JEE Advanced2016ChemistryElectrochemistryActual
For the following electrochemical cell at 289 K, Pt ( s ) | H 2 ( g , 1 bar ) | H + ( aq , 1 M ) | | M 4 + ( aq . ) , M 2+ ( aq . ) | Pt ( s ) E c e l l = 0.092 V when M 2 + a q . M 4 + a q . = 10 x Given: E M 4 + / M 2 + 0 = 0.151 V ; 2.303 R T F = 0.059 The value of x is -
Options
- A-2
- B-1
- C1
- D2
Correct answer
D. 2
Step-by-step solution
At anode : H 2 g ⇌ 2 H + a q + 2 e - At cathode : M 4 + a q + 2 e - ⇌ M 2 + a q Net cell reaction : H 2 g + M 4 + a q ⇌ 2 H + a q + M 2 + ( a q ) E c e l l = E c e l l 0 − 0.059 2 log [ M 2 + ] [ H + ] 2 [ M 4 + ] ( P H 2 ) Now, E c e l l = E M 4 + / M 2 + 0 - E H + / H 2 0 - 0.059 n . log H + 2 M 2 + P H 2 . M 4 + 0.092 = 0.151 - 0 - 0.059 2 . log 1 2 × M 2 + 1 × M 4 + ∴ M 2 + M 4 + = 10 2 ⇒ x = 2