1
JEE Advanced 2016 Paper 2 Offline
+3
-1
For the following electrochemical cell at 298 K

Pt(s) | H2 (g, 1 bar) | H+ (aq, 1 M) || M4+ (aq), M2+ (aq) | Pt (s)

Ecell = 0.092 V when $${{\left[ {{M^{2 + }}(aq)} \right]} \over {\left[ {{M^{4 + }}(aq)} \right]}}$$ = 10x

Give, $$E_{{M^{4+}}/{M^{2 + }}}^o$$ = 0.151 V; 2.303 RT/F = 0.059 V

The value of x is

A
-2
B
-1
C
1
D
2
2
JEE Advanced 2013 Paper 2 Offline
+6
-1.5
The standard reduction potential data at 25oC is given below:
Eo (Fe3+ , Fe2+) = +0.77V;
Eo (Fe2+ , Fe) = -0.44V;
Eo (Cu2+ , Cu) = +0.34V;
Eo (Cu+ , Cu) = +0.52V;
Eo [O2(g) + 4H+ + 4e- $$\to$$ 2H2O] = +1.23V;
Eo [O2(g) + 2H2O + 4e- $$\to$$ 4OH-] = +0.40 V
Eo (Cr3+ , Cr) = -0.74V;
Eo (Cr2+ , Cr) = -0.91V;

Match Eo of the redox pair in List – I with the values given in List – II and select the correct answer using the code given below the lists:

List - I
P. Eo (Fe3+ , Fe)
Q. Eo (4H2O $$\leftrightharpoons$$ 4H+ + 4OH-)
R. Eo (Cu2+ + Cu $$\to$$ 2Cu+)
S. Eo (Cr3+, Cr2+)

List - II
1. -0.18 V
2. -0.4 V
3. -0.04 V
4. -0.83 V
A
P - 4; Q - 1; R - 2; S - 3
B
P - 2; Q - 3; R - 4; S - 1
C
P - 1; Q - 2; R - 3; S - 4
D
P - 3; Q - 4; R - 1; S - 2
3
JEE Advanced 2013 Paper 2 Offline
+6
-1.5
An aqueous solution of X is added slowly to an aqueous solution of Y as shown in List – I. The variation in conductivity of these reactions in List – II. Match List – I with List – II and select the correct answer using the code given below the lists:

List - I
P. $$\mathop {(C{}_2{H_5}){}_3N}\limits_X$$ + $$\mathop {C{H_3}COOH}\limits_Y$$
Q. $$\mathop {KI(0.1M)}\limits_X$$ + $$\mathop {AgN{O_3}(0.01M)}\limits_Y$$
R. $$\mathop {C{H_3}COOH}\limits_X$$ + $$\mathop {KOH}\limits_Y$$
S. $$\mathop {NaOH}\limits_X$$ + $$\mathop {HI}\limits_Y$$

List - II
1. Conductivity decreases then increases
2. Conductivity decreases then does not change much
3. Conductivity increases then does not change much
4. Conductivity does not change much then increases
A
P - 3; Q - 4; R - 2; S - 1
B
P - 4; Q - 3; R - 2; S - 1
C
P - 2; Q - 3; R - 4; S - 1
D
P - 1; Q - 4; R - 3; S - 2
4
IIT-JEE 2012 Paper 2 Offline
+4
-1
The electrochemical cell shown below is a concentration cell. M | M2+ (saturated solution of a sparingly soluble salt, MX2) || M2+ (0.001 mol dm–3) | M The emf of the cell depends on the difference in concentrations of M2+ ions at the two electrodes. The emf of the cell at 298 K is 0.059 V.

The solubility product (Ksp; mol3 dm–9) of MX2 at 298 K based on the information available for the given concentration cell is (take 2.303 $$\times$$ R $$\times$$ 298/F = 0.059 V)
A
1 $$\times$$ 10–15
B
4 $$\times$$ 10–15
C
1 $$\times$$ 10–12
D
4 $$\times$$ 10–12
EXAM MAP
Medical
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