Write the Nernst Equation and Emf of the Following Cells at 298 K Exercise Chapter 3 Electrochemistry Chemistry Class

Question 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).

Answer:

(i)Net reaction 
Mg(s) +Cu²⁺(aq)↔ Mg²⁺(aq) + Cu(s)
Nernst equation

There are two electron are transferring so that n = 2
E°cell = E° right  – E°left
E°cell =+0.34 – (–2.37)V
E°cell =+2.71V
Plug the value we get
Concentration of solid substance is 1 always so that [Mg] = [Cu] =1


(ii) Net reaction 
Fe(s) +2H⁺(aq)↔ Fe²⁺(aq) + H₂(g)
Nernst equation

There are two electron are transferring so that n = 2
E°cell = E° right  – E°left
E°cell =0– (–0.44)V
E°cell =+0.44 V
Plug the value we get
Concentration of solid substance is 1 always so that [Fe] = [H2] =1

(iii) Net reaction 
Sn(s) +2H⁺(aq)↔ Sn²⁺(aq) + H₂(g)
Nernst equation

There are two electron are transferring so that n = 2
E°cell = E° right  – E°left
E°cell =0– (–0.14)V
E°cell =+0.14 V
Plug the value we get
Concentration of solid substance is 1 always so that [Sn] = [H2] =1

(iv) Net reaction 
2Br^–(aq) + 2H⁺(aq)↔ Br₂(l) + H₂(g)
Nernst equation

There are two electron are transferring so that n = 2
E°cell = E° right  – E°left
E°cell =0– (1.08)V
E°cell –1.08V
Plug the value we get
Concentration of solid substance is 1 always so that [Br2] = [H2] =1