Question

What combination of metals together with their 1 mol/L salt solutions would give an electrochemical cell with the highest voltage?

	a.	Mg and Fe
	b.	Mg and Ag
	c.	Mg and Al
	d.	Zn and Cu

How much electrical current will be required to deposit 0.500 g of nickel from an aqueous solution of Ni(NO ₃) ₂ if the available time for electrolysis is 100. minutes?

	a.	0.273 A
	b.	8.20 A
	c.	16.4 A
	d.	0.135 A

What are the major products of the electrolysis of aqueous sodium sulfate?

(1) Na(s) (2) H ₂(g) (3) SO ₂(g) (4) O ₂(g)

	a.	(1), (2) and (3) only
	b.	(1) and (3) only
	c.	(2) and (4) only
	d.	(4) only

Answer 1

Ans 1: ‘b’ Mg & Ag will give highest cell voltage.

Standard reduction potential of the given metals at 25^oC is as follows:

Metal		Standard reduction potential
Mg	Mg2+ + 2e– àMg(s)	-2.36 V
Zn	Zn2+ + 2e– → Zn(s)	-0.76 V
Cu	Cu2+ + 2e– → Cu(s)	0.34 V
Al	Al3+ + 3e– → Al(s)	-1.66 V
Ag	Ag+ + e– → Ag(s)	0.80 V
Fe	Fe2+ + 2e– → Fe(s)	-0.44 V

Standard cell voltage, E^ocell is given by = E^ocathode - E^oanode

One with higher standard reduction potential prefers to undergo reduction and acts as cathode.

One with lower standard reduction potential prefers to undergo oxidation and act as anode

		Anode	Cathode	Cell potential (Eocathode - Eoanode)
a	Mg & Fe	Mg	Fe	-0.44 –(-2.36) = 1.92 V
b	Mg & Ag	Mg	Ag	0.80 - (-2.36) = 3.16 V
c	Mg & Al	Mg	Al	-1.66 –(-2.36) = 0.7 V
d	Zn & Cu	Zn	Cu	0.34 – (-0.76) = 1.1 V

Ans 2: ‘a’ i.e. 0.273 A

According to Faraday’s 1^st law of electrolysis,

W = (M/nF)It

Where, W = mass (g) of the compound liberated for passing ‘I’ ampere of current for ‘t’ seconds

            F = Faraday’s constant = 96500 C/mol

            M = molar mass of the compound gets deposited (here, Ni = 58.693 g/mol)

            n = no. of electrons transferred (Ni²⁺ + 2e^-----> Ni, here n=2)

            t = 100 minute= 100 x 60 = 6000 seconds

           

W = (M/nF)It

0.500 = (58.693/(2 x 96500) x I x 100 x 60

I = 0.273 ampere

Ans 3: ‘c’ i.e. (2) & (4) only

During the electrolysis of aqueous sodium sulfate solution, the products are Na₂SO₄, O₂ and H₂. In this case besides Na⁺ and (SO₄)^2– ions we also have H⁺ and OH^– ions along with the solvent molecules, H₂O.

At cathode: there is competition between the following reduction reactions:

Na⁺ (aq) + e^– → Na (s) E^o = – 2.71 V
H⁺ (aq) + e^– → ½ H2 (g) E^o = 0.00 V {H⁺ comes from water}
The reaction with higher value of E^o is preferred at cathode and hence at cathode

2 H₂O(l) + 2e⁻ ----> H₂(g) + 2 OH⁻(aq) reaction occurs.

At anode: there is competition between the following two oxidation reactions.

2H₂O (l) ----> O₂ (g) + 4H⁺(aq) + 4e^– E^o = 1.23 V

2(SO₄)^2– (aq) -----> (S₂O₈)^2– (aq) + 2e^– E^o= 1.96 V

The reaction with lower value of E^o is preferred at anode and hence at anode

2H₂O (l) ----> O₂ (g) + 4H⁺(aq) + 4e^– reaction occurs

Overall cell reaction is:

At anode:            2H₂O (l) ----> O2 (g) + 4H+(aq) + 4e^–

At cathode:         4 H₂O(l) + 4e⁻ ----> 2H₂(g) + 4 OH⁻(aq)

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Cell reaction: 2 H₂O (l) ------> 2H₂ (g) + O₂ (g) i.e. electrolysis of water.

So, Na₂SO₄ (aq) + 2 H₂O (l) -----> Na₂SO₄ (aq) + 2H₂ (g) + O₂(g)