pH of the anodic solution of the following cell is Pt, H 2 (1 atm) H + xM H + 1 M H 2 (1 atm), Pt if E cell = 0 .2364 V .

Step-by-step solution

A t C a t h o d e : 2 H + 1 M + 2 e - → H 2 g A t A n o d e : H 2 g → 2 H + x M + 2 e - Now: E Cell = E Cell o − 0.0591 2 log [H + ] anode 2 [H + ] cathode 2 E C e l l = E C e l l ο - 0.0591 2 log ⁡ [ H + ] a n o d e [ H + ] c a t h o d e 2 E C e l l = E C e l l ο - 0.0591 2 log ⁡ x 1 2 0.2364 = 0 - 0.0591 2 log ⁡ x 1 2 0.2364 × 2 = − 0.0591   log x 1 2 0.2364 × 2 0.0591 = - log ⁡ x 1 2 - log ⁡ x 1 2 = 0.2364 × 2 0.0591 = 8 − 2 log x 1 = 8 log ⁡ x 1 = - 4 x = 1 0 - 4 p H = - log ⁡ ( x ) p H = 4