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Why is an external e.m.f of more than 2....

Why is an external e.m.f of more than `2.2V` required for the extraction of `Cl_(2)` from brine ?

Text Solution

Verified by Experts

The oxidation reaction involved in the oxidation of `Cl^(-)` ions to `Cl_(2)` is :
`2Cl^(-)(aq)+2H_(2)O(l) to 2OH^(-)(aq)+H_(2)(g)+ Cl_(2)(g)`

The value of `DeltaG^(@)` for the reaction is `+422` kJ (standard value). The `E_("Cell")^(@)` can be calculated as,
`DeltaG^(@)=- nF E_("Cell")^(@) " or " E_("Cell")^(@)=- (DeltaG^(@))/(nF)`
`DeltaG^(@)=422 kJ =422000 J=422000 CV, n=2`
`E_("Cell")^(@)=-((422000 CV))/((2xx96500 C))=-2.19~~-2.20 V`

Since `E_("Cell")^(@)` comes out to be `-2.20` V for the electrolytic cell involving the oxidation of `Cl^(-)` ions, an external emf of more than `2.2` V is needed.
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Knowledge Check

  • Two cells of e.m.f. 2.5 V and 2.0 V having internal resistance of 1Omega and 2Omega respectively are connected in parallel with similar poles connected together so as to send the current in the same direction through an external resistance of 2Omega . The current in the external resistance is

    A
    0.87 A
    B
    1.29 A
    C
    1.00 A
    D
    2.29 A
  • If an external opposing E.M.F. is slightly greater than E.M.F. of Daniel cell the reaction will

    A
    be reversed
    B
    remain same
    C
    move in the forward direction
    D
    all are right
  • Is the work required to be done by an external force on an object on a frictionless , horizontal surface to accelerate it from a speed v to a speed 2v

    A
    equal to the work required to accelerate the object from `t = 0` to `v`.
    B
    twice the work required to accelerate the object from `v = 0` to `v`.
    C
    three time the work required to accelerate the object from `v = 0` to `v`.
    D
    four time the work required to accelerate the object from ` 0` to `v`. Or
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