Redox Reactions Test

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Redox Reactions Test - 64

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Weekly Quiz Competition

Question 1
1 / -0

Which of the following is an intermolecular redox reaction?

A
$$2OCH-CHO\xrightarrow { \quad \overset { \ominus }{ O } H\quad } HOC{ H }_{ 2 }-C{ H }_{ 2 }OH$$

B
$$2{ C }_{ 6 }{ H }_{ 5 }CHO\xrightarrow { Al{ \left( O{ C }_{ 2 }{ H }_{ 5 } \right) }_{ 3 } } { C }_{ 6 }{ H }_{ 5 }COOH+{ C }_{ 6 }{ H }_{ 5 }C{ H }_{ 2 }OH$$

C
$$4Cr{ O }_{ 5 }+6{ H }_{ 2 }S{ O }_{ 4 }\longrightarrow 2{ Cr }_{ 2 }{ \left( S{ O }_{ 4 } \right) }_{ 3 }+6{ H }_{ 2 }O+7{ O }_{ 2 }$$

D
$${ As }_{ 2 }{ S }_{ 3 }+HN{ O }_{ 3 }\longrightarrow { H }_{ 3 }As{ O }_{ 4 }+{ H }_{ 2 }S{ O }_{ 4 }+NO$$
Question 2
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What is the oxidation number of $$Mn$$ in $$KMn{O}_{4},\ {K}_{2}Mn{O}_{4},\ MnS{O}_{4},\ Mn{O}_{2}$$ and $${Mn}_{3}{O}_{4}$$ respectively?

A
$$+7, \:+6,\: +2,\: +4,$$ $$+\frac{8}{3}$$

B
$$+4,\: +6,\: +2,\: +4,$$ $$+\frac{8}{3}$$

C
$$+7,\: +6,\: +2,\: +3,$$ $$+\frac{8}{3}$$

D
None of these

Solution

Let $$x$$ be the oxidation number of $$Mn$$.
Since, the overall charge on the complex is $$0$$, the sum of oxidation states of all elements in it should be equal to $$0$$.
$${K}_{2}Mn{O}_{4}$$; $$2+x+4(-2)=0;\: x=+6$$

$$ MnS{O}_{4}$$; $$x-2=0;\:x=+2$$

$$ Mn{O}_{2}$$; $$x+2(-2)=0;\:x=+4$$

$${Mn}_{3}{O}_{4}$$; $$3x+4(-2)=0;\:x=\frac{8}{3}$$
The oxidation number of $$Mn$$ in $$KMn{O}_{4},\ {K}_{2}Mn{O}_{4},\ MnS{O}_{4},\ Mn{O}_{2}$$ and $${Mn}_{3}{O}_{4}$$ is $$+7,\: +6,\: +2,\: +4$$ and $$+\frac{8}{3}$$ respectively.
Question 3
1 / -0

The oxidation state of $$A, B,$$ and $$C$$ in a compound are $$+2, +5$$ and $$-2$$ respectively. The compound is :

A
$${ A }_{ 2 }{ \left( BC \right) }_{ 2 }$$

B
$${ A }_{ 2 }{ \left( BC \right) }_{ 3 }$$

C
$${ A }_{ 3 }{ \left( { BC }_{ 4 } \right) }_{ 2 }$$

D
$${ A }_{ 2 }{ \left( { BC }_{ 4 } \right) }_{ 3 }$$

Solution

The oxidation state of A, B, and C in a compound are +2, +5 and −2 respectively. The compound is $$\displaystyle A_3(BC_4)_2$$.
Total charge on one $$\displaystyle BC_4 $$ unit will be $$\displaystyle (+5+4(-2))=-3 $$. Total charge on two such units will be $$\displaystyle 2(-3) = -6$$
Total charge on 3 ions of A will be $$\displaystyle 3 \times 2 =+6$$
Thus, the positive charges are balanced with negative charges.
Question 4
1 / -0

Find the oxidation number of elements in each case :

$$ S$$ in $$Na_2S_2O_3$$, $$S_4$$, $$S_8$$ and $$Na_2S_2O_7$$.

