Consider the following electrochemical cell : Pt | O ₂( ~g )(1 bar ) | HCl ( aq ) | M ²⁺( aq , 1.0 M ) M ( ~s ) The pH above which, oxygen gas would start to evolve at anode is _ _ _ _ (nearest integer). [ array ll Given: & E ^ o _ M ²⁺ / M =0.994 ~V & E ^ o _ O ₂ / H ₂ O =1.23 ~V array standard reduction potential ]

Step-by-step solution

In this electrochemical cell, at the anode oxygen evolution occurs from water oxidation. Using Nernst equation for the O₂/H₂O couple: E_ anode = E°_ O₂/H₂O - 0.059 4 1 [H^+]^4 = 1.23 - 0.059 pH At the cathode: E_ cathode = 0.994 V (standard potential for M²⁺/M reduction) For oxygen to start evolving, the cell potential becomes zero: E_ cell = E_ cathode - E_ anode = 0 0.994 = 1.23 - 0.059 pH 0.059 pH = 0.236 pH = 4.0