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Redox Reactions Test - 28

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Redox Reactions Test - 28
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  • Question 1
    1 / -0
    Determine the. oxidation state of chromium in the finfll product formed by the reaction between KI and acidified potassium dichromate solution.
    Solution
    K would not be expected to be in the final NET equation since K+ is a spectator ion; i.e., it does not react. When you put KI into water it dissolves and splits up into K+ and I- ions. Only the I- participates in the redox reaction. It wouldn't matter whether you used KI or NaI. 
    Oxidation: 2II2+2e2I^- \rightarrow I^2 + 2e^-
    Reduction: Cr2O72+14H++6e2Cr3++7H2OCr_2O_7^{2-} + 14H^+ + 6e^- \rightarrow 2Cr^{3+} + 7H_2O

    The final product formed is :
    6I+Cr2O72+14H+>3I2+2Cr3++7H2O6I^- + Cr_2O_7^{2-} + 14H^+ ----> 3I_2 + 2Cr^{3+} + 7H_2O
    Calculating the oxidation state of chromium  we get + 3


  • Question 2
    1 / -0
    Which of the following processes does not involve either oxidation or reduction ?
    Solution
    Here, in this reaction
    CaO+H2OCa(OH)2CaO+ H_2O \rightarrow Ca(OH)_2
    oxidation number doesn't change so its not a redox reaction.
  • Question 3
    1 / -0
    The oxidation number of MnMn is +8+8 in the compound:
    Solution
    In MnO4MnO_4 manganese oxidation number is +8+8
    Oxidation number of oxygen is 2-2. MnO4MnO_4 is neutral molecule so total oxidation number of molecule is zero.
    Oxidation number of Mn+4(2)=0Mn+4(-2)=0
    Oxidation number of Mn=+8Mn =+8.
    So answer is B.
  • Question 4
    1 / -0
    Arrange the following in correct order of oxidation number of Mn
    (i) Mn2+Mn^{2+}                    (ii) MnO2MnO_2
    (III) KMnO4KMnO_4                (iv) K2MnO4K_2MnO_4
    Solution
    i) Oxidation number of Mn in Mn2+=+2Mn^{2+} = +2
    (ii) Let oxidation number of Mn in MnO2=xMnO_2 = x
    x+(2×2)=0,x=+4\therefore x + (-2 \times 2) = 0 , x = +4
    (iii) Let oxidation number of Mn in KMno_4 = x 
    +1+x+(1×4)=0,x+=+7\therefore +1 +x +(-1 \times 4 ) = 0, \therefore x += + 7
    (iv) Let oxidation number of Mn in K2MnO4K_2MnO_4 = x
    (+1×2)+x+(2××4=0\therefore (+1 \times 2 ) + x + (-2 \times \times 4 _ = 0
    or x + 2 - 8 = 0
    \therefore x = +6
    \therefore the order of oxidation state is (i) < (ii) < (iv) < (iii) 
  • Question 5
    1 / -0
    The oxidation number of oxygen in hydrogen peroxide is:
  • Question 6
    1 / -0
    Identify redox reaction among the following:
    Solution
    Metal displacement reactions are redox reactions in which both oxidation and reduction takes place.Example; Mg+ZnSO4MgSO4+ZnMg + ZnSO_{4} \rightarrow MgSO_{4} + Zn
    Here Mg is undergoing oxidation with an increase in oxidation number from 0 to +2 and Zn is undergoing reduction with a decrease in oxidation number.
  • Question 7
    1 / -0
    The oxidation number of chlorine in NaClO3NaCl{O}_{3} is:
    Solution
    The overall charge of molecule is zero (because it's not an ion). Since sodium is a 1A family member, you can assume that the charge is +1+1. The charge of oxygen is almost always 2-2 so you can assume that as well. 
    (+1)+(+1) + oxidation of Cl+3(2)=0Cl + 3(-2) = 0
    So the oxidation of chlorine in that problem is +5+5.
    So answer is C.
  • Question 8
    1 / -0
    Identify X in the equation given below. FeCl3+SO2+H2OFeCl2+HCl+XFeCl_3 + SO_2 + H_2O \rightarrow FeCl_2 + HCl + X
    Solution
    The balanced equation can be written as:
    2FeCl3+SO2+2H2O2FeCl2+2HCl+H2SO42FeCl_{3} + SO_{2} + 2H_{2}O \rightarrow 2FeCl_{2} + 2HCl + H_{2}SO_{4}
    Thus X =  H2SO4 H_{2}SO_{4}
  • Question 9
    1 / -0
    The oxidation state of nitrogen in NaNO3Na{NO}_{3} is :
    Solution
    The overall charge is zero (because it is not an ion). Since sodium is a 1A family member, the charge is +1. The charge of oxygen is almost always -2.
    Hence, 1+OS of N+(3×2)=01+ OS \ of\ N + (3 \times -2)=0
    So, the oxidation of nitrogen in that problem is +5.
  • Question 10
    1 / -0
    Oxidation number of S in SO42SO_4^{2-}:
    Solution
    Let oxidation state of S in SO42=xSO_4^{2-} = x
    x+(4×2)=2\therefore x+ (4 \times - 2) = - 2
    or x8=2x - 8 = - 2    \therefore x=+6x = +6
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