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:)

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^+)

To determine whether the reaction is spontaneous under the given conditions, we can use the Nernst equation and the standard reduction potentials to calculate the cell potential.

First, let's write the half-reactions for the reduction of Br2 and the reduction of O2:

Br2 (l) + 2 e- ↔ 2 Br- (aq)

O2 (g) + 4 H+ (aq) + 4 e- ↔ 2 H2O (l)

The standard reduction potentials provided are the potentials when the concentrations of all species are 1 M and the pressure of gases is 1 atm.

Next, we can calculate the E°cell (standard cell potential) by subtracting the reduction potential of the anode (Br2) from the reduction potential of the cathode (O2):

E°cell = E°cathode - E°anode

E°cell = 1.23 V - 1.07 V = 0.16 V

Now, let's use the Nernst equation to calculate the cell potential under non-standard conditions. The Nernst equation relates the cell potential (E) to the standard cell potential (E°cell), the reaction quotient (Q), the Faraday constant (F), and the temperature (T).

E = E°cell - (RT / nF) * ln(Q)

In this case, we're given the temperature (298 K), and the Faraday constant (F) is 96,485 C/mol. The number of electrons transferred in the cell reaction (n) is 2.

Next, let's determine the concentrations of species in the reaction.

From the given information, we know the buffer is composed of 0.10 M benzoic acid (C6H5COOH) and 0.12 M sodium benzoate (C6H5COONa). Since the pH has been adjusted with this buffer, we can assume that the concentrations of H+ (aq) and OH- (aq) are equal. Therefore, the final concentrations of H+ and OH- will be equal to the pKa of the benzoic acid, which is given as 6.3 x 10^-5.

Now, let's calculate the reaction quotient (Q) using the concentrations of H+ and Br- from the buffer:

Q = [Br-]^2 / [H+]^4

Substituting the given concentrations:

Q = (0.12 M)^2 / (6.3 x 10^-5 M)^4 = 84,642.86

Now, we can substitute the values into the Nernst equation to calculate the cell potential (E):

E = 0.16 V - [(8.314 J/(mol*K))(298 K) / (2 mol)(96,485 C/mol)] * ln(84,642.86)

After calculating this equation, you will find the value of E. If the calculated E is positive, the reaction is spontaneous; if it is negative, the reaction is non-spontaneous.

Note: Be sure to use consistent units throughout the calculation.