Ncert Solutions Chemistry Class 12th

(i) Mg(s)|Mg ²⁺(0.001M)Cu²⁺(0.0001 M)|Cu(s) (ii) Fe(s)|Fe ²⁺(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)|Br ^– (0.010 M)|Br2 (l )H⁺ (0.030 M)| H2 (g) (1 bar)|Pt(s).

A 3.5 Ecell = ?

(i) Mg + Cu²⁺ → Mg²⁺ + Cu (n = 2)

E⁰ Cu²⁺ / Cu⁺ = 0.34V

E⁰ Mg  ²⁺ / Mg = - 2.37 V

E_cell⁰ = E_R⁰-E_L⁰

E_cell⁰  = 0.34 - ( - 2.37) → Equation 1

Using Nernst equation, we get,

 Substituting Equation 1 in equation 2, we get,

∴ E_cell = 0.34 - ( - 2.37) - 0.0591 / 2 log 10 ^-3/ 10^-4

= 2.71 - 0.0591 / 2 log 10

= 2.71 - 0.02955

∴ E_cell = 2.68 V

The e.m.f of the cell, E_cell is 2.68 V

ii) Fe + 2H ⁺ → Fe²⁺ + H₂ (n = 2)

E⁰ H⁺ / H2 = 0V

E⁰ Fe  ²⁺ / Fe = - 0.44 V

E_cell⁰ = E_R⁰-E_L⁰

E_cell⁰  = 0 - ( - 0.44) → Equation 1

Using Nernst equation, we get,

∴ E_cell = 0.5287 V

The e.m.f of the cell, E_cell is 0.5287 V

iii) Sn + 2H ⁺ → Sn²⁺ + H₂ (n = 2)

E_cell⁰ = 0 - ( - 0.14) → Equation 1

Using Nernst equation, we get,

(1) Known - E⁰_Cr³⁺_/Cr = - 0.74 V

E⁰ cd²⁺ = - 0.40 V

? _rG⁰ =? K =?

The galvanic cell of the given reaction is written as - Cr _(s)|Cr³⁺ _(aq)| Cd²⁺ _(aq)|Cd _(s)→ Reaction 1

Hence, the standard cell potential is given as, E⁰ = E_R⁰ - E_L⁰

= - 0.40 - (- 0.74)

∴ E⁰ = + 0.34 V

To calculate the standard Gibb's free energy? _rG⁰, we use,

? _rG⁰ = - nE⁰F → Equation 1

wherenF is the amount of charge passed and E⁰ is the standard reduction electrode potential. Substituting n = 6 (no. of e ^- involved in the reaction 1), F = 96487 C mol^-1,

E⁰ = + 0.34 V in Equation 1, we get, l

? _rG⁰ = - 6*0.34V*96487 C mol^-1

= - 196833.48 CV mol^-1

= - 196833.48 J mol^-1

∴? _rG⁰ = - 196.83348 kJ mol^-1

To find out the equilibrium constant, K, we use the formula,

log K =n E⁰ / 0.0591

=6 X 0.3 V / 0.0591

log K = 34.5177

K = antilog 34.5177

∴ K = 3.294 * 10³⁴

The standard Gibb's free energy? _rG⁰ is - 196.83348 kJ mol ^–1 and equilibrium constant, K is 3.294 * 10³⁴

(2) Known -

E⁰ Fe³⁺ / Fe²⁺ = 0.77V

E⁰ Ag⁺ / Ag = 0.80 V

? _rG⁰ =? K =?

The galvanic cell of the given reaction is written as - Fe²⁺ _(aq)|Fe³⁺ _(aq)| Ag ⁺ _(aq)|Ag _(s)→ Reaction 1

Hence, the standard cell potential is given as, E⁰ = E_R⁰ - E_L⁰

= 0.80 - (0.77)

∴ E⁰ = + 0.03 V

To calculate the standard Gibb's free energy? _rG⁰, we use,

? _rG⁰ = - nE⁰F → Equation 1

wherenF is the amount of charge passed and E⁰ is the standard reduction electrode potential.

Substituting n = 1 (no. of e ^- involved in the reaction 1), F = 96487 C mol^-1, E⁰ = + 0.03V in Equation 1, we get,

? _rG⁰ = - 1*0.03V*96487 C mol^-1

= - 2894.61 CV mol ^-1

= - 2894.61 J mol^-1

∴? _rG⁰ = - 2.89461 kJ mol^-1

To find out the equilibrium constant, K, we use the formula,

log K =n E⁰ / 0.0591

=1 X 0.03 V / 0.0591

log K = 0.5076

K = Antilog 0.5076

∴ K = 3.218

The standard Gibb's free energy? _rG⁰ is - 2.89461 kJ mol ^–1 and equilibrium constant, K is 3.218

14.4 

Amino acids are organic compounds containing amine (basic) and carboxyl (acidic) functional group with a specific side chain. Both acidic and basic group are present in the same molecule. In, aqueous solution carboxyl group can lose a proton (H⁺) and amino group can accept a proton (H⁺) giving rise to the dipolar ion called as zwitter ion. Zwitter ion is shown below:

In this zwitter ion there is the presence of both positive as well as negative charge, so there is the development of strong electrostatic force of attraction between the molecules and the water. For this reason solubility of amino acids is higher. Due to strong electrostatic interaction there is formation of strong bond between them, hence the melting point of amino acid is higher than the halo acids (as they do not exhibit such dipolar behaviour).

In the corrosion reaction, due to the presence of air and moisture, oxidation takes place at a particular point of an object made of iron. That spot behaves as the anode. The reaction at the anode is given by,

Fe (s) ⇒ Fe²⁺ (aq) + 2e-

Electrons released at the anodic spot move through the metal and go to another spot of the object, wherein presence of H⁺ ions, the electrons reduce oxygen. This spot behaves as the cathode. These H⁺ ions come either from H₂CO₃, which are formed due to the dissolution of carbon dioxide from the air into water. The cathodic reaction is given by

O_{2 (air)} + 4H_aq⁺+4e^-⇒ 2H O

The overall reaction is given by,

2Fe (s) + O_{2 (air)} + 4H _(aq)⁺⇒ 2Fe²⁺+2H₂O