Question

CLIOII.
3.5 Write the Nernst equation and emf of the following cells at 298 K:
(I) Mg(s) Mg (0.001M) | Cu? (0.0001 M) Cu(s)
(1) Fe(s)Fe(0.001M) | H'(IM)IH, (g(bar)! Pl(s)
(111) Sn(s) iSn0.050 M (0.020 MTH, (1 bar) | Pt(s)
(iv) Pl(s) 1 Br (0.010 M)1 Br (1) || (0.030 MI H,g (1 bar) Pt(s).​

Answer 1

Answer:

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

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