Question

For the reaction 2Co3+(aq)+2Cl−(aq)→2Co2+(aq)+Cl2(g). E∘=0.483 V2Co3+(aq)+2Cl−(aq)→2Co2+(aq)+Cl2(g). E∘=0.483 V what is the cell potential at 25 ∘C∘C if the concentrations are [Co3+]=[Co3+]= 0.324 MM , [Co2+]=[Co2+]= 0.158 MM , and [Cl−]=[Cl−]= 0.384 MM , and the pressure of Cl2Cl2 is PCl2=PCl2= 5.80 atmatm ?

Answer 1

Answer:

$\large\boxed{\text{0.512 V}}$

Explanation:

We must use the Nernst equation

$E = E^{\circ} - \dfrac{RT}{zF}lnQ$

The equation for the cell reaction is is

2Cl⁻(0.384 mol·L⁻¹) + 2Co³⁺(0.324 mol·L⁻¹) ⇌ Cl₂(5.80 atm) + 2Co²⁺(0.158 mol/L)

Data:

E° = 0.483 V

R = 8.314 J·K⁻¹mol⁻¹

T = 25 °C

n = 2

F = 96 485 C/mol

 

Calculation:  

T = 25 + 273.15 = 298.15 K

$Q = \dfrac{\text{[Cl}^{-}]^{2}[\text{Co}^{3+}]^{2}}{p_{\text{Cl}_{2}}^{2}\text{[Co}^{3+}]^{2}} = \dfrac{0.384^{2} \times 0.324^{2}}{5.80 \times 0.158^{2}} =0.1069\\\\E = 0.483 - \left (\dfrac{8.314 \times 298.15 }{2 \times 96485}\right ) \ln(0.1069)\\\\=0.483 -0.01285 \times (-2.236) = 0.483 + 0.02872 = \textbf{0.512 V}\\\text{The cell potential is } \large\boxed{\textbf{0.512 V}}$