ElectrochemistryHard
Question
A solution containing one mole per litre of each Cu(NO3)2, AgNO3, Hg2 (NO3)2 is being electrolysed by using inert electrodes. The values of standard electrode potentials in volts (reduction potential) are :
Ag+ | Ag = + 0.80, Hg22+ |2Hg = + 0.79
Cu2+ | Cu = + 0.34, Mg2+ | Mg = - 2.37
With increasing voltage, the sequence of deposition of metals on the cathode will be :
Ag+ | Ag = + 0.80, Hg22+ |2Hg = + 0.79
Cu2+ | Cu = + 0.34, Mg2+ | Mg = - 2.37
With increasing voltage, the sequence of deposition of metals on the cathode will be :
Options
A.Ag, Hg, Cu, Mg
B.Mg, Cu, Hg, Ag
C.Ag, Hg, Cu
D.Cu, Hg, Ag
Solution
A metal having highest EoR.P. will have highest tendency to remain as cation. On the basis of EoO.P. values given. The order of ions to remain in the solution is :
Ag+ < Hg22+ < Cu2+ < Mg2+
EoR.P. + 0.80 + 0.79 + 0.34 - 2.37
The reverse will be the order of deposition on the cathode during electrolysis. Thus the correct order of deposition will be :
Ag → Hg → Cu → Mg
Ag+ < Hg22+ < Cu2+ < Mg2+
EoR.P. + 0.80 + 0.79 + 0.34 - 2.37
The reverse will be the order of deposition on the cathode during electrolysis. Thus the correct order of deposition will be :
Ag → Hg → Cu → Mg
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