1
JEE Main 2019 (Online) 8th April Morning Slot
MCQ (Single Correct Answer)
+4
-1
Change Language
Given that $${E^\Theta }_{{O_2}/{H_2}O} = 1.23\,V$$ ;

$${E^\Theta }_{{S_2}O_8^{2 - }/SO_4^{2 - }} = 2.05\,V$$

$${E^\Theta }_{B{r_2}/B{r^ - }} = 1.09\,V$$

$${E^\Theta }_{A{u^{3 + }}/Au} = 1.4\,V$$

The strongest oxidizing agent is :
A
O2
B
Au3+
C
Br2
D
$${S_2}O_8^{2 - }$$
2
JEE Main 2019 (Online) 12th January Evening Slot
MCQ (Single Correct Answer)
+4
-1
$$ \wedge _m^ \circ $$ for NaCl, HCl and NaA are 126.4, 425.9 and 100.5 S cm2 mol–1, respectively. If the conductivity of 0.001 M HA is-

5 $$ \times $$ 10–5 S cm–1, degree of dissociation of HA is -
A
0.50
B
0.125
C
0.25
D
0.75
3
JEE Main 2019 (Online) 12th January Morning Slot
MCQ (Single Correct Answer)
+4
-1
The standard electrode potential $${E^o }$$ and its temperature coefficient $$\left( {{{d{E^o }} \over {dT}}} \right)$$ for a cell are 2V and $$-$$ 5 $$ \times $$ 10$$-$$4 VK$$-$$1 at 300 K respectively.
The cell reaction is
Zn(s) + Cu2+ (aq) $$\buildrel \, \over \longrightarrow $$ Zn2+ (aq) + Cu(s)

The standard reaction enthalpy ($$\Delta $$rH$${^o }$$) at 300 K in kJ mol–1 is, [Use R = 8 JK–1 mol–1 and F = 96,000C mol–1]
A
$$-$$ 412.8
B
$$-$$ 384.0
C
192.0
D
206.4
4
JEE Main 2019 (Online) 11th January Evening Slot
MCQ (Single Correct Answer)
+4
-1
Given the equilibrium constant:

KC of the reaction :

Cu(s) + 2Ag+ (aq) $$ \to $$ Cu2+ (aq) + 2Ag(s) is

10 $$ \times $$ 1015, calculate the E$$_{cell}^0$$ of this reaciton at 298 K

[2.303 $${{RT} \over F}$$ at 298 K = 0.059V]
A
0.4736 mV
B
0.04736 V
C
0.4736 V
D
0.04736 mV
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