What is the single electrode potential of a half-cell for zinc electrode dipping in 0.01 M `ZnSO_(4)` solution at `25^(@)C` ? The standard electrode potential of `Zn//Zn^(2+)` system is 0.763 volt at `25^(@)C`.
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For the cell prepared from electrode A and B, electrode A:`(Cr_2O_7^(2-))/(Cr^(3+)), E_("red")^(0)=+1.33V` and electrode B :`(Fe^(3+))/(Fe^(2+)), E_("
Correct Answer - D A acts as cathode and B acts as anode because reduction potential of A is more. Anode is B `implies` is negative electrode. `implies` electrons flow from...
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A piece of zinc metal is dipped in a 0.1 M solution of zinc salt. The salt is dissociated to the extent of 20%. Calcualte the electrode potential of `
Correct Answer - `-0.8102` volt `[Zn^(2+)]=0.1 xx 0.2 =0.02 M` `Zn^(2+)+2e^(-) rarr Zn" (Reduction)"` `E_(Zn^(2+)//Zn)=E_(Zn^(2+)//Zn)^(@)+0.0591/2 log_(10) [Zn^(2+)]`
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What is the standard electrode potential for the electrode `MnO_(4)^(-)//MnO_(2)` in solution ? `("Given: "E_(MnO_(4)^(-)//Mn^(2+))^(@)=1.51" volt," E
Correct Answer - 1.7 volt `{:(MnO_(4)^(-)+8H^(+)+5e^(-) rarr Mn^(2+)+4H_(2)O," "E^(@)=1.51" volt"),(MnO_(2)+4H^(+)+2e^(-) rarr Mn^(2+)+2H_(2)O," "E^(@)=1.23" volt"),("Subtracting"),(bar(MnO_(4)^(-)+4H^(+)+3e^(-) rarr MnO_(2)+2H_(2)O," "E^(@)=?" ")):}` `E^(@)=(5xx1.51-2xx1.23)/(3)=(7.55-2.46)/(3)=5.09/3=1.70` volt
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Calculate the electrode potential at a copper electrode dipped in a 0.1 M solution of copper sulphate at `25^(@)C`. The standard electrode potential o
We know that, `E_("red")=E_("red")^(@)+0.0591/n log_(10) ["ion"]` Putting the value of `E_("red")^(@)=0.34 V, n=2` and `[Cu^(2+)]=0.1 M` `E_("red")=0.34+0.0591/2 log_(10) [0.1]` `=0.34+0.02955xx(-1)` `=0.34-0.02955=0.31045` volt
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The standard oxidation potential potential of zinc is 0.76 volt and of silver is - 0.80 volt. Calculate the emf of the cell : `Zn|underset(0.25 M)(Zn(
The cell reaction is `Zn+2Ag^(+) rarr 2Ag+Zn^(2+)` `E_(o x)^(@)` of `Zn =0.76` volt `E_("red")^(@)` of `Ag=0.80` volt `E_(cell)^(@)=E_(o x)^(@)` of `Zn+E_("red")^(@)` of `Ag=0.76+0.80=1.56` volt We know that, `E_(cell)=E_(cell)^(@)-0.0591/n"log" (["Products"])/(["Reactants"])` `=E_(cell)^(@)-0.0591/2"log"...
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To find the standard potential of `M^(3+)//M` electrode, the following cell is constituted : `Pt|M|M^(3+) (0.0018 mol^(-1) L)||Ag^(+) (0.01 mol^(-1) L
The cell reaction is `M+3Ag^(+) rarr 3Ag+M^(3+)` Applying Nernst equation, `E_(cell)=E_(cell)^(@)-0.0591/3"log" ([M^(3+)])/([Ag^(+)]^(3))` `0.42=E_(cell)^(@)-0.0591/3"log"((0.0018))/((0.01)^(3))=E_(cell)^(@)-0.064` `E_(cell)^(@)=(0.42+0.064)=0.484` volt `E_(cell)^(@)=E_("Cathode")^(@)-E_("Anode")^(@)` or `E_("Anode")^(@)=E_("Cathode")^(@)-E_(Cell)^(@)` `=(0.80-0.484)=0.32` volt
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The normal oxidation potential of zinc referred to the standard hydrogen electrode is 0.76 volt and that of copper is -0.34 volt at `25^(@)C`. When ex
The reaction is, `Zn + CuSO_(4) rarr Cu+ZnSO_(4)` or `Zn+Cu^(2+) rarr Cu+Zn^(2+)` `E_(cell)=E_(cell)^(@)-0.0591/2"log" ([Zn^(2+)])/([Cu^(2+)])` At equilibrium, `E_(cell)=0` `E_(cell)^(@)=0.0591/2"log"([Zn^(2+)])/([Cu^(2+)])` or `"log"([Zn^(2+)])/([Cu^(2+)])=(2xxE_(cell)^(@))/(0.0591)` `(E_(cell)^(@)=0.76+0.34=1.10" volt")` `=(2xx1.10)/0.0591=37.225` `([Zn^(2+)])/([Cu^(2+)])=1.679xx10^(37) : 1`
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Two students use same stock solution of `ZnSO_(4)` and a solution of `CuSO_(4)`. The `EMF` of one cell is `0.03` higher than the other. The concentrat
`Zn(s)|ZnSO_(4) (C_(1))||0.5 M CuCO_(4)|Cu(s)` `E_(1)=E^(@)-0.06/2"log" C_(1)/0.5` ...(i) student II. `Zn(s)|ZnSO_(4) (C_(1))||CuSO_(4) (C_(2))|Cu(s)` `E_(2)=E^(@)-0.06/2"log" C_(1)/C_(2)` ...(ii) `E_(1)-E_(2)=0.06/2 ["log"C_(1)/C_(2)-"log"C_(1)/0.5]` `0.03=0.06/2[log (0.5/C_(2))]` `1="log" 0.5/C_(2)` `C_(2)=0.05 M`
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The electrode potential becomes equal to standard electrode potential when reactants and products concentration ratio is:
Correct Answer - A `E=E^(0)-(0.0591)/(n)log((["product"])/(["reactant"]))` if `(["product"])/(["reactant"])=1` then `E=E^(0)`
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The potential of the Daniel cell, `Zn |(ZnSO_(4))||(CuSO_(4)|Cu` was reported by Buckbee, Surdzial and Metz as `E^(@) = 1.1028 - 0.641 xx 10^(-3)T` +
Correct Answer - D `E^(@)=1.1028-0.641xx10^(-3)T+0.72xx10^(-5)T^(2)` `((dE^(@))/(dT))_(2S)=-0.641xx10^(-3)+2xx0.72xx10^(-5)T=(-0.641+0.36)xx10^(-3)=-0.281xx10^(-3)` `triangleS^(@)=nF(dE)/(dT)=2xx96500xx(-281xx10^(3))=-54.23EU`
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