JEE Main202229 Jun 2022Evening ShiftChemistryElectrochemistryActual
The cell potential for the given cell at 298   K   Pt H 2 g , 1   bar H + aq ‖ Cu 2 + aq Cu s is 0 . 31   V . The pH of the acidic solution is found to be 3 , whereas the concentration of Cu 2 + is 10 x   M . The value of x is _________. (Given: E Cu 2 + / Cu Θ = 0 . 34   V and 2 . 303 RT F = 0 . 06   V )
Correct answer
0
Step-by-step solution
H 2 g + Cu 2 + aq . → 2 H + aq . + Cu s Applying nernst equation E cell = E cell 0 - 0 . 06 n log products Reactants E cell = Cell   potential = 0 . 31 V E cell 0 = 0 . 34 p H = 3 , Hence , H + = 10 - 3 0 . 31 = 0 . 34 - 0 . 06 2 log H + 2 Cu 2 + Cu 2 + = 10 - 7 M x = 7