ECr2O72-/Cr3+=1.33V  ;  ECl2/Cl-=1.36VEMn04-/Mn2+=1.51V    ;  ECr3+/Cr=-0.74V

Using the data given above find out in which option the order of reducing power is correct.

1.  Cr3+ < Cl- < Mn2+ < Cr

2.  Mn2+ < Cl- < Cr3+ < Cr

3.  Cr3+ < Cl- < Cr2O72- < MnO4-

4.  Mn2+ < Cr3+ < Cl- < Cr

 

Subtopic:  Electrochemical Series |
 77%
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ECr2O72-/Cr3+=1.33V;  ECl2/Cl-=1.36VEMnO4-/Mn2+=1.51V;  ECr3+/Cr=-0.74V

Use the data given above to find out the most stable ion in its reduced form.

1. Cl-  2. Cr3+ 
3. Cr 4. Mn2+ 
Subtopic:  Electrochemical Series |
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The most stable oxidized species among the following is: 
\(E_{{\mathrm{Cr}_2 \mathrm{O}_7^2}/ \mathrm{Cr}^{3+}}^{o} =1.33 \mathrm{~V} ; E_{\mathrm{Cl}_2 / \mathrm{Cl}^{-}}^{o}=1.36 \mathrm{~V} \)
\( E_{\mathrm{MnO_{4}}^{-} / \mathrm{Mn}^{2+}}^{o}=1.51 \mathrm{~V} ; E_{\mathrm{Cr}^{3+} / \mathrm{Cr}}^{o}=-0.74 \mathrm{~V}\)

1. Cr3+  2. MnO4-
3. Cr2O72- 4. Mn2+ 
Subtopic:  Electrochemical Series |
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The quantity of charge required to obtain one mole of aluminium from Al2O3 is :

1.  1 F

2.  6 F

3.  3 F

4.  2 F

Subtopic:  Faraday’s Law of Electrolysis |
 68%
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The cell constant of a conductivity cell-

1. Changes with the change of electrolyte.
2. Changes with the change of concentration of electrolyte.
3. Changes with the temperature of the electrolyte.
4. Remains constant for a cell.

Subtopic:  Conductance & Conductivity |
 80%
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The correct statement about charging of the lead storage battery is:

1. PbSO4 anode is reduced to Pb.
2. PbSO4 cathode is reduced to Pb.
3. PbSO4 cathode is oxidised to Pb.
​​​​​​
4. PbSO4 anode is oxidised to PbO2.

Subtopic:  Batteries & Salt Bridge |
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\(\Lambda _{m(NH_{4}OH)}^{o}\) is equal to -
1. \(\Lambda _{m(NH_{4}OH)}^{o} \ + \ \Lambda _{m(NH_{4}Cl)}^{o} \ - \ \Lambda _{m(HCl)}^{o}\)
2. \(\Lambda _{m(NH_{4}Cl)}^{o} \ + \ \Lambda _{m(NaOH)}^{o} \ - \ \Lambda _{m(NaCl)}^{o}\)
3. \(\Lambda _{m(NH_{4}Cl)}^{o} \ + \ \Lambda _{m(NaCl)}^{o} \ - \ \Lambda _{m(NaOH)}^{o}\)
4. \(\ \Lambda _{m(NaOH)}^{o} \ + \ \Lambda _{m(NaCl)}^{o}\ - \ \Lambda _{m(NH_{4}Cl)}^{o}\)

Subtopic:   Kohlrausch Law & Cell Constant |
 88%
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The half-cell reaction at the anode during the electrolysis of aqueous sodium chloride solution  is represented by : 

1. Na+(aq) + e- ⟶ Na(s) ; \(E_{cell}^{o} \ = \ -2.71 \ V \)

2. 2H2O(l) ⟶ O2(g) + 4H+(aq) + 4e; \(E_{cell}^{o} \) = 1.23 V

3. H+(aq) + e-\(\frac{1}{2}\)H2(g) ; \(E_{cell}^{o} \) = 0.00 V

4. Cl-(aq) ⟶ \(\frac{1}{2}\)Cl2(g) + e- ; \(E_{cell}^{o}\) 1.36 V

Subtopic:  Electrolytic & Electrochemical Cell |
 68%
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The positive value of the standard electrode potential of Cu2+ / Cu indicates that-

(a) This redox couple is a stronger reducing agent than the H+ /H2 couple
(b) This redox couple is a stronger oxidising agent than H+ /H2
(c) Cu can displace H2 from acid.
(d) Cu cannot displace H2 from acid.


1. (a, b)
2. (b, c)
3. (c, d)
4. (b, d)

Subtopic:  Electrochemical Series | Electrolytic & Electrochemical Cell |
 73%
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ECell° of some half cell reactions are given below. 

I.  H+aq+e-12H2g ;    E°cell =0.00V

II.  2H2OlO2g+4H+aq+4e-;            Ecell =1.23V

III.  2SO42-aqS2O82-aq+2e-;           Ecell =1.96V

The correct choice among the given is -
 

a. In dilute sulphuric acid solution, hydrogen will be reduced at cathode
b. In concentrated sulphuric acid solution, water will be oxidised at anode
c. In dilute sulphuric acid solution, water will be oxidised at anode
d. In dilute sulphuric acid solution, SO42- ion will be oxidised to tetrathionate ion at anode

Choose the correct option
1. (a, b)
2. (b, c)
3. (c, d)
4. (a, c)

Subtopic:  Electrolytic & Electrochemical Cell |
 60%
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