Ecell = Eocell - (0.0592/n)*log(Q) where Q is Keq expression.
You will need to calculate the pH of the solution which can be done with the Henderson-Hasselbalch equation. That's
pH = pKa + log (base)/(acid). Use 0.10M for acid and 0.12M for base. Solve for pH and convert to H^+ with pH = -log(H^+)
under standard conditions, the following reaction is spontaneous at 298K.
O2 + 4H+ + 4Br- --> 2H2O + 2 Br2
Will the reaction be spontaneous if the PH value is adjusted using a buffer composed of 0.10 M benzoic acid and 0.12M sodium benzoate?
Given:Given: Ka for benzoic acid = 6.3 x 10-5
Br2 (l) + 2 e- ¡÷ 2 Br- (aq) Eo = 1.07 V
O2 (g) + 4 H+ (aq) + 4 e- ¡÷ 2 H2O (l) Eo = 1.23 V
thank you:)
1 answer
1 answer