Test: MCQs (One or More Correct Option): Electrochemistry | JEE Advanced - JEE MCQ

More negative is the value of reduction potential, higher will be the reducing property, i.e., the power to give up electrons.

The reduction potentials (as given) of the ions are in the order :

Mg²⁺ (aq.) will not be reduced as its reduction potential is much lower than that of water (–0.83 V). Hence the sequence of deposition of the metals will be Ag, Hg, Cu.

Charge of one mole of electrons = 96500 C ∴ 1 mole gram equivalent of substance will be deposited by one mole of electrons.

NOTE : Oxidation is loss of electron and in a galvanic cell it occurs at anode. Reduction is gain of electron and in a galvanic cell it occurs at cathode.
Cell representation :
Anode / Anodic electrolyte || Cathodic electrolyte / Cathode
Reaction at Anode : H₂ → 2H⁺ + 2e^–
Reaction at Cathode : AgCl + e^– → Ag + Cl^–

Water is reduced at the cathode and oxidized at the anode instead of Na+ and  

TIPS/FORMULAE :
(i) In a galvanic cell oxidation occurs at anode and reduction occurs at cathode.
(ii) Oxidation occurs at electrode having higher oxidation potential and it behaves as anode and other electrode acts as cathode.
(iii) E_Cell = E_C – E_A
(substitute reduction potential at both places).

Fe²⁺ + Zn →  Zn²⁺ + Fe
∵ Zn →  Zn⁺⁺ + 2e^- and Fe²⁺ + 2e^- →  Fe
∴ Zn is anode and Fe is cathode.
E_cell = E_C – E_A = –0.41 – (–0.76) = 0.35V.

H₂O is more readily reduced at cathode than Na⁺. It is also more readily oxidized at anode than  . Hence, the electrode reactions are

We have 

From the half-cell reactions, it follows that

The given order of reduction potentials is Z > Y > X. A
spontaneous reaction will have the following
characteristics
Z reduced and Y oxidised
Z reduced and X oxidised
Y reduced and X oxidised
Hence, Y will oxidise X and not Z.

For M⁺ + X^– → M + X,  = 0.44 – 0.33 = 0.11V is positive, hence reaction is spontaneous

The salt used to make ‘salt-bridge’ must be such that the ionic mobility of cation and anion are of comparable order so that they can keep the anode and cathode half cells neutral at all times. KNO₃ is used becasue velocities of K⁺ and ions are nearly same

As we go down the group 1 (i.e. from Li⁺ to K⁺), the ionic radius increases, degree of solvation decreases and hence effective size decreases resulting in increase in ionic mobility.Hence equivalent conductance at
infinite dilution increases in the same order.

will oxidise Cl^– ion according to the following equation

The cell corresponding to this reaction is as follows : 

Pt, Cl₂ (1 atm) | Cl^– || , Mn²⁺, H⁺ | Pt

  being +ve, ΔG° will be -ve and hence the above reaction is feasible.   will not only oxidise Fe²⁺ ion but also Cl^– ion simultaneously. So the quantitative estimation of aq Fe(NO₃)₂ cannot be done by this.

NOTE : In an electrolytic cell, electrons do not flow themselves. It is the migration of ions towards oppositely charged electrodes that indirectly constitutes the flow of electrons from cathode to anode through internal supply.

TIPS/FORMULAE :
Use Nernst's equation;
Cell reaction : Zn + Fe²⁺ → Zn²⁺ + Fe
Using Nernst equation

Give : I = 10 milliamperes ; IF = 96500 C mol^–1
t = ? ; Moles of H₂ produces = 0.01 mol
From the law of electrolysis, we have
Equivalents of H₂ produces = 

Substituting given values, we get

i.e. (b) is the correct answer

AgNO₃ (aq) + KCl(aq)  →  AgCl(s) + KNO₃(aq)
Conductivity of the solution is almost compensated due to formation of KNO₃(aq). However, after at end point, conductivity increases more rapidly due to addition of excess AgNO₃ solution.

Here n = 4, and [H⁺] = 10^–pH = 10^–3
Applying Nernst equation

At anode : H₂(g) 2H⁺ (aq) + 2e^–
At cathode : M⁴⁺ (aq) + 2e^– M²⁺ (aq)
Net cell reaction : H₂(g) + M⁴⁺ (aq) 2H⁺ (aq) + M²⁺ (aq)

NOTE : More negative or lower is the reduction potential , more is the reducing property. Thus the reducing power of the corresponding metal will follow the reverse order, i.e. Y > Z > X.

The species having less reduction potential with respect to (E° = + 0.96 V) will be oxidised by . These species are V, Fe and Hg.

Salt bridge is introduced to keep the solutions of two electrodes separate, so that the ions in electrodes do not mix freely with each other. Salt bridge maintains the diffusion of ions from one electrode to another.