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.

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.

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

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 electrode potential of the Cu²⁺/Cu half-cell change when the concentration of Cu²⁺ ions is reduced to one-hundredth of its initial value at 298 K?

Find the relationship between E₁ and E₂ for the following Daniell cell:

Zn | ZnSO₄ (0.01 M) || CuSO₄ (1.0 M) | Cu

The emf of this cell is E₁.

If the concentration of ZnSO₄ is changed to 1.0 M and that of CuSO₄ is changed to 0.01 M, the emf becomes E₂.

Given:

RT/F = 0.059 V

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: