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

The cell potential will be:

Molten sodium chloride conducts electricity due to the presence of: 

1. Free ions.

2. Free molecules.

3. Free electrons.

4. Atoms of sodium and chlorine.

The metal that cannot be produced upon reduction of its oxide by aluminium is :

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:

The correct statement among the following options is: 

Consider the change in the oxidation state of Bromine corresponding to different emf values as shown in the diagram below:

\(\small{BrO_4^-\ \overset{1.82V}{\longrightarrow}\ BrO_3^-\ \overset{1.5V}{\longrightarrow} HBrO\ \overset{1.595V}{\longrightarrow}\ Br_2 \overset{1.0652V}{\longrightarrow}\ Br^-}\) 
Then the species undergoing disproportionation is:-

The standard reduction potential for Fe²⁺|Fe and Sn²⁺|Sn electrodes are -0.44 V and -0.14 V respectively. For the cell reaction,

Fe²⁺ + Sn  →  Fe + Sn²⁺, the standard Emf is - 

1. +0.30 V

2. 0.58 V

3. +0.58 V

4. -0.30 V

Calculate the emf of the given cell: 

Zn(s) | Zn⁺² (0.1M) || Sn⁺² (0.001M) | Sn(s)

(Given ${\mathrm{E}}_{{\mathrm{Zn}}^{+2}/\mathrm{Zn}}^{\mathrm{o}}=-0.76$ $\mathrm{V},$ ${\mathrm{E}}_{{\mathrm{Sn}}^{2+}/\mathrm{Sn}}^{\mathrm{o}}=-0.14$ $\mathrm{V)}$

1. 0.62 V

2. 0.56 V

3. 1.12 V

4. 0.31 V