Electrochemistry

The conductivity of a solution is defined as the conductance of one unit volume of solution kept between two platinum electrodes with a unit area of cross-section and at a distance of unit length.The molar conductivity of the solution is defined as the conducting power of all the ions produced by 
one gram mole of an electrolyte in a solution. It is denoted by ∧m. 

The conductivity of a solution (both for strong and weak electrolytes) always decreases with the decrease in concentration of the electrolyte i.e., on dilution. This pattern is seen because the number of ions per unit volume that carry the current in the solution decreases on dilution. The molar conductivity of the solution increases with the decrease in concentration of the electrolyte. This is because both the number of ions as well as mobility of ions increases with dilution.

(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