Practice free Redox Reactions (Chemistry) NEET multiple-choice questions online with instant answers and detailed explanations. No login required.
A metallic oxide contains 20% oxygen. The equivalent weight of metal is
In metallic oxide, 80% will be metal Suppose, long wt. of metal = 80g
wt. of oxygen = 20g
Equivalents of metal = Equivalents of oxygen
EM = 32
2 gm limestone (CaCO3 + Impurities ) is reacted with 100ml N/2 HCl. The excess acid requires 60ml N/2 NaOH. The % purity of limestone is
Total milliequivalents of HCl = 100x1/2=50
Left milliequivalents of HCl = 60x1/2=30
Used milliequivalents of HCl = 20
Miliequivalents of pure CaCO3 = Milli w/50x1000=20
A 2 gm mixture of K2CO3 and KCl is completely reacted with 100 ml N/10 HCl. The % of K2CO3 in the mixture is (K = 39, Cl =35.5, C=12, O=16)
2 gm mixture of K₂CO₃ and KCl is treated with 100 ml N/10 HCl. K₂CO₃ neutralizes 2 equivalents of HCl while KCl remains unreacted. Using the mole concept, it can be calculated that the % of K₂CO₃ in the mixture is 34.5%.
400 ml M/10 H2SO4 is mixed with 600 ml N/10 NaOH then normality and nature of solution respectively will be
When 400 ml M/10 Hâ‚‚SOâ‚„ (0.4 mol) is mixed with 600 ml N/10 NaOH (0.6 mol), the excess NaOH neutralizes some Hâ‚‚SOâ‚„, leaving the solution slightly acidic with a normality of 0.02N.
The oxidation number of sulphur in S8, S2F2 and H2S respectively are :
In S8, S2F2 and H2S, the oxidation number of S is 0, +1 and -2 respectively.
When Cl2 is converted into Cl- & Cl then n-factor of Cl2 will be:
When Cl2 is converted to Cl- and ClO3-, the change in oxidation state is from 0 to -1 and +5 respectively. The overall change is 5 units. Therefore, the n-factor for Cl2 is 5/3 according to the formula n = (highest O.S. - lowest O.S.)/number of atoms undergoing change.
MnO2 + 4HCl MnCl2+2H2O+Cl2, the equivalent wt. of HCl will be (MMol wt of HCl)
Which acts as a reducing agent only?
IN H2S, S has lowest oxidation number i.e. -2
Standard electrode potential data are useful for understanding the suitability of an oxidant in a redox titration. Some half cell reactions and their standard potential are given below
Mn(aq) + 8H+(aq) + 5e- Mn2+(aq) + 4H2O(l);=1051V
Cr2O72-(aq) +14H+(aq) + 6e 2Cr3+(aq) + 7H2O(l);=1.38V
Fe3+(aq) + e- Fe2+(aq);=0.77V
Cl2(g) + 2e- 2Cl-(aq); = 1.40V
Identify the only incorrect statement regarding the quantitative estimation of aqueous Fe(NO3)2
From reduction electrode potential values, Mn can oxidise Cl- as well as Fe2+. Hence, Mn cannot be used in aqueous HCl for quantitative estimation of Fe(NO3)2.
The equivalent weight of H3PO2, when it disproportionates into PH3 and H3PO3 is
The disproportionation reaction of H3PO2 is: 2H3PO2 → PH3 + H3PO3. In this reaction, 2 moles of H3PO2 produce 1 mole of PH3 and 1 mole of H3PO3. Therefore, the equivalent weight of H3PO2 is (2 × molecular weight of H3PO2) / 2 = 49.5.
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