faraday is required to produce 100100g of Ca from molten CaCl2CaCl_2?
    Solution
    We have to note that one mole of electron charge is equivalent to one Faraday.
    Given reaction,
    CaCl2Ca2++2ClCaCl_2\rightarrow Ca^{2+}+2Cl^{-}
    Ca is undergoing oxidation i.e, Ca2++2eCaCa^{2+}+2e^{-}\rightarrow Ca
    We can observe that 40g of Ca takes 2e2e^{-} charge,
    So,1 mole of Ca(40g)2FCa(40g)\equiv 2F
    Now, 100g of Ca5FCa\equiv 5F
  • Question 6
    1 / -0
    Electrical conductance through metals is called metallic or electronic conductance and is due to the movement of electrons. The electronic conductance depends on:
    Solution
    Electronic conductance depends upon-
    • Nature of electrolyte : The conductance of an electrolyte depends upon the number of ions present in the solution. Therefore, the greater the number of ions in the solution the greater is the conductance.
    •  Concentration of the solution : The molar conductance of electrolytic solution varies with the concentration of the electrolyte. In general, the molar conductance of an electrolyte increases with decrease in concentration or increase in dilution.nature of the 
    •  Temperature : The conductivity of an electrolyte depends upon the temperature. With increase in temperature, the conductivity of an electrolyte increases.
    • Valence electrons per atom present in metals describes the electric conductance of it.
    • Nature and structure is also a significant factors which influences electrical conductance.
  • Question 7
    1 / -0
    Which of the following is the correct cell representation for the given cell reaction?
    Zn+H2SO4ZnSO4+H2Zn+H_2SO_4\rightarrow ZnSO_4+H_2
    Solution
    In the given reaction the oxidation state of ZnZn is changing from 0 to +2 so it is undergoing oxidation. Whereas the oxidation state of H+H^{+} is changing from +1 to 0 which implies it is undergoing reduction. And in cell representation oxidation is represented in left part and the reduction in right. It has to be remembered that H2H_2 electrode has Pt too in it.
    So, the correct representation of the cell is,
    ZnZn2+H+,H2PtZn|Zn^{2+}||H^{+},H_2|Pt
    Hence option (B) is correct
  • Question 8
    1 / -0
    A galvanic cell has electrical potential of 1.11.1 V. If an opposing potential of 1.11.1 V is applied to this cell, what will happen to the cell reaction and current flowing through the cell?
    Solution
    We can study the effect of opposing potential applied to a galvanic cell as follows.
    1) When Eexternal<1.1VE_{external}< 1.1V
    In a galvanic cell, if an external opposite potential is applied and increased slowly, we find that the reaction continues to take place till the opposing voltage reaches the value 1.1 V.
    2) When  Eexternal=1.1VE_{external}= 1.1V
     When the external voltage is 1.1V the reaction stops altogether and no current flows through the cell. 
    3) When  Eexternal>1.1VE_{external}> 1.1V
    Any further increase in the external potential again starts the reaction but in the opposite direction. It now functions as an electrolytic cell, a device for using electrical energy to carry non-spontaneous chemical reactions.

    Hence option (A) is correct.
  • Question 9
    1 / -0
    How long would it take to deposit 5050g of AlAl from an electrolytic cell containing Al2O3Al_2O_3 using a current of 105105 ampere?
    Solution
    Faraday's First Law of Electrolysis states that only, According to this law, the chemical deposition due to flow of current through an electrolyte is directly proportional to the quantity of electricity (coulombs) passed through it.
    The reaction of Al is,
    Al3++3eAlAl^{3+}+3e^{-}\rightarrow Al
    Now as n=3, Equivalent weight of Al=273=9\dfrac{27}{3}=9
    As per Faraday's first law,
    W=Z×I×t    Z=Eq.wt96500W=Z\times I\times t\implies Z=\dfrac{Eq. wt}{96500}
        t=W×96500Eq.wt×1=50×965009×105=5102.2\implies t=\dfrac{W\times96500}{Eq. wt\times1}=\dfrac{50\times96500}{9\times105}=5102.2 or 1.421.42 hourshours
  • Question 10
    1 / -0
    If a current of 1.51.5 ampere flows through a metallic wire for 33 hours, then how many electrons would flow through the wire?
    Solution
    According to this Faraday's first law, the chemical deposition due to the flow of current through an electrolyte is directly proportional to the quantity of electricity (coulombs) passed through it.

    So, Q=I×t=1.5×3×60×60=16200CQ=I \times t=1.5\times3\times 60 \times 60=16200 C

    Charge on one electron =1.601×1019C=1.601\times10^{-19}C

    Now,
    16200C charge is on 1×162001.601×1019\dfrac{1\times16200}{1.601\times10^{-19}} electrons =1.01×1023=1.01\times10^{23}.  
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