Calculate the cell potential for the reaction as written at 25.00 C given that [Cr2+ ]=0.862 M and [Fe2+ ]=0.0140M Use the standard reduction potentials in this table.

0.497 V

Explanation:

We need to apply the Nernst equation here. According to the Nernst equation;

Ecell= E°cell - 0.0592/n log Q

Where;

Ecell= emf of the cell under the given conditions

E°cell= standard emf of the cell

n= number of electrons transferred

Q= reaction quotient= [products]/[Reactants]= [Cr^2+]/[Fe^2+]

Balanced redox reaction equation; Cr(s)+Fe2+(aq)>Cr2+(aq)+Fe(s)

Values of standard electrode potential

Fe II: -0.44 V

Cr II: -0.91 V

E°cell= (-0.44) - (-0.99)

E°cell= 0.55V

[Fe2+ ]=0.0140M

[Cr2+ ]=0.862 M

Number of electrons transferred (n)= 2

Substituting into the Nernst's equation;

Ecell= 0.55- 0.0592/2 log [0.862]/[0.0140]

Ecell= 0.55 - 0.053

Ecell= 0.497 V

the answer to this question is 3.