The cell potential for the following cell Pt H 2 g H + aq | | Cu 2 + 0 .01 M Cu s is 0 . 576 V at 298 K . The pH of the solution is (Nearest integer) (Given : E Cu 2 + / Cu o = 0 . 34 V and 2 . 303 RT F = 0 . 06 V )

Step-by-step solution

Anode ⇒ H 2 g ⟶ 2 H + aq + 2 e - Cathode ⇒ Cu 2 + aq + 2 e - ⟶ Cu s Overall ⇒ H 2 q + Cu 2 + aq ⟶ 2 H + aq + Cu s ¯ E cell ° = E Cu 2 + Cu ° - E H + H 2 ° = 0 . 34   V E cell = E cell ° - 0 . 06 2 log H + 2 Cu 2 + 0 . 576 = 0 . 34 + 0 . 03 - log H + 2 + log Cu 2 + 0 . 576 = 0 . 34 + 0 . 03 2 pH + log Cu 2 + 0 . 236 = 0 . 03 2 pH - 2 7 . 866 = 2 pH - 2 pH = 4 . 93 ≈ 5