H2(g)+Cu2+(aq.)→2H+(aq.)+Cu(s)
Applying nernst equation
Ecell=Ecell0−n0.06log[Reactants][products]
Ecell=Cellpotential=0.31VEcell0=0.34pH=3,Hence,[H+]=10−3
0.31=0.34−20.06log[Cu2+][H+]2
[Cu2+]=10−7M
x=7
JEE Main 2022 — Chemistry Physical Chemistry
The cell potential for the given cell at 298KPt∣H2(g,1bar)∣∣H+(aq)∥Cu2+(aq)∣Cu(s) is 0.31V. The pH of the acidic solution is found to be 3, whereas the concentration of Cu2+ is 10xM. The value of x is _________.
(Given: ECu2+/CuΘ=0.34V and F2.303RT=0.06V)
Held on 29 Jun 2022 · Verified 6 Jul 2026.
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