Q. 53.8( 46 Votes )

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

(i) Mg(s)|Mg^{2+} (0.001M)||Cu^{2+} (0.0001 M)|Cu(s)

(ii) Fe(s)|Fe^{2+} (0.001M)||H^{+} (1M)|H_{2}(g)(1bar)| Pt(s)

(iii) Sn(s)|Sn^{2+} (0.050 M)||H^{+} (0.020 M)|H_{2}(g) (1 bar)|Pt(s)

(iv) Pt(s)|Br_{2}(l)|Br^{–}(0.010 M)||H^{+} (0.030 M)| H_{2}(g) (1 bar)|Pt(s).

Answer :

E_{cell} = ?

(i) Mg + Cu^{2 +} → Mg^{2 +} + Cu (n = 2)

E^{0}_{Cu}^{2 +} _{/Cu} ^{+} = 0.34V

E^{0}_{Mg}^{2 +} _{/Mg} = - 2.37 V

E^{0}_{cell} = E^{0}_{R} - E^{0}_{L}

E^{0}_{cell} = 0.34 - ( - 2.37) → Equation 1

Using Nernst equation, we get,

→ Equation 2

Substituting Equation 1 in equation 2, we get,

∴ E_{cell} = 0.34 - ( - 2.37) -

= 2.71 -

= 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 ^{+} → Fe^{2 +} + H_{2} (n = 2)

E^{0}_{H} ^{+} _{/H2} = 0 V

E^{0}_{Fe}^{2 +} _{/Fe} = - 0.44 V

E^{0}_{cell} = E^{0}_{R} - E^{0}_{L}

E^{0}_{cell} = 0 - ( - 0.44) → Equation 1

Using Nernst equation, we get,

→Equation 2

Substituting Equation 1 in equation 2, we get,

∴ E_{cell} = 0 - ( - 0.44) -

= 0.44 -

= 0.44 - 0.0887

∴ E_{cell} = 0.5287 V

__The e.m.f of the cell, E _{cell} is 0.5287 V__

iii) Sn + 2H ^{+} → Sn^{2 +} + H_{2} (n = 2)

E^{0}_{H} ^{+} _{/H2} = 0 V

E^{0}_{Sn}^{2 +} _{/Sn} = - 0.14V

E^{0}_{cell} = E^{0}_{R} - E^{0}_{L}

E^{0}_{cell} = 0 - ( - 0.14) → Equation 1

Using Nernst equation, we get,

→Equation 2

Substituting Equation 1 in equation 2, we get,

∴ E_{cell} = 0 - ( - 0.14) -

= 0.14 -

= 0.14 - 0.0295×2.0969

∴ E_{cell} = 0.08 V

__The e.m.f of the cell, E _{cell} is 0.08 V__

iv) 2Br ^{-} + 2H ^{+} → Br_{2} + H_{2} (n = 2)

E^{0}_{H} ^{+} _{/H2} = 0 V

E^{0}_{Br2/Br} ^{-} = 1.08V

E^{0}_{cell} = E^{0}_{R} - E^{0}_{L}

E^{0}_{cell} = 0 - (1.08) → Equation 1

Using Nernst equation, we get,

→Equation 2

Substituting Equation 1 in equation 2, we get,

∴ E_{cell} = 0 - (1.08) -

= - 1.08 -

= - 1.08 - 0.208

∴ E_{cell} = 1.288 V

__The e.m.f of the cell, E _{cell} is 1.288 V__

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