Home
Class 12
CHEMISTRY
Two metals M(1) and M(2) have E^(c-).(re...

Two metals `M_(1)` and `M_(2)` have `E^(c-)._(red)=-0.76 V` and `0.80V`, respectively. Which will liberate `H_(2)(g)` from `H_(2)SO_(4)` ?
Given `:`
`{:(M_(1)^(n+)+n e^(-)rarr M_(1),,,,E^(c-)._(red)=-0.76V),(M_(2)^(n+)+n e^(-)rarr M_(2),,,,E^(c-)._(red)=0.80V),(2H^(o+)+2e^(-) rarr H_(2)(g),,,,E^(c-)._(red)=0.0V):}`

Text Solution

Verified by Experts

`a.` Now constructing a cell of metal `M_(1)` with `H_(2)SO_(4)`,
`2M_(1)+nH_(2)SO_(4) rarr (M_(1))_(2)(SO_(4))_(n)+nH_(2). ,,,,,(i)`
`[M_(1)` undergoes oxidation to `M_(1)^(2+)` and `2h^(o+)` undergoes reduction to `H_(2)]`
`:. E^(c-)._(cell)=E^(c-)._(red)+E^(c-)._(o x i d )`
`=E^(c-)._(2H^(o+)|H_(2))+E^(c-)._(M_(1)|M_(1)^(2+))`
`=0+0.76=0.76V`
Since `E^(c-)._(cell)` is positive and reaction `(i)` is spontaneous, therefore, `M_(1)` will liberate `H_(2)` from `H_(2)SO_(4)`.
`b.` Now constructing a cell of metal `M_(2)` with `H_(2)SO_(4)`.
`:. E^(c-)._(cell)=E^(c-)._(red)+E^(c-)._(o x i d)`
`=E^(c-)._(2H^(o+)|H_(2))+E^(c-).M_(2)|M_(2)^(2+)`
`=0-0.80=-0.80V`
Since `E^(c-)._(cell)` is negative and reaction `(ii)` is non`-` spontaneous, thus, `M_(2)` will not liberate `H_(2)` from `H_(2)SO_(4)`.
Doubtnut Promotions Banner Mobile Dark
|

Topper's Solved these Questions

  • ELECTROCHEMISTRY

    CENGAGE CHEMISTRY|Exercise Solved Examples(Electrolysis And Electrolytic Cells)|12 Videos

  • ELECTROCHEMISTRY

    CENGAGE CHEMISTRY|Exercise Ex 3.1 (Objective)|28 Videos

  • ELECTROCHEMISTRY

    CENGAGE CHEMISTRY|Exercise Archieves Subjective|35 Videos

  • D AND F BLOCK ELEMENTS

    CENGAGE CHEMISTRY|Exercise Archives Subjective|29 Videos

  • GENERAL PRINCIPLES AND PROCESS OF ISOLATION OF ELEMENTS

    CENGAGE CHEMISTRY|Exercise Archives (Subjective)|14 Videos

Similar Questions

Explore conceptually related problems

Two metals A and B have E_(RP)^(@) = +0.76V and -0.80V respectively. Which will liberated H_(2) from H_(2)SO_(4) ?

E^(c-) of some elements are given as : {:(I_(2)+2e^(-)rarr 2I^(c-),,,,E^(c-)=0.54V),(MnO_(4)^(c-)+8H^(o+)+5e^(-)rarr Mn^(2+)+4H_(2)O,,,,E^(c-)=1.52V),(Fe^(3+)+e^(-)rarr Fe^(2+),,,,E^(c-)=0.77V),(Sn^(4+)+2e^(-)rarr Sn^(2+),,,,E^(c-)=0.1V):} a. Select the stronges reductant and weakes oxidant among these elements. b. Select the weakest reductant and strongest oxidant among these elements.

Knowledge Check

  • Two metals A and B have E_(RP)^(0) = – 0.76 V and + 0.80 V respectively. Which of these will liberate H_(2) from H_(2) SO _(4) ?

