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Redox Reactions Test - 31

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Redox Reactions Test - 31
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  • Question 1
    1 / -0
    Your group was asked to identify which types of reactions can be represented as redox reactions. Everyone has given an examples of what they think is correct.
    Check their answer and decide who is right.
    Solution
    $$Crystal$$ $$Lynn$$ is correct because, 
    $$2Na_{(s)}+{Cl_2}_{(g)}\longrightarrow 2NaCl_{(s)}$$ is a redox reaction. 
    Oxidation state of $$Na$$ and $$Cl$$ in reactant side is is $$0$$. 
    Because they are in free state.
    But in right side, $$Na$$ is in $$+1$$ state and $$Cl$$ is in $$-1$$ state.
    $$Na$$ changed from $$0$$ to $$+1\Longrightarrow Oxidised$$
    $$Cl$$ changed from $$0$$ to $$-1\Longrightarrow Reduced$$
    Oxidation and reduction, both occured in a reaction. So, it is a redox reaction.
  • Question 2
    1 / -0
    $$I$$. The oxidation state of carbon in the compounds $${CH}_{4},{C}_{2}{H}_{6}$$ and $${C}_{2}{H}_{2}$$ is $$+4$$.
    $$II$$. Carbon has four valence electrons.
    Solution
    $$I.$$ For covalent compounds, there is no oxidation state.
    $$II.$$ Carbon has $$4$$ valence electrons. see the figure.

  • Question 3
    1 / -0
    What is the oxidation number of $$Br$$ in the compound $$RbBr{O}_{4}$$?
    Solution
    $$RbBrO_4$$
    Total charge$$=0.$$
    $$Rb$$ is a $$group$$ $$I$$ element. So, its oxidation no. $$=+1$$.
    Oxidation no. of oxygen $$=-2$$.
    Let, Oxidation no. of $$Br$$ be $$x$$.
    So, $$1+x+4(-2)=0\\1+x-8=0\\x-7=0\\x=7 $$
    So, oxidation state of $$Bromine$$ is $$+7$$.
  • Question 4
    1 / -0
    Balanced chemical equation imply which of the following?
    Solution
    In a balanced decimal equation, no. of atoms are conversed. Means, no. of atoms before reaction are exactly same as no. of atoms after reaction.
  • Question 5
    1 / -0
    In which of the following compounds does nitrogen have the highest oxidation number?
    Solution
    In Nitrogen gas, oxidation no. of Nitrogen is $$0$$. Because , it exist in free state.
    In Aluminium Nitrate, it is $$-3\Longrightarrow AlN\Longrightarrow +3+x=0=-3.$$
    In Potassium Nitrate, $$KNO_3\Longrightarrow 1+x+3(-2)=0=+5$$ .
    In $$HNO_2$$ $$\Longrightarrow 1+x+2(-2)=0,\\x-3=0,\\x=3. $$
    So, highest oxidation no. is in Potassium nitrate.
  • Question 6
    1 / -0
    Which of the following is not a redox reaction?
    Solution

  • Question 7
    1 / -0
    $$2Pb{ O }_{ \left( s \right)  }+{ C }_{ \left( s \right)  }\rightarrow 2{ Pb }_{ \left( s \right)  }+C{ O }_{ 2\left( g \right)  }$$
    Which of the following statements are correct for the above chemical reaction?
    a) Lead is reduced
    b) Carbon dioxide is oxidized
    c) Carbon is oxidized
    d) Lead oxide is reduced
    Solution
    $$2PbO_{(s)}+C_{(s)}\longrightarrow 2Pb_{(s)}+CO_{2(g)}$$
    $$\therefore$$ $$C$$ is oxidized from $$0$$ to $$+4$$ and $$Pb$$ is reduced from $$+2$$ to $$0.$$
    Correct answer $$(a)$$ and $$(c).$$

  • Question 8
    1 / -0
    Which expression gives the correct value for the standard potential for a gold-rhodium voltaic cell?
    Half-reaction$$E^{\circ}(V)$$
    $$Rh_{(aq)}^{+3} + 3e^{-} \rightarrow Rh_{(s)}$$$$0.76\ V$$
    $$Au_{(aq)}^{+} + e^{-} \rightarrow Au_{(s)}$$$$1.69\ V$$
    Solution
    We know that, metal having higher reduction potential undergoes reduction;
    so, Au will undergo reduction and Rh will undergo oxidation.
    $$E^0_{cell}$$ = 1.69 V - 0.76 V
  • Question 9
    1 / -0
    Which of the following represents the oxidation half-reaction for the unbalanced redox equation?
    $$Cu + HNO_{3} \rightarrow Cu (NO_{3})_{2} + NO_{2}$$
    Solution
    $$Cu+HNO_3 \longrightarrow Cu(NO_3)_2+NO_2$$
    As Oxidation of $$Cu$$ is taking place in the reaction.
    Thus the oxidation half-cell reaction will be $$Cu \longrightarrow Cu^{+2}+2e^-$$
  • Question 10
    1 / -0
    The given figure represents a $$Zn-Cu$$ electrochemical cell:
    Based on the figure, which of the following represents the balanced chemical reaction?

    Solution
    $$Zn_{(s)}+Cu^{+2}_{(aq.)} \longrightarrow Zn^{+2}_{(aq.)}+Cu_{(s)}$$
    $$\longrightarrow$$ This is also called Galvanic cell reaction.
    $$\longrightarrow$$ In this $$Cu^{+2}$$ converts to $$Cu$$ due to high standard reduction potentials than $$Zn$$.
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