For a cell 2 M _ ( S ) + O ₂( ~g )+4 H ⁺ 2 M ²⁺( aq )+2 H ₂ O _ ( l ) the E ^0 cell =+1.67 ~V . When [ M ²⁺ ] is 1.0 10⁻³ M and p ( O ₂ ) is 0.1 atm , the EMF of the cell becomes +1.57 V . Calculate the pH of the electrochemical cell.

Step-by-step solution

The cell reaction is 2M_ (s) + O_ 2(g) + 4H^+_ (aq) 2M²⁺_ (aq) + 2H₂O_ (l) . The Nernst equation for the cell is E_ cell = E^0_ cell - 0.0591 n Q , where n=4 electrons are transferred. The reaction quotient Q is given by Q = [M²⁺]^2 p(O₂) [H^+]^4 . Substituting the given values: 1.57 = 1.67 - 0.0591 4 ( (10⁻³)^2 0.1 [H^+]^4 ) . -0.10 = - 0.0591 4 ( 10⁻⁶ 0.1 [H^+]^4 ) . 0.10 4 0.0591 = ( 10⁻⁵ [H^+]^4 ) . 6.768 = (10⁻⁵) - ([H^+]^4) = -5 - 4 [H^+] . Since pH = - [H^+] , we have 6.768 = -5 + 4(pH) . 4(pH) = 11.768 . pH