Pt ( s ) H 2 ( g ) ( 1 bar ) H + ( aq ) ( 1 M ) M 3 + ( aq ) , M + ( aq ) Pt ( s ) The E cell for the given cell is 0 . 1115 V at 298 K When M + ( aq ) M 3 + ( aq ) = 10 a The value of a is ___________ Given : E θ = M 3 + / M + 0 . 2 V 2 . 303 RT F = 0 . 059 V

Step-by-step solution

Overall cell reaction :- H 2 (   g ) + M ( aq ) 3 + ⟶ M ( aq ) + + 2 H ( aq ) + Using Nernst equation: E Cell  = E Cathode  o - E anode  o - 0 . 059 2 log M + × 1 2 M + 3 1 ⇒ 0 . 1115 = 0 . 2 - 0 . 059 2 log M + M + 3 ⇒ 3 = log M + M + 3 ∴ a = 3