Explanation
Graphite is a good conductor.
H2 undergoes oxidation and AgCl(Ag+) undergo reduction.
Ag+ + e- →Ag, Cu2+ + 2e- → Cu. Thus same quantity of electricity wil deposit double the no. of Ag atoms than Cu i. e. x = 2y.
At cathode, reduction takes place.
Eq. conductance increases with dilution and ultimately becomes constant.
In galvanic cell, cathode is positive but in electrolytic cell, cathode is negative.
Less the reduction potential, greater is the reducing power.
In electrolytic conductors, ions flow and not the electrons.
Conductance of electrolytic solution increases with temperature because dissociation increases
A large negative reduction potential means a high oxidation potential so that A- is easily oxidized
Higher the reduction potential, stronger is the oxidizing agent.
Standard electrode potential for a half- reaction is fixed.
At equilibrium, Q = Kc and E = 0.
Reduction half reaction is Ag+ + e‒ → Ag
Greater the reduction potential, less is the reducing power.
Similar to Answer 41 Greater the reduction potential, less is the reducing power.
Greater the reduction potential stronger is the oxidizing agent.Hence Y is stronger oxidizing agent than X but weaker than Z.
Greater the reduction potential, stronger is the oxidizing agent or weaker is the reducing agent
More voltage is required to reduce H+ at Hg than at Pt.
Impure sample is made the anode and pure copper as the cathode.
Zinc can lose electrons to form Zn2+ and carbon can gain electron to form carbide ion (C4-). Hence zinc is oxidized by carbon.
Electrons flow from cathode to anode through the internal supply.
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