Consider the change in oxidation state of Bromine corresponding to different emf values as shown in the diagram below: 
 
Then the species undergoing disproportionation is:-

1. 

2. 

3. 

4. HBrO

In the electrochemical cell: 
  $Zn | ZnS{O}_4 (0.01 M) || CuS{O}_4(1.0 M) | C\mathrm{u, }$the emf of this Daniel cell is E₁. When the concentration of ZnSO₄ is changed to 1.0 M and that of CuSO₄ is changed to 0.01 M, the emf changes to E₂. From the following, which one is the relationship between E₁ and E₂ ? 
(Given, \(\frac{RT}{F}\) = 0.059)

1. $E_1<E_2$

2. $E_1>E_2$

3. $E_2=0\ne E_1$

4. $E_1=E_2$

If half-cell potentials are:

The cell potential will be:

1. 0.42 V

2. 0.84 V

3. 1.34 V

4. 1.10 V

Standard electrode potential for Sn⁴⁺/Sn²⁺ couple is +0.15 V and that for Cr³⁺/Cr couple is -0.74. These two couples in their standard state are connected to make a cell. The cell potential will be:

1. +0.89 V

2. +0.18 V

3. +1.83 V

4. +1.199 V

Consider the following relations for emf of an electrochemical cell:

The correct relation among the given options is : 

1. (a) and (b)

2. (c) and (d)

3. (b) and (d)

4. (c) and (a)

If ${\mathrm{E}}_{{\mathrm{Fe}}^{2+}/\mathrm{Fe}}^{\mathrm{o}}$ = -0.441 V and  ${\mathrm{E}}_{{\mathrm{Fe}}^{3+}/{\mathrm{Fe}}^{2+}}^{\mathrm{o}}$ = 0.771 V, the standard emf of the reaction : 

Fe + 2Fe³⁺→ 3Fe²⁺ will be :

1. 0.330 V

2. 1.653 V

3. 1.212 V

4. 0.111 V

On the basis of the information available from the reaction :

$\frac{4}{3}\mathrm{Al}+{\mathrm{O}}_2\to \frac{2}{3}{\mathrm{Al}}_2{\mathrm{O}}_3, ∆\mathrm{G}=-827 \mathrm{KJ} {\mathrm{mol}}^{-1}$

The minimum e.m.f. required to carry out electrolysis of Al₂O₃ is :
(F = 96500 C mol^–1)

1. 2.14 V

2. 4.28 V

3. 6.42 V

4. 8.56 V

The EMF of a Daniel cell at 298 K is E₁  Zn|ZnSO₄(0.01 M) || CuSO₄(1.0 M)|Cu .
When the concentration of ZnSO₄ is 1.0 M and that of CuSO₄ is 0.01 M, the EMF is changed to E₂. The correct relationship between E₁ and E₂ is :

1. E₁ > E₂

2. E₁ < E₂

3. E₁ = E₂

4. E₂ = 0 ≠ E₁