Test: Electrolysis & Faraday's Law(12 Oct) - JEE MCQ

In preference to reduction of Zn²⁺ at cathode, O^2- (aq) is oxidised to O₂ at anode.

When impure metal (M) is anode and pure metal is made cathode, then 

Thus, impure metal from the anode is deposited at the cathode.  

Manganate (VII) is MnO₄^- with ON of Mn = + 7
1 mole required 3F of electricity thus, 1 F will reduce only (1/3) mole MnO₄^- t o MnO₂.
Thus, MnO₂ formed = 0.33 mol.

(d)

In RHS, [NH₃] increases hence, pOH decreases
pH = 14 - pOH
Hence, pH of RHS increases.
In LHS, [NH₄⁺] increases, hence pOH increases and pH decreases.

Cl₂(g) is formed at anode and NaOH and H₂ are formed at cathode. Since, cathode and anode are separated. Hence. there is no reaction between the products formed at the cathode and anode. 

Note: but these reaction do not occur

Note: NaCl is neutral pH = 7 initially. After electrolysis , NaOH is formed

Hence, solution become, basic pH >7

(a) 

Oxidation takes place at anode - correct

2H₂O(l) →4H⁺ (aq) + O₂(g) + 4e^-

(b) Reduction take place at cathode - correct

since, E⁰_cell < 0, in electrolysis.

Cell reaction is spontaneous.

(c) Copper is deposited at the cathode, hence the meted to be electroplated (say Fe), should be made cathode - correct (d) Since. Cu²⁺(aq)- (blue) is reduced Cu(s), hence blue colour fades - correct. 
 

a(Cu So₄ and CuCl₂ changes to Cu and solution becomes colourless

(b) KCl changes to Cl₂ (yellow)
KI changes to I₂ (voilet)

Thus, distinguished.

(c) AgNo₃ No change

CuSo₄ (blue) changes to colourless

(d) CuBr₂ (blue) both changes to colourles.

NiBr₂ (Blue)