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+] = 0.383 m, [co2+] = 0.369 m, [cl-] = 9.00*10-2 m, and [cl2] = 0.150 m?

The working equation for this problem would be:

E = E° - RTlnK/nF

For n, we must know the number of electrons involved. The half-reactions are:

CO³⁺ - - > CO²⁺ + e⁻

Cl⁻ - - > Cl₂ + e⁻

So, n=1. Then, F is the Faraday's constant = 96,500 C/mol. For K, the expression would be:

K = [Cl₂][CO²⁺]²/[Cl⁻]²[CO³⁺]² = [0.15][0.369]²/[0.09]²[0.383]²

K = 17.19

Now,

E = 0.483 V - (8.314 J/mol-K) (25+273 K) ln (17.19) / (1 mol e) (96,500 C/mol e)

E = 0.41 V