For a cell involving one electron $$E_{cell}^\Theta $$ = 0.59 V at 298 K, the equilibrium constant for the cell reaction is :
[Given that $${{2.303RT} \over F}$$ = 0.059 V at T = 298 K ]
A
1.0 $$ \times $$ 1030
B
1.0 $$ \times $$ 1010
C
1.0 $$ \times $$ 102
D
1.0 $$ \times $$ 105
2
NEET 2019
MCQ (Single Correct Answer)
+4
-1
For the cell reaction
2Fe3+(aq) + 2I–
(aq) $$ \to $$ 2Fe2+(aq) + I2(aq)
$${E_{cell}^\Theta }$$ = 0.24 V at 298 K. The standard Gibbs energy ($$\Delta $$rGo) of the cell reaction is :
[Given that Faraday constant F = 96500 C mol–1]
A
46.32 kJ mol–1
B
23.16 kJ mol–1
C
–46.32 kJ mol–1
D
–23.16 kJ mol–1
3
NEET 2018
MCQ (Single Correct Answer)
+4
-1
Consider the change in oxidation state of
bromine corresponding to different emf values
as shown in the given diagram :
Then the species undergoing disproportionation
is :
A
BrO4–
B
BrO3–
C
Br2
D
HBrO
4
NEET 2017
MCQ (Single Correct Answer)
+4
-1
In the electrochemical cell :
$$Zn\left| {ZnS{O_4}\left( {0.01\,M} \right)} \right|$$$$\left| {CuS{O_4}\left( {1.0M} \right)} \right|Cu,$$
the emf of this Daniell cell is E1. When the concentration of ZnSO4 is changed to 1.0 M and that of CuSO4 changed to 0.01 M, the emf changes to E2. From the followings, which one is the relationship between E1 and E2? (Given, RT/F = 0.059)
