Question

BACKROUND:

E=E∘−(0.0592/n)logQ

The reaction quotient has the usual form

Q=[products]^x/[reactants]^y

A table of standard reduction potentials gives the voltage at standard conditions, 1.00 M for all solutions and 1.00 atm for all gases. The Nernst equation allows for the calculation of the cell potential E at other conditions of concentration and pressure.

-------------------------------------

For the reaction

2Co3+(aq)+2Cl−(aq)→2Co2+(aq)+Cl2(g).  E∘=0.483 V

what is the cell potential at 25 ∘C if the concentrations are [Co3+]= 0.592 M , [Co2+]= 0.866 M , and [Cl−]= 0.704 M and the pressure of Cl2 is PCl2= 6.00 atm ?

Express your answer with the appropriate units.

Answer 1

2Co3+(aq)+2Cl−(aq)→2Co2+(aq)+Cl2(g).  E∘=0.483 V

Ecell = E° - 0.0592/n * log(Q)

note that n = number of electrons transferred:

Co3+ + e- --> Co+2; 1 electron, but it has 2 coeff so 1x2 = 2 electrons, the same is true for Cl

n = 2 electrons

Ecell = E° - 0.0592/n * log(Q)

Ecell = 0.483 - 0.0592/2 * log(Q)

Calculate Q

Q = [products]/[reactants]

Q = [Co+2]^2P-Cl2 / ([Co+3]^2[Cl-]^2)

substitute data

Q = (0.866^2)(6)/((0.592^2)(0.704^2))

Q = 25.905

substitute

Ecell = 0.483 - 0.0592/2 * log(25.905)

Ecell = 0.4411 V