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Zinc rod is dipped in 0.1 M solution of ...

Zinc rod is dipped in 0.1 M solution of `ZnSO_4`. The salt is 95% dissociated at this dilution at 298 K. Calculate the electrode potential.
[Given, `E_(Zn^(2+)//Zn)^(@) = -0.76 V`]

Text Solution

Verified by Experts

Applying Nernst Equation
`E_(Zn^(2+)//Zn) = E_(Zn^(2+)//Zn)^(@) -0.0591/2 log 1/([Zn^(2+)])`
`[Zn^(2+)] =0.1 xx 95/100 = 0.095 M`
`Zn^(2+) + 2e^(-) to Zn` or n=2
`E_(Zn^(2+)//Zn) = -0.76 V - 0.0591/2 [log1000- log 95]=-0.76 V - (0.0591)/2 [ 3.000 - 1.9777]`
`=-0.76 V - (0.0591)/2 xx 1.0223 = -0.76 V - 0.00604/2`
`=-0.76 V - 0.0302 = -0.7902` V
Electrode potential `=-0.7902` V
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A zinc rod is dipped in 0.1 M solution of ZnSO_(4) . The salt is 95% dissociated at this dilution at 298K. Calculation the electrode potential (E_(Zn^(2+)//Zn)^(@)=-0.76V)

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Knowledge Check

  • Standard reduction electrode potential of Zn^(2+)//Zn is -0.76V . This means:

    A
    `ZnO` is reduced to `Zn` by `H_(2)`
    B
    `Zn` can't liberates `H_(2)` with conentrated acid
    C
    `Zn` is generally the anode in an electrochemical cell
    D
    `Zn` is generally the cathode in an electrochemical cell.

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