1
JEE Main 2019 (Online) 11th January Evening Slot
+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
2
JEE Main 2019 (Online) 11th January Morning Slot
+4
-1
For the cell Zn(s) |Zn2+ (aq)| |Mx+ (aq)| M(s), different half cells and their standard electrode potentials are given below :

Mx+ (aq)/M(s) Au3+(aq)/Au(s) Ag+(aq)/Ag(s) Fe3+(aq)/Fe2+ (aq) Fe2+(aq)/Fe(s)
E0Mx+/M/(V) 1.40 0.80 0.77 $$-$$0.44

If $$E_{z{n^{2 + }}/zn}^0$$ = $$-$$ 0.76 V, which cathode will give maximum value of Eocell per electron transferred?
A
Ag+/Ag
B
Fe3+/Fe2+
C
Au3+/Au
D
Fe2+/Fe
3
JEE Main 2019 (Online) 10th January Evening Slot
+4
-1
In the cell

Pt$$\left| {\left( s \right)} \right|$$H2(g, 1 bar)$$\left| {HCl\left( {aq} \right)} \right|$$AgCl$$\left| {\left( s \right)} \right|$$Ag(s)|Pt(s)

the cell potential is 0.92 V when a 10–6 molal HCl solution is used. The standard electrode potential of (AgCl/ AgCl ) electrode is :
$$\left\{ {} \right.$$Given,  $${{2.303RT} \over F} = 0.06V$$  at  $$\left. {298} \right\}$$
A
0.94 V
B
0.40 V
C
0.76 V
D
0.20 V
4
JEE Main 2019 (Online) 10th January Morning Slot
+4
-1
Consider the following reduction processes :
Zn2+ + 2e $$\to$$ Zn(s) ; Eo = – 0.76 V
Ca2+ + 2e $$\to$$ Ca(s); Eo = –2.87 V
Mg2+ + 2e $$\to$$ Mg(s) ; Eo = – 2.36 V
Ni2 + 2e $$\to$$ Ni(s) ; Eo = – 0.25
The reducing power of the metals increases in the order :
A
Ca < Mg < Zn < Ni
B
Ni < Zn < Mg < Ca
C
Zn < Mg < Ni < Ca
D
Ca < Zn < Mg < Ni
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