Dr. Sangeeta Khanna Ph.D 1 CHEMISTRY  

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Electrochemistry (Subjective Assignment) Answers

1. (a) 0

oxidationE of Cu+ , Fe2+, Hg+, Br– are given.

0oxidationE of Fe2+ Fe3+ is highest.

Hence Fe2+ is more oxidized & Fe2+ is strong reducing agent. Since Fe2+ is better reducing agent than Cu+, therefore Fe2+ will reduce Cu2+ → Cu+

(b) I– is strong reducing agent. It reduces Cu2+ into Cu+. Therefore reaction of Cu2+ & I– produces Cu2I2 rather than [CuI4]2–. Cl– does not reduce Cu2+ into

Cu+. Hence, reaction of Cu2+ & Cl– leads to formation of [CuCl4]2– 2. (i) Cl2 & H2 are given off at anode & cathode respectively 2Cl– –2e– ⎯→ Cl2 2H+ + 2e– ⎯→ H2 (ii) H2 & O2 are given off when Cl– conc. is very low, water molecules begin to discharge at anode, while

solution remain acidic, H+ are likely to be discharged in preference to H2O molecules but eventually letter molecules will be discharged.

At anode H2O (l) ⎯→ 2H+ + 21 O2(g) + 2e–

At cathode 2H2O + 2e– ⎯→ H2 + 2OH– (iii) H2 & O2 are given off at anode & cathode. 3. (a) Fe3+ + Ce3+ ⎯→ Ce4+ + Fe2+ V60.176.0E0cell += = -0.8 4 V So Fe3+ will not oxidize Ce3+ (b) V0.82- V 1.36 - V 54.0E0

cell == I2 does not displace Cl2 from KCl (c) 2FeCl3 + SnCl2 → SnCl4 + 2FeCl2

r 0.61, 15.076.0E0cell =+= so this reaction occur

4. (a) H2 ⎯→2H+ + 2e–

2

20

H2H pH

]H[log2059EE

+

+

−= =0.0–0.0591 log [H+]

0.25 = 0.059 pH pH = 0.25/0.0591 = 4.23 (b) ]H[ log 059.0EE 0

H2H

+

+

−=

0.3 = 0.0591 pH pH = 5.076 5. When MnO2 will be used up in cathodic process, the dry cell will stop to produce current. Cathodic process 2MnO2+Zn2+ + 2e–→ ZnMn2O4 Eq. mass of MnO2 = 87

187

state oxidation in changemass molecular

==

From 1st law of electrolysis

96500

tiE ××=ω

87104896500t

3 ××

×=

−

Dr. Sangeeta Khanna Ph.D

Dr. Sangeeta Khanna Ph.D 2 CHEMISTRY  

COACHING  CIRCLE Macintosh:Users:sahil:Desktop:Sub-Electrochemistry-Answers.doc

t = 2218390.805 s

days 67.25606024805.2218390t =××

=

6. Fe2+ ⎯→Fe3+ + e–

]Fe[]Fe[log

1059.0EE

2

30

+

+

−=

10log1059.077.0 −−=

= - .83 V OH4Mne5H8MnO 2

24 +⎯→⎯++ +−+−

8

4

20

]H][MnO[]Mn[log

5059.0EE

+−

+−=

= 25.1)10(01.0

10log5059.051.1

83

4=

×−

−

−

Ecell = 1.25 – 0.83 = 0.42 V 7. (a) Cl– from salt bridge will react with Ag+ to make AgCl and with Pb2+ to make PbCl2. Both of these are

insoluble, so concentration of solution in two half cells will change. The cell will no longer be at standard conditions. She should use potassium nitrate solution.

(b) Consider half reaction Cu+ + e– Cu E0 = 0.52 V Cu2+ + e– Cu+ E0 = 0.18 V Red. Pot. Of Cu+ Cu is greater 2Cu+⎯→Cu+2 + Cu

Cu/Cu2Cu/CuEEE +++ +=

= - 0.18 + 0.52 = 0.34 V EMF is positive so this reaction will be spontaneous.

So, its Disproportionation into Cu2+ & Cu is true statements. 8. 3m3mmm NaNOAgNO aClN )AgCl( ∞∞∞∞ λ−λ+λ=λ = 110.3 + 116.5 – 105.2 = 121. 6

S1000k

m×

=λ S = solubility in mole/lit

S10001026.121

6 ××=

−; gm/lit 5.143

6.1211000102S

6×

××=

−

S = 1.463 x 10–3 g/lit

9. no. of moles of H+ deposited = F nt I×

×

= 965001

6060110×

×××

= 0.373 moles Moles of H+ left = 1 – 0.373 = 0.627 pH = - log 0.627 = 0.18

2 2H Cl of moles 21 H moles

21n +=+ =

2373.0

2373.0

+

V = 0.373 x 22.4 lit. = 8360 ml 10. The degree of dissociation ‘g’ is given by

∞λ

λ=α v

Dr. Sangeeta Khanna Ph.D

Dr. Sangeeta Khanna Ph.D 3 CHEMISTRY  

COACHING  CIRCLE Macintosh:Users:sahil:Desktop:Sub-Electrochemistry-Answers.doc

0133.076.3902.5

==α or 1.33%

CH3COOH CH3COO– + H+ Before dissociation 1 0 0 After dissociation (1 - α) α α

[H+] = Cα = 0.1 x 0.0133 = 0.00133 M