During electrolysis of `NaCl (aq)` using inert electrodes, the product at anode is oxygen

A solution of sodium sulphate in water is electrolysed using inert electrodes, The products at the cathode and anode are respectively.

During the electrolysis of NaCl solution, the gas liberated at the anode is

A dilute aqueous solution of CuSo_4 is electrolysed using platinum electrods. The products at the anode and cathode are:

An aqueous solution of Na_2SO_4 is electrolysed using Pt electrodes. The products at the cathode and anode are respectively:

An aqueous solution containing 1M each of Au^(3+),Cu^(2+),Ag^(+),Li^(+) is being electrolysed by using inert electrodes. The value of standard potentials are : E_(Ag^(+)//Ag)^(@)=0.80 V,E_(Cu^(+)//Cu)^(@)=0.34V and E_(Au^(+3)//Au)^(@)=1.50,E_(Li^(+)//Li)^(@)=-3.03V will increasing voltage, the sequence of deposition of metals on the cathode will be :

In the electrolysis of aq. NaCl solution which half-cell reaction will occurs at anode

Assertion: Electrolysis of NaCl solution gives chlorine at anode instead of O_2 . Reason: Formation of oxygen at anode requires overvoltage.

During electrolysis of molten ICl3, both I_2 and Cl_2 are liberated at anode /cathode.

During electrolysis of H_2SO_4 (aq) with high charge density, H_2S_2O_8 formed as by product. In such electrolysis 22.4L H_2(g) and 8.4 L O_2(g) liberated at 1 atm and 273 K at electrode. The moles of H_2S_2O_8 formed is : a) 0.25 b) 0.5 c) 0.75 d) 1

When a solution of AgNO_3 (1 M) is electrolysed using platinum anode and copper cathode, what are the products obtained at two electrodes? Given : E_(cu^(2+)|cu)^(@)=+0.34 V , E_(H_2O|O_2,H^+)^(@)=-1.23 V E_(H^+|H_2)^(@)=+0.0V , E_(ag^+|Ag)^(@)=+0.8 V