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For Zn^(2+)|Zn, E^(@)=-0.76V then EMF of...

For `Zn^(2+)|Zn, E^(@)=-0.76V` then EMF of the cell `Zn|Zn^(2+)(1M)||2H^(+)(1M)|H_(2)(1atm)` will be -

A

`-0.76V`

B

`0.76V`

C

`0.38V`

D

`-0.38V`

Text Solution

Verified by Experts

The correct Answer is:
B
`E_("cell")^(@)=E_("right")^(@)-E_("left")^(@)`
`E_("cell")^(@)=0-(-0.76)=0.76`
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Knowledge Check

  • At 25^(@)C tempertaure, E_("cell")^ of the galvanic cell Zn|Zn^(2+)(aq, 1M)||H^(+)(aq)|H_(2)("g, 1 atm")|Pt is +0.583V . If E_(Zn^(2+)|Zn)^(@)=-0.76V , then pH of the aqueous solution present in the reduction half - cell is -

    A
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    C
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    D
    4.5

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    A
    equilibrium of the cell reaction will shift towards left
    B
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    C
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    D
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  • At 25^(@)C, E_("cell") of the galvanic cell Zn|Zn^(2+)(aq, 0.01M)||Cu^(2+)(aq, 0.01M)|Cu is +1.07V . The equilibrium constant of the cell reaction occurring in this cell is -

    A
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    B
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    C
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    D
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