The change in reduction potential of a hydrogen electrode when its solution initialy at pH = 0 is neutralised to pH = 7, is a/an-

1.	Increase by 0.059 V	2.	Decrease by 0.059 V
3.	Increase by 0.41 V	4.	Decrease by 0.41 V

When a lead storage battery is discharged, then:

1. SO₂ is evolved.

2. Lead is formed.

3. Lead sulphate is consumed.

4. Sulphuric acid is consumed.

What happens to the voltage in a galvanic cell when the salt bridge is removed?

1. The voltage drops to zero.
2. The voltage remains the same.
3. The voltage gradually increases.
4. The voltage rapidly increases.

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

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 voltage of the cell given below increases with: 

Cell: Sn(s) + 2Ag⁺(aq) → Sn²⁺(aq) + 2Ag(s)

1. Increase in size of the silver rod.

2. Increase in the concentration of Sn²⁺ ions.

3. Increase in the concentration of Ag⁺ ions.

4. None of the above.

By how much will the potential of half cell Cu²⁺| Cu change if the solution is diluted to 100 times at 298 K. It will -

In the electrochemical cell:

Zn|ZnSO₄(0.01 M) || CuSO₄(1.0M),Cu, 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. The relationship between E₁ and E₂ is : 
( Given, \(\frac{R T}{F}\)= 0.059)

1. E₁ = E₂

2. E₁ < E₂

3. E₁ > E₂

4. E₂ = 0 \(\neq\)E₁

The  number of electrons involved in the deposition of 63.5 g of Cu from a solution of $CuS{O}_4$ is:

1 $6.022$ $\times$ ${10}^{23}$ $$

2 $3.011$ $\times$ ${10}^{23}$ $$

3 $12.044$ $\times$ ${10}^{23}$ $$

4 $$ $6.022$ $\times$ ${10}^{22}$

The $k=$ $4.95$ $\times$ ${10}^{-5}S$ $c{m}^{-1}$ for a 0.00099 M solution. The reciprocal of the degree of dissociation of acetic acid, if ${\wedge }_m^0$ for acetic acid is 400 S $c{m}^{2}mo{l}^{-1}$ will be: