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Write the Nernst equation and emf of the following cells at 298 K: (i) Mg_((S))|Mg^(2+)(0.001M)||Cu^(2+)(0.0001M)|Cu_((S)) (ii) Fe_((S))|Fe^(2+)(0.001M)||H^(+)(1M)|H_(2(g))(1" bar")|Pt_((S)) (iii) Sn_((S))|Sn^(2+)(0.050M)||H^(+)(0.020M)|H_(2(g))(1" bar")|Pt_((S)) (iv) Pt_((S))|Br^(-)(0.010M)|Br_(2(l))||H^(+)(0.030M)|H_(2(g))(1" bar")|Pt_((S)). |
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Answer» Solution :For the given reaction, the Nernst equation can be given as: `E_(Mg^(2+)|Mg)^(THETA)=-2.37V` `E_(Cu^(2+)|Cu)^(Theta)=+0.34V,E_(Sn^(2+)|Sn)^(Theta)=-0.14V` `E_(Br^(2+)|Br^(-))^(Theta)=+1.08V,E_(FE^(2+)|Fe)^(Theta)=-0.44V`. (i) Standard electrode POTENTIAL calculation: Cell reactioin : `Mg_((S))+underset((0.0001M))(Cu^(2+))toMg^(2+)+underset((0.001M))(Cu_((S)))` * In above cell reaction the concentration are not standard and its Nernst equation will be `therefore E_(cell)^(Theta)=E_(Cu^(2+)|Cu)^(Theta)-E_(Mg^(2+)|Mg)^(Theta)` `=0.34-(-2.37V)` `=2.71V` Where `[Mg^(2+)]0.001M and [Cu^(2+)]=0.0001M` `therfore E_(cell)=2.71V+((0.0591)/(2))xxlog((0.0001M)/(0.001M))` `=2.71+((0.0591)/(2))(LOG" "0.1)` `=2.71+0.02955xx(-1.0)` `=2.71-0.02955` `=2.68045V~~2.68V` `E_(cell)=E_(cell)^(Theta)-(0.0591)/(N)log(([Mg^(2+)])/([Cu^(2+)]))` where, n=2 `E_(cell)=E_(cell)^(Theta)+(0.0591)/(2)log(([Cu^(2+)])/([Mg^(2+)]))` |
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