Chapter 3 Electrochemistry Solutions
Question - 11 : - Suggest a list of metals that areextracted electrolytically.
Answer - 11 : -
Metals that are on the top of thereactivity series such as sodium, potassium, calcium, lithium, magnesium,aluminium are extracted electrolytically.
Question - 12 : - What is the quantity ofelectricity in coulombs needed to reduce 1 mol of
Answer - 12 : - 
? Consider the reaction:
Answer
The given reaction is as follows:
Therefore, to reduce 1 mole of , the required quantity ofelectricity will be:
=6 F
= 6 × 96487 C
= 578922 C
Write the chemistry of rechargingthe lead storage battery, highlighting all the materials that are involvedduring recharging.
Answer - 13 : -
A lead storage battery consistsof a lead anode, a grid of lead packed with lead oxide (PbO2) as thecathode, and a 38% solution of sulphuric acid (H2SO4) asan electrolyte.
When the battery is in use, thefollowing cell reactions take place:
Question - 14 : - Suggest two materials other thanhydrogen that can be used as fuels in fuel cells.
Answer - 14 : -
Methane and methanol can be usedas fuels in fuel cells.
Question - 15 : - Explain how rusting of iron isenvisaged as setting up of an electrochemical cell.
Answer - 15 : -
In the process of corrosion, dueto the presence of air and moisture, oxidation takes place at a particular spotof an object made of iron. That spot behaves as the anode. The reaction at theanode is given by,
Electrons released at the anodicspot move through the metallic object and go to another spot of the object.
There, in the presence of H+ ions,the electrons reduce oxygen. This spot behaves as the cathode. These H+ ionscome either from H2CO3, which are formed due to thedissolution of carbon dioxide from air into water or from the dissolution ofother acidic oxides from the atmosphere in water.
The reaction corresponding at thecathode is given by,
The overall reaction is:
Also, ferrous ions are further oxidized by atmosphericoxygen to ferric ions. These ferric ions combine with moisture, present in thesurroundings, to form hydrated ferric oxide
i.e., rust.
Hence, the rusting of iron isenvisaged as the setting up of an electrochemical cell.
Question - 16 : - Arrange the following metals inthe order in which they displace each other from the solution of their salts. Al, Cu, Fe, Mg and Zn
Answer - 16 : -
The following is the order inwhich the given metals displace each other from the solution of their salts.
Mg, Al, Zn, Fe, Cu
Question - 17 : - Given the standard electrode potentials,
Answer - 17 : -
K+/K = −2.93V, Ag+/Ag= 0.80V,
Hg2+/Hg = 0.79V
Mg2+/Mg = −2.37 V, Cr3+/Cr= − 0.74V
Arrangethese metals in their increasing order of reducing power.
Answer
The lower the reductionpotential, the higher is the reducing power. The given standard electrodepotentials increase in the order of K+/K < Mg2+/Mg< Cr3+/Cr < Hg2+/Hg < Ag+/Ag.
Hence, the reducing power of thegiven metals increases in the following order:
Ag < Hg < Cr < Mg < K
Depict the galvanic cell in whichthe reaction Zn(s) + 2Ag+(aq) → Zn2+(aq) + 2Ag(s) takesplace. Further show:
Answer - 18 : -
(i) Which ofthe electrode is negatively charged?
(ii) Thecarriers of the current in the cell.
(iii) Individualreaction at each electrode.
Answer
The galvanic cell in which thegiven reaction takes place is depicted as:
(i) Znelectrode (anode) is negatively charged.
(ii) Ionsare carriers of current in the cell and in the external circuit, current willflow from silver to zinc.
(iii) Thereaction taking place at the anode is given by,
The reaction taking place at thecathode is given by,
Question - 19 : - Calculate the standard cellpotentials of galvanic cells in which the following reactions take place:
Answer - 19 : -
(i) 2Cr(s) +3Cd2+(aq) → 2Cr3+(aq) + 3Cd
(ii) Fe2+(aq)+ Ag+(aq) → Fe3+(aq) + Ag(s)
Calculatethe ΔrGθ and equilibrium constant of thereactions.
Answer
(i)
The galvanic cell of the givenreaction is depicted as:
Now, the standard cell potentialis
In the given equation,
n = 6
F = 96487 C mol−1
= +0.34 V
Then, 
= −6 × 96487 C mol−1 ×0.34 V= −196833.48 CV mol−1
= −196833.48 J mol−1
= −196.83 kJ mol−1
Again, = −RT ln K = 34.496
K = antilog (34.496)
= 3.13 × 1034
(ii)
The galvanic cell of the givenreaction is depicted as:
Now, the standard cell potentialis
Here, n = 1.
Then,
= −1 × 96487 C mol−1 ×0.03 V
= −2894.61 J mol−1
= −2.89 kJ mol−1
Again, 
= 0.5073
K = antilog (0.5073)
= 3.2 (approximately)
Answer - 20 : -
(i) Mg(s) |Mg2+(0.001M) || Cu2+(0.0001 M) | Cu(s)
(ii) Fe(s) |Fe2+(0.001M) || H+(1M)|H2(g)(1bar) | Pt(s)
(iii) Sn(s) |Sn2+(0.050 M) || H+(0.020 M) | H2(g) (1 bar) |Pt(s)
(iv) Pt(s)| Br2(l) | Br−(0.010 M) || H+(0.030 M)| H2(g) (1 bar) | Pt(s).
Answer
(i) Forthe given reaction, the Nernst equation can be given as:
