Given the standard electrode potentials:

K+/K = –2.93 V
Ag+/Ag = 0.80 V
Hg2+/Hg = 0.79 V
Mg2+/Mg = –2.37 V
Cr3+/Cr = – 0.74 V
The correct increasing order of reducing power of the metals is:  
1. Cr < Mg < K < Ag < Hg 2. Mg < K < Ag < Hg < Cr
3. K < Ag < Hg < Cr < Mg 4. Ag < Hg < Cr < Mg < K
Subtopic:  Electrode & Electrode Potential |
 86%
Level 1: 80%+
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The correct statement about the given galvanic cell equation is:

Zn(s) + 2Ag+­­­(aq) → Zn2+(aq) + 2Ag(s)

1. The current will flow from silver to zinc in the external circuit.
2. The current will flow from zinc to silver in the external circuit.
3. The current will flow from silver to zinc in the internal circuit.
4. The current will flow from zinc to silver in the internal circuit.

Subtopic:  Electrolytic & Electrochemical Cell |
 64%
Level 2: 60%+
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 2Cr(s) + 3Cd2+(aq)  2Cr3+(aq) + 3Cd

ECr3+/Cr= -0.74 VECd2+/Cd=-0.40 V

The value of Gro in the above reaction will be-

1. -196.83 kJ

2. 196.83 kJ

3. 186.83 kJ

4. -186.83 kJ

Subtopic:  Relation between Emf, G, Kc & pH |
 78%
Level 2: 60%+
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Mg(s) | Mg2+(0.001M) || Cu2+(0.0001 M) | Cu(s)

EMg2+/Mgo=-2.36V;ECu2+/Cuo=0.34V

The value of Ecell for the above reaction is -

1. 3.46 V  2. 3.15 V
3. 2.67 V 4. 1.24 V
Subtopic:  Nernst Equation |
 87%
Level 1: 80%+
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The value of Ecell in the reaction below will be:

\(\small{Pt(s)|Br^{-}(0.010 \ M)|Br_{2}(l) \ ||H^{+}(0.030 \ M)|H_{2}(g)(1 \ bar)|Pt(s)}\)

\(E_{Br^{-}/Br_{2}}^{o} \ = \ -1.09 \ V\)

1. +1.298 V

2. –1.398 V

3. –1.298 V

4. –1.198 V
Subtopic:  Nernst Equation |
Level 3: 35%-60%
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The value of  ∆G°  in the reaction below would be:
\(\small{\mathrm{Zn}(\mathrm{s})+\mathrm{Ag}_2 \mathrm{O}(\mathrm{s})+\mathrm{H}_2 \mathrm{O}(\mathrm{l}) \rightarrow \mathrm{Zn}^{+2}(\mathrm{aq})+2 \mathrm{Ag}(\mathrm{s})+2 \mathrm{OH}^{-}(\mathrm{aq})}\)


Given: \(E_{cell}^{\circ} = 1.04V\)

1. 2.13 kJ

2. 21.3 kJ

3. 201 kJ

4. 31.12 kJ

Subtopic:  Relation between Emf, G, Kc & pH |
 80%
Level 1: 80%+
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The incorrect statement(s) among the following is/are:

(a) The unit of conductivity is S cm–2.
(b) Specific Conductivity of weak and strong electrolytes always decreases with a decrease in concentration.
(c) The unit of molar conductivity is S cm2 mol–1.
(d) Molar conductivity increases with an increase in concentration.


1. Only (a)

2. (a) and (d)

3. (b) and (d)

4. (a) and (c)

Subtopic:  Conductance & Conductivity |
 57%
Level 3: 35%-60%
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The conductivity of 0.20 M solution of KCl at 298 K is 0.0248 S cm–1. The molar conductivity will be -

1. 124 S cm2 mol-1 2. 134 S cm2 mol-1
3. 128 S cm2 mol-1 4. 136 S cm2 mol-1
Subtopic:  Conductance & Conductivity |
 90%
Level 1: 80%+
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The resistance of a cell containing 0.001 M KCl solution at 298 K is 1500 . The conductivity is 0.146 × 10–3 S cm–1The cell constant would be:

1. 0.12 cm-1 2. 0.56 cm-1
3. 0.22 cm-1 4. 1.36 cm-1
Subtopic:   Kohlrausch Law & Cell Constant |
 88%
Level 1: 80%+
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Find the dissociation constant (Kₐ) of 0.00241 M acetic acid if its conductivity is 7.896 × 10⁻⁵ S cm⁻¹
 and its molar conductivity at infinite dilution (Λₘ°) is 390.5 S cm² mol⁻¹.

1. 2.45 × 10⁻⁵ mol L⁻¹
2. 1.86 × 10⁻⁵ mol L⁻¹
3. 3.72 × 10⁻⁵ mol L⁻¹
4. 2.12 × 10⁻⁵ mol L⁻¹


 

Subtopic:  Conductance & Conductivity |
 68%
Level 2: 60%+
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