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=E∘cell−0.05912log[Mg2+]Cu2+ EMF of the cell Ecell=[0.34−(−2.37)]−0.05912log[10−3]10−4 =2.71−0.02955=2.68V EMF = 2.68V (ii) Cell equation : Fe(s)+2H+(aq)→Fe2+(aq)+H2(g) (n=2) Nernst equation : Ecell=E∘cell−0.05912log[Fe2+][H+]2 EMFof the cell, Ecell=[0−(−0.44)]0.05912log[10−3][1]2 =0.44−0.5912×(−3) =0.44+0.0887 =0.5287V≈0.53V EMF = 0.53V (iii) Cell Equation:Sn(s)+2H+(aq)→Sn2+(aq)+H2(g) (n=2) Nernst equation ; Ecell=E∘cell−0.05912logSn2+[H+]2 EMF of the cell, Ecell=[0−(−0.14)−0.05912log[0.05][0.02]2] =0.14−0.05912×(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=E∘cell−0.05912log1[Br−]2[H+]2 EMF of the cell, Ecell=[0−1.08]−0.05912log1(0.01)2×(0.03)2 =−1.08−0.05912log(1.11.1×107) =−1.08−0.05912(7.0457) =−1.08−0.208=−1.288 V EMF=−1.288V (∴ Oxidation will occur at hydrogen electrode and reduction at Br2 electrode) ChemistryNCERT TextbookStandard XII

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