Home
Class 12
CHEMISTRY
The emf values of the cell reactions Fe^...

The emf values of the cell reactions `Fe^(3+)+e^(-)rarr Fe^(2+) and Ce^(2+)rarr Ce^(3+)+e^(-)` are 0.61 V and `-0.85V` respectively. Construct the cell such that the free energy change of the cell is negative. Calculate the emf of the cell.

Text Solution

Verified by Experts

`E_("cell")=0.24V`
Promotional Banner

Similar Questions

Explore conceptually related problems

The relationship between free energy change and e.m.f. of a cell is

The e.m.f. of a cell reaction is positive when the free energy of the cell reaction is :

The standard electrode potential of Zn^(2+)|ZnandCu^(2+)|Cu are - 0.76 and 0.34 V respectively . The standard e.m.f. of the cell is :

Calculate the standard emf of the cell having the standard free energy change of the cell reaction is -64.84 kJ for 2 electrons transfer.

The standard free energy change of the reaction M_((aq))^(+)+e rarrr M_((s)) is -23.125kJ . Calculate the standard emf of the half cell.

For the cell reaction Zn(s) + Cu^(2+)(aq) rarr Cu(s) + Zn^(2+)(aq) E_("cell")^(@) = 1.10 V, DeltaG^(@) is :

The cell potential (E_("cell")^(@)) is related to free energy change (Delta G^(@)) as :

The standard reduction potential of Fe^(3+), Fe^(2+)//Pt is + 0.771V . This half cell is connected with another half cell such that e.m.f. of the cell is 0.771V. What is the other half cell ?

The emf of a galvanic cell constituded with the electrodes Zn^(2+)//Zn (-0.76V) and Fe^(2+)//Fe (-0.41V) is