(a) What type of a battery is lead storage battery? Write the anode and cathode reactions and the overall cell reaction occurring in the operation of a lead storage battery.

(b) Calculate the potential for half-cell containing 0.10 M K₂Cr₂O₇ (aq), 0.20 M Cr³⁺ (aq) and 1.0 x 10^-4 M H⁺ (aq) The half-cell reaction is and the standard electrode potential is given as E° = 1.33 V

OR

(a) How many moles of mercury will be produced by electrolyzing 1.0 M Hg(NO₃)₂ solution with a current of 2.00 A for 3 hours? [Hg(NO₃)₂ = 200.6 g mol^–1]

(b) A voltaic cell is set up at 25°C with the following half-cells Al³⁺ (0.001 M) and Ni²⁺ (0.50 M). Write an equation for the reaction that occurs when the cell generates an electric current and determine the cell potential.
(Given: E⁰_{Ni2+| Ni} = -0.25V, E⁰_Al3+|Al = -1.66V) [5]

(a) Lead storage battery is an example of secondary cells that is they are rechargeable.

Reaction at anode:

Reaction at cathode:

∴ Overall cell reaction:

(b)

From Nernst equation, we know,

E_cell = E⁰_cell – ()

E_cell = E⁰_cell - () [n=6 as 6 e^- are accepted]

Given concentration of K₂Cr₂O₇ is 0.10 M and concentration of Cr³⁺ is 0.20 M and 1.0 x 10^-4 M H⁺ (aq). Also, E° = 1.33 V.

E_cell = 1.33 - ()

= 1.33V – 0.55V

E_cell = 0.78V

OR

(a)

Here, n=2 as 2 e^- are accepted.

We know, quantity of electricity (Q) = I(current) × t(time)

I = 2 A and t = 3hrs =(3×60×60)seconds.

∴ Q = 2×3×60×60 = 21600C

1 mole Hg is deposited by (nF)C

As, n = 2 and F = 96500C

∴ From Faraday’s law of electrolysis,

(2×96500)C deposits 1 mole Hg

So, 21600C deposits () moles of Hg = 0.1119 moles of Hg.

(b)The cell reaction is:

Now, from Nernst equation,

E_cell = E⁰_cell – ()

E_cell = E⁰_cell – ()

Here n=6 as 6 e^- were needed to balance this redox reaction.

Given,

Concentration of Al³⁺ is 0.001 M and for Ni²⁺ it is 0.50 M.

E⁰_{Ni2+| Ni} = -0.25V, E⁰_Al3+|Al = -1.66V

E⁰_cell = E⁰_{Ni2+| Ni} - E⁰_Al3+|Al [As Ni²⁺ gets reduced]

= -0.25V-(-1.66V) = 1.41V

Applying this is in Nernst equation,

E_cell = 1.41V– ()

= 1.41V – (0.00985)

= 1.41V – (0.00985)

= 1.41V – (0.00985)

E_cell = 1.46V