Identify the product and the colour when `MnO_(2)` is fused with solid KOH in the presence of oxygen `(O_(2))`
A. `KMnO_(4)`(purple)
B. `K_(2)MnO_(4)`(dark green)
C. `MnO_(2)` (Colourless)
D. `Mn_(2)O_(3)` (brown)
1 Answers
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`MnO_(4)^(-) + 8H^(+) + 5e^(-) rightarrow Mn^(2+) + 4H_(2)O`, `E^(@) = 1.51V` `MnO_(2) + 4H^(+) + 2e^(-) righarrow Mn^(2+) + 2H_(2)O` `E^(@) = 1.23V`
Correct Answer - D Eq(i) - Eq(ii) `MnO_(4)^(-) + 4H^(+) + 3e^(-) to Mn underset(+4)(O_(2)) + 2 H_(2)O , E_(3)^(@) = ?` `Delta G_(3)^(@) = DeltaG_(1)^(@) - DeltaG_(2)^(@) - n_(3) E_(3)^(@)...
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The charge required for the reduction of 1 mol of `MnO_(4)^(-)` to `MnO_(2)` is
Correct Answer - B `overset(+7)(MnO_(4)^(-)rarroverset(+4)(MnO_(2))` `i.e. MnO_(4)^(-)+3e^(-)rarrMnO_(2)` `MnO_(4)^(-)+2H_(2)O+3e^(-) rarrMnO_(2)+4OH^(-)` Thus for reduction of 1 mol of `MnO_(4)^(-)` to `MnO_(2)` charge required =3F
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How would you prepare (i) `K_(2)MnO_(4)` and `MnO_(2)`? (ii) `Na_(2)Cr_(2)O_(7)` from `Na_(2)CrO_(4)` ?
(i)`MnO_(2)` is fused with KOH in the presence of oxygen (air). `4KOH+2MnO_(2)+O_(2) overset(Heat) to 2K_(2)MnO_(4)+2H_(2)O` (ii) `2Na_(2)CeO_(4)` is treated with dilute `H_(2)SO_(4)`. `2Na_(2)CrO_(4)+H_(2)SO_(4) to Na_(2)Cr_(2)O_(7)+Na_(2)SO_(4)+H_(2)O`
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Given : (i)`MnO_(4)^(-)+8H^++5e^(-)toMn^(2+)+4H_2O , E^0=x_1V` (ii)`MnO_(2)+4H^++2e^(-)toMn^(2+)+2H_2O , E^0=x_2V` Find `E^(@)` for the following reac
Correct Answer - C `DeltaG_1^@=-5x_1F , " " DeltaG_2^@=-2x_2F` `DeltaG_3^@=DeltaG_1^@-DeltaG_2^@ , " " DeltaG_3^@=2x_2F-5x_1F` `therefore x=(5x_1-2x_2)/3`
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A black compound (A) is solid state is fused with KOH with `KClO_3`.The aqueous extract of fused mass is green colour solution (B).On passing `CO_2` g
Correct Answer - 7 `underset((A))(3MnO_2)+6KOH+KCIO_3oversetDeltatounderset((B))(3K_2MnO_4)+KCl+3H_2O` In presence of `CO_2`, the medium becomes acidic. `3MnO_4^(2-)+4H^(+)toMnO_2 + 2MnO_4^(-)+2H_2O` `underset((C ))(MnO_4^-)+5Fe^(2+)+8H^(+)toMn^(2+)+5Fe^(3+)+4H_2O`
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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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Upon titrating `50mL` of `1.96%(w//v) H_(2)SO_(4)` with a `KOH` solution (containing `11.2g KOH` per litre of solution on adding `50mL KOH` solution.
`[H_(2)SO_(4)]=(19.6)/(98)=0.2 M` and `KOH=(11.2)/(56)=0.2M` so for equivalent point, `(0.2xx50xx2)/(0.2)=100mL` of `KOH`solution should be added. `[H^(+)]=(0.2xx2-0.2)/(2)=0.1M` so `pH=1`.
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Chlorine gas can be produced by reacting sulphuric acid with a mixutre of `MnO_(2)` and NaCl. The reacting follows the equations: `2NaCl+MnO_(2)+3H_(2
Correct Answer - d `therefore "2 moles NaCl(117 g")-=1"mole of "Cl_(2)(22.4"at STP")` `therefore "1 g NaCl will give"=(22.4)/(117)"L "Cl_(2)" at STP"` =0.191 L `Cl_(2)` at STP
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`E_(Mn^(2+)//MnO_(4)^(-))^(0)=-1.51V` `E_(MnO_(2)//Mn^(+2))^(0)=+1.23V` `E_(MnO_(4)^(-)//MnO_(2)=?` (All in acidic medium)
`4H_(2)O+Mn^(2+)toMnO_(4)^(-)+8H^(+)+5e^(-)" triangleG_(1)` (i). `lArrMnO_(4)^(-)+8H^(+)+5e^(-)to4H_(2)O+Mn^(2+)" " -triangleG_(1)` `2e^(-)+MnO_(2)+4H^(+)toMn^(2+)+2H_(2)O" "triangleG_(2)` (ii). `lArr2H_(2)O+Mn^(2+)toMnO_(2)+4H^(+)+2e^(-)" "-triangleG_(2)` (iii). `lArr4H^(+)+MnO_(4)^(-)+3e^(-)toMnO_(2)+2H_(2)O" "triangleG_(3)` `(i)+(ii)=(iii)` `triangleG_(3)=-triangleG_(2)-triangleG_(2)` `-3E_(3)F=5E_(1)^(0)F+2E_(2)^(0)F` `E=(-[5E_(1)+2E_(2)])/(3)=(-[5(-1.51)+2(1.23)])/(3)=(-[-7.55+2.46])/(3)=(+5.09)/(3)=1.69V`
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One g equivalent of Na metal is formed from electrolyis of fused NaCl. No of "mole" of Al from the fused `Na_(3)AlF_(6)` with the same current passed
Correct Answer - C `underset(1" mole")(Na^(+))+underset(1" faraday")(e^(-)to)Na(s)` `AP^(+)+underset(1" faraday")(3e^(-)to)Al(s)` No. of mole of `Al=(1)/(3)` mole
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