A
$$+2,\: 0,\: 0,\: +6$$

B
$$+1,\: 0,\: +1,\: +6$$

C
$$+1,\: 0,\: 0,\: +4$$

D
None of the above

Solution

The oxidation number of $$ S$$ in $$Na_2S_2O_3$$, $$S_4$$, $$S_8$$ and $$Na_2S_2O_7$$ are $$+2,\: 0,\: 0,\: +6$$ respectively.
Let X be the oxidation number of $$S$$ in $$Na_2S_2O_3$$,
$$\displaystyle 2+2X+3(-2)=0 $$
$$\displaystyle X = +2 $$
Let X be the oxidation number of $$S$$ in $$S_4$$,
$$\displaystyle 4X=0 $$
$$\displaystyle X=0 $$
Let X be the oxidation number of $$S$$ in $$S_8$$
$$\displaystyle 8X = 0 $$
$$\displaystyle X = 0$$
Let X be the oxidation number of $$S$$ in $$Na_2S_2O_7$$
$$\displaystyle 2+2x+7(-2) = 0 $$
$$\displaystyle X=+6 $$.
Question 5
1 / -0

Two half-cells have potentials $$-0.44$$ and $$0.799$$ volt respectively. These two are coupled to make a galvanic cell. Which of the following will be true?

A
Electrode of half-cell potential $$-0.44\ V$$ will act as anode

B
Electrode of half-cell potential $$-0.44\ V$$ will act as cathode

C
Electrode of half-cell potential $$0.799\ V$$ will act as anode

D
Electrode of half-cell potential $$-0.44\ V$$ will act as a positive terminal

Solution

Galvanic Cell:-
$$\rightarrow$$ In Galvanic cell,left hand side (anode), oxidation occurs. At right hand side electrode(cathode) , reduction occurs.
$$\rightarrow$$ The sum of the oxidation half reaction at anode and reduction half reaction at cathode is the overall cell reaction.
$$\rightarrow$$ Electrode with positive value of electrode potential suggest forward reaction i.e. reduction at cathode. So, electrode with 0.799 volt will act as cathode.
$$\rightarrow$$Electrode with negative value of electrode potential suggest reverse reaction i.e. oxidation at anode. Hence, electrode with -0.44 volt will act as anode.
Question 6
1 / -0

For the redox process,
$$Zn(s) + Cu^{2+} \rightleftharpoons Zn^{2} + Cu(s)\ E_{cell}^{\circ} = + 1.10\ V$$
which graph correctly represents $$E_{cell}$$ (Y-axis) as a function of $$\log \dfrac {[Zn^{2+}]}{[Cu^{2+}]}$$ (X-axis)?

A

B

C

D

Solution

At the initial stages voltage is greater than $$e^o$$ because the reaction quotient is less than $$1(Q<1)$$.
As the reaction proceeds
the ratio Q= $$\cfrac {[Zn^{+2}]}{[Cu^{+2}]}$$ steadily increases and cell voltage steadily decreases then graph looks like
(Refer to Image)
Question 7
1 / -0

A steady current of $$10.0$$ Amps is passed through a nickel production cell of $$15$$ minutes.
Which of the following is the correct expression for calculating the number of grams of nickel produced?
Note:
$$\bullet 1\ faraday = 96,500\ Coulombs$$
$$\bullet$$ The electroytic cell involves the following half-reaction:
$$Ni_{(aq)}^{2+} + 2e^{-} \rightarrow Ni_{(s)}$$

A
$$\dfrac {(10.0)(15)(96,500)}{(59)(60)} g$$

B
$$\dfrac {(10.0)(15)(59)}{(60)(96,500)} g$$

C
$$\dfrac {(10.0)(15)(60)(59)}{(96,500)(2)} g$$

D
$$\dfrac {(96,500)(59)}{(10)(15)(60)(2)} g$$

Solution

According to Faraday's Law
$$W= \cfrac {Z \times I \times t}{F}$$
$$W$$= mass of compound
$$Z$$= Equivalent mass $$F$$= 96500
$$I$$= Current in Ampere
$$t$$= time in seconds
$$\Rightarrow W= \left(\cfrac {59}{2}\right)\cfrac {\times 15\times 60 sec\times 10A}{96500}$$
$$\Rightarrow W= \cfrac {(10.0)(15)(60)(59)}{(96500)(2)}g$$

Question 8

1 / -0

Consider the table of standard reduction potentials shown below.

