Avoltaic cell utilizes the following reaction: 4fe2+(aq)+o2(g)+4h+(aq)→4fe3+(aq) +2h2o(l).
the emf under standard contions is .46 v.
what is the emf of this cell when [fe2+]= 2.0m , [fe3+]= 1.9

The cell EMF is obtained from Nernst equation as 0.451 V.

The reaction equation is given by;

4Fe2+ (aq) + O2(g) + 4H+ (aq)>4Fe3+(aq)+2H2O(l)

We have the following information;

EMF under standard conditions = 0.46 V

[Fe2+]= 2.0M

[Fe3+]= 1.9 M

From Nernst equation;

Ecell = E°cell - 0.0592/n logQ

Where;

E°cell = 0.46 V

n = 4

Q =  [Fe3+]^4/ [Fe2+]^4

Q = [1.9]^4/[2.0]^4

Q = 0.8

Substituting values;

Ecell =  0.46 - 0.0592/4 log (0.8)

Ecell = 0.451 V

Learn more:link

The Emf of the given cell at [Fe2+] = 2.0 M and [Fe3+] = 1.9 M is 0.48 V

Explanation:

The half cell reaction can be written as :

Anode-Half (oxidation) :

......E = 0.77 V

(multiply this equation by 4 to balance the electrons)

Cathode-half (reduction)

....E= 1.23 V

According to Nernst Equation

n = number of electron transferred in the cell reaction = 4

The balanced equation is :

Insert the value of log Q in Nernst  Equation:

(using :

)

it means there are 3 atoms of oxygen