Electrochemistry - 1 - JEE 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 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 B
Al3+ + 3e− → Al
w = zQ
where w = amount of metal
=5.12kg = 5.12 × 103g
z = electrochemical equivalent

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