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Electrochemistry Test - 49

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Electrochemistry Test - 49
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  • Question 1
    1 / -0
    According to electrochemical theory of corrosion involves:
    Solution
    In corrosion $$Fe^{2+}$$ get oxidised to $$Fe^{3+}$$ which takes place on anode surface.
    Anode: $$Fe(s)\rightarrow { Fe }^{ 2+ }+2{ e }^{ - }$$
    Cathod:$${ O }_{ 2 }+4{ H }^{ + }+4{ e }^{ - }\rightarrow { 2H }_{ 2 }O$$
    Overall:$$2Fe(s)+{ O }_{ 2 }+4{ H }^{ + }\rightarrow 2{ Fe }^{ 2+ }+{ H }_{ 2 }O$$
                 $${ 4Fe }^{ 2+ }+{ O }_{ 2 }+(2+4x){ H }_{ 2 }O\rightarrow { 2Fe }_{ 2 }{ O }_{ 3 }.x{ H }_{ 2 }O+4{ H }^{ + }$$
    So,$$Fe^{3+}$$ get deposited on metal surface.
  • Question 2
    1 / -0
    $$10800C$$ of electricity on passing through the electrolyte solution deposited $$2.977$$ of metal with atomic mass $${ 106.4gmol }^{ -1 }$$ the charge on the metal cation is:
    Solution
    $$10800C\rightarrow 2.977g$$ of metal deposited.
    $$96500C\rightarrow \cfrac { 2.977 }{ 10800 } \times 96500\\  \quad \quad \quad =26.600g$$
    Charge on metal cation is $$=\cfrac { 106.4 }{ 26.6 } $$
                                                $$=+4$$
  • Question 3
    1 / -0
    Number of coulombs required to liberate 0.5 mol of $${ O }_{ 2 }$$ is:
    Solution
    Formation of $${ O }_{ 2 }$$ is $${ 4e }^{ - }$$ transfer.

    i.e. $$4\times 96500C\rightarrow 1 \ mol$$ of $${ O }_{ 2 }$$

    $$\therefore  0.5\ mol{  O }_{ 2 }\rightarrow \dfrac { 1 }{ 2 } \times 4\times 96500$$

    $$=193000C$$
  • Question 4
    1 / -0
    If the density of copper is $$8.94 g/{ cm }^{ 3 }$$, the number of faradays required to plate an area $$(10cm\times 10cm)$$ of thickness of $${ 10 }^{ -2 }$$ cm using $${ CuSO }_{ 4 }$$ solution as electrolyte is:
    Solution
    Vol. of plate $$=10\times 10\times \dfrac { 1 }{ 100 } =1\ cm^3$$

    $$\therefore $$  density $$=\dfrac { mass }{ volume } $$

    $$\therefore $$  mass of $$Cu$$ to be deposited $$=8.94g$$

    $${ Cu }^{ 2+ }+2{ e }^{ - }\rightarrow Cu$$

    moles of $$Cu=\dfrac { 8.94 }{ 63.5 } $$

    Faraday required $$=\dfrac { 2\times 8.94 }{ 63.5 } =0.28F$$
  • Question 5
    1 / -0
    The ratio of volume of $$H_2$$ and $$Cl_2$$ evolved at NTP by electrolysis of aqueous solution of $$HCl$$ is:
    Solution
    Answer is option $$C.$$
    Electrolysis of $$HCl-$$
    At cathode, 
                 $$2H^++2e^-\longrightarrow H_{2(g)}$$
    At anode,
                  $$2Cl^-\longrightarrow Cl_{2(g)}+2e^-$$
    Theoretically the gas volume ratio is $$1:1$$
    as overall reaction is given by,
    $$2H^+_{(aq)}+2Cl^-_{(aq)}\longrightarrow H_{2(g)}+Cl_{2(g)}$$
  • Question 6
    1 / -0
    How many faradays are needed for reduction of 2.5 moles of $$Cr_2O_7^{2-}$$ into $$Cr^{3+}$$ ?
    Solution
    $${ n }_{ factor }=2\times (3)=6\left( for2{ Cr }^{ +6 }\rightarrow 2{ Cr }^{ +3 } \right) $$
    $$\therefore$$ Number of equivalent $$=\left( { n }_{ factor }\times no. of\quad moles \right) $$
                                             $$=(6\times2.5)$$
                                             $$=15\quad equivalent$$
    Or, $$15 faraday$$.
  • Question 7
    1 / -0
    A current $$9.65$$ ampere is passed through the aqueous solution $$NaCl$$ using suitable electrodes for $$1000\ s$$. The amount of $$NaOH$$ formed during electrolysis is:
    Solution

  • Question 8
    1 / -0
    The standard electrode potential of the half cells is given below:
    $$Zn^{2+}+2e^{-}\rightarrow Zn; E=-0.76 V$$
    $$Fe^{2+}+2e^{-}\rightarrow Fe; E=-0.44 V$$
    The emf of the cell $$Fe^{2+}=Zn^{-}\rightarrow Zn^{2+}+Fe $$
    Solution
    $${ E }_{ cell }^{ 0 }=\left( { E }_{ { Fe }^{ 2+ }|Fe }^{ 0 }+{ E }_{ Zn|{ Zn }^{ 2+ } }^{ 0 } \right) $$
             $$=\left( -0.44 V + 0.76 \right)$$
       $$E^0=+0.32 Volt$$
  • Question 9
    1 / -0
    Amount of charge is required to convert $$17\ g\ H_{2}O_{2}$$ into $$O_{2}$$.
    Solution
    $$2{ H }_{ 2 }\overset { \left( -2 \right)  }{ { O }_{ 2 } } \longrightarrow 2{ H }_{ 2 }O+{ \overset { \left( 0 \right)  }{ O }  }_{ 2 }$$
    Change in electrons $$=2$$
    No. of moles of $$17$$ gms of $${ H }_{ 2 }{ O }_{ 2 }=\dfrac { 17 }{ 34 } =0.5$$
    Charge of one electrons $$=1.607\times { 10 }^{ -9 }$$ coloumbs.
    Total no. of electrons transferred $$=0.5\times 2\times 6.023\times { 10 }^{ 23 }$$
    Total charged required $$=0.5\times 2\times 6.023\times { 10 }^{ 23 }\times 1.607\times { 10 }^{ -19 }$$
                                            $$=9.6\times { 10 }^{ 4 }$$
                                            $$=1$$ Faraday.
    Ans :- Option A.
  • Question 10
    1 / -0
    Given,
    $$E_{Ag^{+}/Ag}^{\circ}  = +0.80V, E_{Co^{2+}/Co}^{\circ} = -0.28 V, E_{Cu^{2+}/Cu}^{\circ} = +0.34 V, E_{Zn^{2+}/Zn}^{\circ} = -0.76V$$, 
    Which metal will corrode to greater extent?
    Solution
    Metal having much more positive potential will Oxidise less easily and coorrode slowly and vice-versa. So, Zinc will corrode fastly.
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