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

Must be negative.substance increases

Ais stronge