We are asked to calculate for the **standard potential (E****° _{cell})** of the reaction. We will use the

$\overline{){\mathbf{E}}{\mathbf{\xb0}}{\mathbf{cell}}{\mathbf{}}{\mathbf{=}}{\mathbf{}}\frac{\mathbf{RT}}{\mathbf{nF}}{\mathbf{lnK}}}$

E°_{cell} = cell potential, V

R = gas constant = 8.314 J/(mol·K)

T = temperature, K

n = mole e^{-} transferred

F = Faraday’s constant, 96485 C/mol e^{-}

K_{ }= equilibrium constant

**Let’s first determine how many electrons were transferred in the reaction:**

Calculate the cell potential for the following reaction when the pressure of the oxygen gas is 2.50 atm, the hydrogen ion concentration is 0.10 mol/L, and the bromide ion concentration is 0.25 mol/L.

O_{2}(g) + 4H^{+}(aq) + 4Br ^{-}(aq) → 2H_{2}O(l) + 2Br_{2}(l)

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