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+]= 0.190 m , [co2+]= 0.205 m , and [cl−]= 0.144 m , and the pressure of cl2 is pcl2= 7.30 atm ?

Ecel =0,04 V

Explanation:

Apply the Nerst equation,

Ecel= Ecelº - (RT/nF)*lnQ

where R=8,314 J/molK, T=25ºC=298K and F =96 485 Coulombs/mol e- and n=number of moles of electrons transferred in the balanced equation. Q is cocient of products and reactives power to respective coefficients, if is a gas apply partial pressure

Write the semiequation redox and verify the numbers of electron for balance. In this case you don't need to change nothing

2Cl−(aq)→Cl2(g) + 2e-

2CO3+(aq) + 2e-→2CO2+(aq)

2Cl−(aq) + 2CO3+(aq) →2CO2+(aq) + Cl2(g)

Hence

Ecel= 0.483 V -  0.013Ln ([CO2+]^2*PCl2] / [CO3+]^2*[Cl-]^2)

Ecel= 0.483 V -  0.013Ln ([0.205]^2 * 7.3] / [0.19]^2*[0.144]^2)

Ecel =0,04 V

20 drops hope this

the answer is a and have a great day