Half-reaction	$$E^{\circ}$$
$$Cl_{2} + 2e^{-}\rightarrow 2Cl^{-}$$	$$1.36\ V$$
$$O_{2} + 4H^{+} + 4e^{-} \rightarrow 2H_{2}O$$	$$1.23\ V$$
$$2H_{2}O + 2e^{-} \rightarrow H_{2} + 2OH^{-}$$	$$-0.83\ V$$
$$Rb^{+} + e^{-} \rightarrow Rb$$	$$-2.93\ V$$

Use the information from the table and your knowledge of electrochemistry to predict the CORRECT net ionic equation for the reaction that will occur when an aqueous solution of rubidium chloride undergoes electrolysis.

$$2Rb^{+} + 2Cl^{-} \rightarrow 2Rb + Cl_{2}$$

$$2H_{2}O \rightarrow 2H_{2} + O_{2}$$

$$H_{2} + 2OH^{-} + 2Rb^{+} \rightarrow 2Rb + 2H_{2}O$$

$$4Cl^{-} + O_{2} + 4H^{+} \rightarrow 2H_{2}O + 2Cl_{2}$$

Solution

From the given table, we have;

Rb as the lowest reduction potential therefore it undergo oxidation

and $$Cl_2$$ have the highest oxidation potential therefore it undergo reduction.

so, the net reaction becomes

$$2Rb^+ + 2Cl^- \rightarrow 2Rb + Cl_2$$

Question 9

1 / -0

Consider a galvanic cell using solid $$Cu$$ and $$Fe$$ metals with their corresponding solutions.
What is the $$E^{\circ}_{cell}$$?

Standard Potential (V)	Reduction Half-Reaction
$$2.87$$	$$F_{2}(g) + 2e^{-} \rightarrow 2F^{-}(aq)$$
$$1.51$$	$$MnO_{4}^{-}(aq) + 8H^{+}(aq) + 5e^{-}\rightarrow Mn^{2+}(aq) + 4H_{2}O(l)$$
$$1.36$$	$$Cl_{2}(aq) + 3e^{-} \rightarrow 2Cl^{-}(aq)$$
$$1.33$$	$$Cr_{2}O_{7}^{2-} (aq) + 14H^{+}(aq) + 6e^{-} \rightarrow 2Cr^{3+}(aq) + 7H_{2}O(l)$$
$$1.23$$	$$O_{2}(g) + 4H^{+}(aq) + 4e^{-}\rightarrow 2H_{2}O(l)$$
$$1.06$$	$$Br_{2}(l) + 2e^{-} \rightarrow 2Br^{-}(aq)$$
$$0.96$$	$$NO_{3}^{-}(aq) + 4H^{+}(aq) + 3e^{-}\rightarrow NO(g) + H_{2}O(l)$$
$$0.80$$	$$Ag^{+}(aq) + e^{-} \rightarrow Ag(s)$$$
$$0.77$$	$$Fe^{3+} (aq) + e^{-} \rightarrow Fe^{2+}(aq)$$
$$0.68$$	$$O_{2}(g) + 2H^{+}(aq) + 2e^{-}\rightarrow H_{2}O_{2}(aq)$$
$$0.59$$	$$MnO_{4}^{-}(aq) + 2H_{2}O(l) + 3e^{-}\rightarrow MnO_{2}(s) + 4OH^{-}(aq)$$
$$0.54$$	$$I_{2}(s) + 2e^{-}\rightarrow 2I^{-}(aq)$$
$$0.40$$	$$O_{2}(g) + 2H_{2}O(l) + 4e^{-} \rightarrow 4OH^{-}(aq)$$
$$0.34$$	$$Cu^{2+}(aq) + 2e^{-} \rightarrow Cu(s)$$
$$0$$	$$2H^{+}(aq) + 2e^{-}\rightarrow H_{2}(g)$$
$$-0.28$$	$$Ni^{2+}(aq) + 2e^{-}\rightarrow Ni(s)$$
$$-0.44$$	$$Fe^{2+}(aq) + 2e^{-}\rightarrow Fe(s)$$
$$-0.76$$	$$Zn^{2+}(aq) + 2e^{-}\rightarrow Zn(s)$$
$$-0.83$$	$$2H_{2}O(l) + 2e^{-}\rightarrow H_{2}(g) + 2OH^{-}(aq)$$
$$1.66$$	$$Al^{3+}(aq) + 3e^{-}\rightarrow Al(s)$$
$$-2.71$$	$$Na^{+}(aq) + e^{-} \rightarrow Na(s)$$
$$-3.05$$	$$Li^{+}(aq) + e^{-}\rightarrow Li(s)$$

