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Question

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

(i) Mg(s)|Mg2+(0.001M)||Cu2+(0.001M)|Cu(s)

(ii) Fe(s)|Fe2+(0.001M)||H+(1M)|H2(g)(1 bar)|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(0.010M)||H+(0.030M)|H2(g)(1bar|Pt(s))

E(Mg2+Mg)=2.37V;E(Cu2+Cu)=+0.34V;E(Fe2+Fe)=0.44V;E(Sn2+Sn)=0.14V;E(1Br2Br)=+1.08V,


Solution

(i) Cell equation:Mg(s)+Cu2+(aq)Mg2+(aq)+Cu(s)

(n=2)

Nernst equation : Ecell=Ecell0.05912log[Mg2+]Cu2+

EMF of the cell Ecell=[0.34(2.37)]0.05912log[103]104

=2.710.02955=2.68V

EMF = 2.68V

(ii) Cell equation : Fe(s)+2H+(aq)Fe2+(aq)+H2(g)

(n=2)

Nernst equation : Ecell=Ecell0.05912log[Fe2+][H+]2

EMFof the cell, Ecell=[0(0.44)]0.05912log[103][1]2

=0.440.5912×(3)

=0.44+0.0887

=0.5287V0.53V

EMF = 0.53V

(iii) Cell Equation:Sn(s)+2H+(aq)Sn2+(aq)+H2(g)

(n=2)

Nernst equation ; Ecell=Ecell0.05912logSn2+[H+]2

EMF of the cell, Ecell=[0(0.14)0.05912log[0.05][0.02]2]

=0.140.05912×(2.097)

= 0.14 - 0.0620 = 0.08V

EMF = 0.08 V

(iv) Cell equation : 2Br(l)+2H+(aq)Br2(l)+H2(g)

(n=2)

Nernst equation : Ecell=Ecell0.05912log1[Br]2[H+]2

EMF of the cell,

Ecell=[01.08]0.05912log1(0.01)2×(0.03)2

=1.080.05912log(1.11.1×107)

=1.080.05912(7.0457)

=1.080.208=1.288 V

EMF=1.288V

( Oxidation will occur at hydrogen electrode and reduction at Br2 electrode)


Chemistry
NCERT Textbook
Standard XII

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