Write the Nernst equation and emf of the following cells at 298 K

Q.5:Write the Nernst equation and emf of the following cells at 298 K:

(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).

Solution:

(i)Write the Nernst equation and emf of the following cells at 298 K

Net reaction 

Mg(s) +Cu⁺²(aq)↔ Mg⁺²(aq) + Cu(s)

Nernst equation

There are two electron are transferring so that n = 2

E^o_cell = E^o _right  – E^o_left

E^o_cell =+0.34 – (–2.37)V

 E^o_cell =+2.71V

Plug the value we get

Concentration of solid substance is 1 always so that [Mg] = [Cu] =1

(ii)  Write the Nernst equation and emf of the following cells at 298 K

Net reaction 

Fe(s) +2H⁺(aq)↔ Fe⁺²(aq) + H₂(g)

Nernst equation

There are two electron are transferring so that n = 2

E^o_cell = E^o _right  – E^o_left

E^o_cell =0– (–0.44)V

 E^o_cell =+0.44 V

Plug the value we get

Concentration of solid substance is 1 always so that [Fe] = [H₂] =1

(iii) Write the Nernst equation and emf of the following cells at 298 K

 Net reaction 

Sn(s) +2H⁺(aq)↔ Sn⁺²(aq) + H₂(g)

Nernst equation

There are two electron are transferring so that n = 2

E^o_cell = E^o _right  – E^o_left

E^o_cell =0– (–0.14)V

 E^o_cell =+0.14 V

Plug the value we get

Concentration of solid substance is 1 always so that [Sn] = [H₂] =1

(iv) ) Write the Nernst equation and emf of the following cells at 298 K

  Net reaction 

2Br^–(aq) +2H⁺(aq)↔ Br₂(l)+ H₂(g)

Nernst equation

There are two electron are transferring so that n = 2

E^o_cell = E^o _right  – E^o_left

E^o_cell =0– (1.08)V

 E^o_cell –1.08V

Plug the value we get

Concentration of solid substance is 1 always so that [Br₂] = [H₂] =1