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The standard reduction potential E^(@) ...

The standard reduction potential `E^(@) ` for half reaction are `Zn→Zn^(2+)+2e^−` ;`E^o` =+0.76V `Fe→Fe^(2+)+2e^−` ;`E^o`=+0.41V The EMF of the cell reaction is: `Fe^(2+)+Zn→Zn^(2+)+Fe`

A

` -0.35 V`

B

` +0.35 V `

C

` +1.17 V `

D

` -1.17V `

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Given the standard half - cell potentials (E^(@)) of the following as : Zn rarr Zn^(2+)+2e , E^(@)=+0.76V Fe rarr Fe^(2+)+2e, E^(@)=0.41V The the standard EMF of the cell with the reaction Fe^(2+)+Zn rarr Zn^(2+)+Fe is -

Why standard reduction potential (E^(0)) of Zn^(2+)|Zn system is more negative than expected ?

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Write Nernst equation for electrode potential for the following electrodes : (a) Zn^(2+)+2e rarr Zn

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

A few half - cell reactions and their standard reduction potentials are given belwo : Pb^(2+)+2e rarr Pb," "E^(@)=-0.126V," "Al^(3+)+3e rarrAl, E^(@)=-1.66V Zn^(2+)+2e rarr Zn, E^(@)=-0.76V , Cd^(2+)+2e rarr Cd, E^(@)=-0.402V Arrange Zm Pb, Cd and Al in ascending order of their ability to act as reducing agent.

A half - cell is constructed by dipping a Zn- rod into a solution of ZnSO_(4) at 25^(@)C . If the concentration of Zn^(2+) ion in the solution be 0.01 (M), calculate the oxidation potential of the half- cell, Given : E_(Zn^(2+)|Zn)^(@)=-0.76V .

See whether the cell representation given below is in accordance with the conventation or not. If it is not, then write the correct representation and reaction of the cell. Calculate the standard EMF of the cell: Cu|Cu^(2+)||Zn|Zn^(2+) . ["Given : "E_(Cu^(2+)|Cu)^(@)=+0.34V, E_(Zn^(2+)|Zn)^(@)=-0.76V]

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