Tag: redox reactions

Quizzes (4) Questions (15)

Questions Related to redox reactions

Multiple choice redox and stoichiometry applications of redox reaction oxidation- reduction reactions redox reactions chemistry

A volume of $100 ml$ of $ H _{2}O _{2}$ is oxidized by 100 ml of $ 1 M \ KMnO _{4}$ in acidic medium $ (MnO _{4}^{-}$ reduced to $ Mn^{2+}).$ A volume of $100 ml$ of same $H _{2}O _{2}$ is oxidized by $'V'$ ml of $1 M \ KMnO _{4}$ in basic medium ($ MnO _{4}^{-}$ reduced to $ MnO _{2}).$ The value of $'V'$ is

  1. $500$
  2. $100$
  3. $33.33$
  4. $166.67$
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

In acidic medium, 

$ N _{1}V _{1} = N _{2}V _{2} $

$ n _1\times M _{1}\times V _{1} = n _2\times M _{2}\times V _{2} $

$ 5\times 1\times 100 =  2\times M _{2} \times 100$ 

$ M _{2} = 2.5\,M $

Now in basic medium,

$ N _{1}V _{1} = N _{2}V _{2} $

$ 2\times 2.5 \times 100=  3\times 1\times V $

$ \therefore V = \dfrac{200\times 2.5}{3} $

$V = 166.66\,ml $

Multiple choice redox and stoichiometry applications of redox reaction oxidation- reduction reactions redox reactions chemistry

In a solid $AB$ of $NaCl$ structure, A atoms occupy the corners of the cubic unit cell. If all the corner atoms are removed then the formula of the unit cell will be

  1. $A _{4}B _{4}$
  2. $B$
  3. $A _{3}B _{4}$
  4. $AB$
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

In an NaCl structure, A atoms are at corners and face centers (total 4). If all corner atoms (8 corners * 1/8 = 1 atom) are removed, 3 A atoms remain. B atoms are at edge centers and body center (total 4). The formula becomes A3B4.

Multiple choice redox and stoichiometry applications of redox reaction oxidation- reduction reactions redox reactions chemistry

Number of electrons involved in the electrodeposition of 63.5 g of Cu from solution of $ CuSO _{4} $ is:

  1. $ 6.022\times 10^{23} $
  2. $ 3.011\times 10^{23} $
  3. $ 12.044\times 10^{23} $
  4. $ 6.022\times 10^{23} $
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation
$ Cu^{2+}+2e^-\rightarrow Cu $ 

1 mole Cu = 63.5 g
 
For deposition of 1 mole of Cu, 2 moles of $e^-$ is required.

$ \therefore$ Number of electrons $ =12.044\times 10^{23} $      
                                    
Hence, option C is correct.
Multiple choice redox and stoichiometry applications of redox reaction oxidation- reduction reactions redox reactions chemistry

Write balanced half reactions for the following redox reaction:
$Cr _2O _7^{2-} + Fe^{2+}\rightarrow  Cr^{3+} + Fe^{3+}$

  1. Reduction: $6e^-+ 14H^+ + Cr _2O _7^{2-} \rightarrow Cr^{3+} + 7H _2O $

    Oxidation: $ \\ Fe^{2+} \rightarrow Fe^{3+} + e^{-}$
  2. Oxidation: $6e^-+ 14H^+ + Cr _2O _7^{2-} \rightarrow Cr^{3+} + 7H _2O $

    Reduction: $ \\ Fe^{2+} \rightarrow Fe^{3+} + e^{-}$
  3. Reduction: $4e^-+ 14H^+ + Cr _2O _7^{2-} \rightarrow Cr^{3+} + 7H _2O $

    Oxidation: $ \\ Fe^{2+} \rightarrow Fe^{3+} + e^{-}$
  4. Oxidation: $6e^-+ 14H^+ + Cr _2O _7^{2-} \rightarrow Cr^{3+} + 6H _2O $

    Reduction: $ \\ Fe^{2+} \rightarrow Fe^{3+} + e^{-}$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

Given Redox reaction is

$Cr _2O _7^{2-}+Fe^{+2} \longrightarrow Cr^{+3}+Fe^ {3+}$
Oxidation half reaction:- $Fe^{+2} \longrightarrow Fe^{+3}+e^-$
                                          [as oxidation number increases from $2$ to $3$]
Reduction half reaction:- $Cr _2O _7^{2-}+6e^-+14H^+ \longrightarrow 2Cr^{+3}+7H _2O$
                                          [as oxidation number decreases from $6$ to $3$]

Multiple choice redox and stoichiometry applications of redox reaction oxidation- reduction reactions redox reactions chemistry

Which particles are gained and lost during a redox reaction?

  1. Protons

  2. Neutrons

  3. Electrons

  4. Positrons
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

  • Indirect redox reactions only take place if in between the electrodes is an environment that is conducive for electric current (meaning that charged particles are present that can freely move around, like the ions of dissolved or molten salts). There is no direct contact between the particles of the oxidation and the reduction. The transfer of the electrons is realised via-via, mostly via metal wiring or other conductive material. 
Exteriorly the conducting wires take care for the electron transport from RED to OX, without direct contact between the reactants.

Multiple choice redox and stoichiometry applications of redox reaction oxidation- reduction reactions redox reactions chemistry

Two half reactions are given as follows:
$2e^{-} + H^{+} + H _5IO _6 \rightarrow IO^{3-} + 3H _2O $
$Cr \rightarrow Cr^{3+} + 3e^{-}$
Final balanced reaction is:

  1. $3H^{+} + 3H _5IO _6 + 2Cr \rightarrow 2Cr^{3+} + 3IO^{3-} + 9H _2O$
  2. $5H^{+} + 3H _5IO _6 + 2Cr \rightarrow 2Cr^{3+} + 3IO^{3-} + 10H _2O$
  3. $3H^{+} + 3H _5IO _6 + 4Cr \rightarrow 4Cr^{3+} + 3IO^{3-} + 9H _2O$
  4. $3H^{+} + 3H _5IO _6 + 3Cr \rightarrow 3Cr^{3+} + 3IO^{3-} + 9H _2O$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

Multiply each given half reactions by a number to balance electrons on both sides.

$[2e^-+H^++H _5IO _6 \longrightarrow IO^{-3}+3H _2O] \times 3 \longrightarrow (1)$
$[Cr \longrightarrow Cr^{+3}+3e^- ] \times 2 \longrightarrow (2)$
Now adding equation (1) and (2) we get balanced equation
$3H^++3H _5IO _6+2Cr \longrightarrow 2Cr^{+3}+3IO^{3-}+9H _2O$ .

Multiple choice redox and stoichiometry applications of redox reaction oxidation- reduction reactions redox reactions chemistry

In the reaction, $FeS {2} + KMnO _{4} + H^{+} \rightarrow Fe^{3+} + SO _{2} + Mn^{2+} + H _{2}O$, the equivalent mass of $FeS _{2}$ would be equal to__________.

  1. $\text{molar mass}$
  2. $\dfrac {\text {molar mass}}{10}$
  3. $\dfrac {\text {molar mass}}{11}$
  4. $\dfrac {\text {molar mass}}{13}$
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

$Fe^{2+} \rightarrow Fe^{3+} + e^{-}; S _{2}^{2-} \rightarrow 2S^{4+} + 10e^{-}$
$\therefore FeS _{2} \rightarrow 2S^{4+} + Fe^{3+} + 11e^{-}$
Equivalent mass of $FeS _{2} = \dfrac {\text {Molar mass}}{11}$.