JEE Previous Year Questions- Electrochemistry with solutions - Free MCQ

The correct answer is Option C.

The anode is the electrode where oxidation (loss of electrons) takes place; in a galvanic cell, it is the negative electrode, as when oxidation occurs, electrons are left behind on the electrode. 
 
The cathode is the electrode where reduction (gain of electrons) takes place; in a galvanic cell, it is the positive electrode, as less oxidation occurs, fewer ions go into solution, and fewer electrons are left on the electrode.
So, in the reaction 
2Cr³⁺ +7H₂O→Cr₂O₇²⁻+14H⁺
 
Cr is getting oxidised as the oxidation state is changing from +3 to +6 and hence this reaction is possible at anode.
 

The correct answer is Option C.

The left portion of the salt bridge(| |)  is anode and the right portion of the salt bridge(| |) is cathode.

Since, Ag | Ag⁺ is half-cell oxidation and Cu²⁺ | Cu is half-cell reduction. Thus,
E^o _cell = E _cathode - E _anode
           = y - x

The correct answer is Option B.
E^o_cell = (0.0591/n) log K_c
0.259 = (0.0591/2) log K_c    
0.259 = 0.0295 log K_c
K_c = antilog 10
K_c = 1 x 10¹⁰

The correct answer is option C       
         A       B        C
E°_red +0.5V –3.0V –1.2V
More is the value of reduction potential, more is the tendency to get reduced, i.e. less is the reducing power. The reducing power follows the following order: B > C > A.

The correct answer is Option A.
E^o_cell =  E_cell – (0.0591/n) log (Zn²⁺/ Cu²⁺)
= 1.10 – (0.0591/n) log (1 / 0.1)
= 1.10 – 0.02955
= 1.070 Volt

The correct answer is Option D.
Ag+ + e- ---> Ag
So, no. of the moles of Ag = Q / F
                                           = 9650 / 96500
                                           =1 / 10 moles
∴ Mass of Ag produced :->
                                  (1/10) × 108
                                        = 10.8 g

The correct answer is option C
E⁰_cell ​= E^o_oxd ​+ E^o_red​
         = E^o _Sn/Sn2+ ​+ E^o_Fe3+/Fe_2+​
Given : E^o​_Sn2+/Sn = −0.14 ⇒E^o_Sn/Sn2+​=−(−0.14)
=0.14V
∴ E^o_cell ​= 0.14 + 0.77 = 0.91V

The correct answer is option C
For a cell reaction in equilibrium at 298 K,
E^o_cell =0.0591/nlogK_c
(K_e = equilibrium constant)
Give,     E^o_cell = 0.591V
Now, log K_c = (E^o_cell x n )/ 0.0591
                   =(0.0591 x n )/ 0.0591
log K_c = 10
K_c = antilog 10

1. K_c = 1 x 10¹⁰

The correct answer is Option A.

The limiting molar conductivities of NaCl, KBr and KCl are 126, 152 and 150 Scm²mol⁻¹ respectively. 
The limiting molar conductivity  Λo for NaBr is calculated by the following expression.
λ^∞_NaBr = λ^∞_NaCl + λ^∞_KBr + λ^∞_KCl
λ^∞_NaBr = 126+152−150=128 Scm.²mol⁻¹

The correct answer is Option C.

Zn(s) +2H⁺ (aq) →Zn²⁺(aq)+H₂(g)

So as we add sulphuric acid, conc. of hydrogen ion increases and equilibrium shifts to right.
Also according to the equation emf of the cell will be increased by adding acid.

The correct answer is option B
Al3+ + 3e− → Al
w = zQ
where w = amount of metal
=5.12kg = 5.12 × 103g
z = electrochemical equivalent

The correct answer is option D

 = 91.0 +426.2 - 126.5
= 390.7 S cm² mol^-1 .

The correct answer is option B
we know  ΔG=−RTlnK=−nFE∘
We know for a spontaneous reaction
ΔG<0.
∴−RTlnK<0
∴lnK>0
∴K>1
and  −nFE∘<0
∴Ecell∘​>0
 

The correct answer is option C
AgI (s) + e⁻ ⇌ Ag(s) + I⁻;
E∘ = −0.152 + Ag(s) → Ag+ + e⁻
E∘ = −0.8
Agl (s) → Ag⁺ + I⁻
E∘ = −0.952

The correct answer is option C
If it is allowed to completely discharge, Ecell​=0

The correct answer is option D
Cr∣Cr³⁺(0.1 M)∣∣Fe²⁺(0.01 M)∣Fe
Oxidation half-cell
Cr→Cr³⁺+3e⁻]×2
Reduction half-cell
Fe²⁺+2^e−→Fe]×3
Net cell reaction
2Cr+3Fe²⁺→2Cr³⁺+3Fe, n=6
E^∘​_cell​= E^∘​_oxi​+E^∘_​red
=0.72−0.42 = 0.30 V

 

The correct answer is option C
The elements with high negative value of standard reduction potential are good reducing agents and can be easily oxidised. ltbRgt Thus X should have a high negative value of standard potential then Y, so that it will be oxidised to X²⁺ by reducing Y²⁺ to Y.
X= Zn, Y= Ni
Zn + Ni2+ → Zn2+ + Ni
Alternatively, for a spontaneous reaction E∘ must be positive.
E∘=E∘reduced−E∘oxidised
=−0.23−(0.76)
implies E∘=+0.53V
 

Higher the SRP, better is oxidizing agent hence 
MnO⁻⁴
MnO^4- is the strongest oxidizing agent.
 

Cathode is always taken of pure substance because cathode is for reduction. at anode oxidation takes place and the metal oxidises itself into cation and moves to cathode, and gets deposited there. Whereas the impurities settle down below anode as anode mud.