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Calculate the emf of the given Fe ^(2+...

Calculate the emf of the given
`Fe ^(2+) + Zu to Zn ^(2+) + Fe `
The standard reduction potential `E^(@)` for half reactions are
`An = Zn ^(2+) + Ze , E ^(@) = + 0. 76 V`
`Fe = Fe ^(2+) + Ze , E^(@) = + 0. 41 V`

A

`-0.35 V`

B

`+ 0.35`

C

`+ 1.17 V`

D

`-1.17 V`

Text Solution

Verified by Experts

The correct Answer is:
B
`E _(cell) =E _("anode(op)")^(@) - E _("cathode (op)") ^(@)`
`= 0.76 -0.41`
`= + 0.35 V`
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The standared reduction potntial E^o for the half-reactions are as. Zn=Zn^(2+) + 2 e^- , E^o = + 76 V Fe = Fe^(2+) + 2e^- , E^o = + 0.41 V . The EMF for cell recaction Fe^(2+) = Zn rarr Zn^(2+) + Fe is.

The standard oxidation potential E^@ for the half cell reaction are Zn rarr Zn^2+2e^- E^@=+ 0.76 V Fe rarr Fe^2+ + 2e^- E^@=+ 0.41 V EMF of the cell rection is Zn+Fe^(2+) rarr Zn^(2+)+Fe

Given that the standard reduction potentials E ^(0) of Fe^(+2)|Feis 0.26V and Fe ^(+3) |Fe is 0.76V respectively. The E ^(@)of Fe ^(+2)|Fe^(+3) is:

E^(o) for the reaction Fe + Zn^(2+) rarr Zn + Fe^(2+) is – 0.35 V. The given cell reaction is :

The standard oxidation potentials, E^(@) , for the half reactions are as, Zn rarr Zn^(2+) + 2e^(-), " " E^(@) = + 0.76 volt Fe rarr Fe^(2+) + 2e^(-), " " E^(@) = +0.41 volt The emf of the cell, Fe^(2+) + Zn rarr Zn^(2+) + Fe is:

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