    A
    A
    B
    B
    C
    Both A and B
    D
    None of these

  • If {:(Sn^(2+) + 2e^(-) rarr Sn(s), E^(o) = - 0.14 V),(Sn^(4+)+2e^(-) rarr Sn^(2+),E^(o)= + 0.13 V):} then which of these is true?

    A
    `Sn^(2+)` is unstable and disproportionates to form `Sn^(4+)` and Sn
    B
    `Sn^(2+)` is stable and disproportionation reaction is not spontaneous
    C
    `Sn^(4+)` is easily reduced to Sn
    D
    None of these is correct

  • Given : A^(2+)+2e^(-) rarr A(s)" "E^(c-)=0.08V B^(o+)+e^(-)rarr B(s)" "E^(c-)=-0.64V X_(2)(g)+2e^(-) rarr 2X^(c-)" "E^(c-)=1.03V Which of the following statements is // are correct ?

    A
    `X_(2)(g)` will oxidize both `(A)` and `(B)`.
    B
    `A^(2+)` will oxidize both `(A)` and `(B)`
    C
    The reaction
    `2X^(c-)(1.0M)+A^(2+)(1.0M) rarr X_(2)(1 atm)+A(s)`
    will be spontaneous.
    D
    The oxidizing power of `A^(2+),B^(o+),` and `X_(2)(g)` is in the order `X_(2)gtA^(2+)gtB^(o+)`

    • Similar Questions

    Explore conceptually related problems

    An aqueous solution containing 1 M NiSO_(4) & 1 M S_(2) O_(8)^(2-) is electrolysed using palladium electrode, at 25^(@)C {:(Ni^(+2)+2e^(-) rarrNi,,E^(@)=-0.25 V),(2H^(+)+2e^(-)rarrH_(2),,E^(@)=0.00 V ),(O_(2)+4H^(+)+4e^(-)rarr2H_(2)O,,E^(@)=1.23 V),(Pd^(+2)+2e^(-) rarrPd,,E^(@) = 0.92 V),(S_(2)O_(8)^(2-)+2e^(-) rarr 2SO_(4)^(2-),,E^(@)=2 V):} pH of solution is 7. Select the correct statement on the basis of above given information. (Ignore over voltage & neglect variation of E_(Pd^(+2)//Pd)^(@) with concentration)

    For a given reaction :M^((x+n))+n e^(-)rarr M^(c+),E^(c-).(red) is known along with M^(x+n) and M^(x+) ion concentrations. Then

    The standard reduction potential at 298 K for the following half reactions are given below: Fe^(3+)(1M)+e^(-)rarrFe^(2+)(aq),E^(o)=0.770 V, Zn^(2+)(1M)+2e^(-)rarrZn(s),E^(o)=-0.762 V Cd^(2+)(1M)+2e^(-)rarrCd(s),E^(o)=-0.402 V, 2H^(+)(1M)+2e^(-)rarrH_(2)(g),E^(o) = 0.00 V. The strpongest reducing agent is:

    Given : Oxidation H_(2)O_(2) rarr O_(2)+2H^(o+)+2e^(-)" "E^(c-)=-0.69V, 2F^(c-)rarr F_(2)+2e^(-)" "E^(c-)=-287V, Reduction : H_(2)O_(2)+2H^(o+)+2e^(-) rarr 2H_(2)O" "E^(c-)=1.77V, 2I^(c-)rarr I_(2)+2e^(-)" "E^(c-)=-0.54V, Which of the following statements is // are correct ?

    Use the standard reduction potentials to determine what is observed at the cathode during the electrolysis of a 1.0 M solution of KBr that contains phenolphthalein. O_(2)(g) + 4H^(+)(aq) + 4e^(-) rightarrow 2H_(2)O(l) E^(@) = 1.23V Br_(2)(l) + 2e^(-) rightarrow 2Br^(-)(aq) E^(@) = 1.07V 2H_(2)O(l) + 2e^(-) rightarrow H_(2)(g) + 2OH^(-) E^(@0 = -0.80V K^(+) (aq) + e^(-) rightarrow K(s) E^(@) = -2.92V

    Home
    Profile