$$-0.78 V$$

$$0.10 V$$

$$0.78 V$$

$$-0.10 V$$

Solution

$$E^0$$ for Cu = 0.34 V

$$E^0$$ for Fe = -0.44V

so, $$E^0_{cell} = 0.34 - (-0.44) = 0.78 V$$

Question 10
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Oxidation number of 'S' in $$Na_2S_4O_6$$ is:

A
$$+0.5$$

B
$$2.5$$

C
$$+4$$

D
$$+6$$

Solution

Let the oxidation number of $$S$$ in $${ Na }_{ 2 }{ S }_{ 4 }{ O }_{ 6 }$$ be $$x$$. The oxidation states of $$Na$$ and $$O$$ are $$+1$$ and $$-2$$ respectively. Since the atom as whole is neutral, following equation holds true:
$$2\times (+1)\quad +\quad 4\times (x)\quad +\quad 6\times (-2)\quad =\quad 0\\ \Longrightarrow \quad 2\quad +\quad 4x\quad -12\quad =\quad 0\\ \Longrightarrow \quad 4x\quad =\quad 12-2\\ \Longrightarrow \quad 4x\quad =\quad 10\\ \Longrightarrow \quad x\quad =\quad \dfrac { 5 }{ 2 } \quad =\quad 2.5$$

Hence, answer is $$B$$.

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Redox Reactions Test - 64

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Redox Reactions Test - 64

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Question 1

$$2OCH-CHO\xrightarrow { \quad \overset { \ominus }{ O } H\quad } HOC{ H }_{ 2 }-C{ H }_{ 2 }OH$$

$$2{ C }_{ 6 }{ H }_{ 5 }CHO\xrightarrow { Al{ \left( O{ C }_{ 2 }{ H }_{ 5 } \right) }_{ 3 } } { C }_{ 6 }{ H }_{ 5 }COOH+{ C }_{ 6 }{ H }_{ 5 }C{ H }_{ 2 }OH$$

$$4Cr{ O }_{ 5 }+6{ H }_{ 2 }S{ O }_{ 4 }\longrightarrow 2{ Cr }_{ 2 }{ \left( S{ O }_{ 4 } \right) }_{ 3 }+6{ H }_{ 2 }O+7{ O }_{ 2 }$$

$${ As }_{ 2 }{ S }_{ 3 }+HN{ O }_{ 3 }\longrightarrow { H }_{ 3 }As{ O }_{ 4 }+{ H }_{ 2 }S{ O }_{ 4 }+NO$$

Question 2

$$+7, \:+6,\: +2,\: +4,$$ $$+\frac{8}{3}$$

$$+4,\: +6,\: +2,\: +4,$$ $$+\frac{8}{3}$$

$$+7,\: +6,\: +2,\: +3,$$ $$+\frac{8}{3}$$

None of these

Solution

Question 3

$${ A }_{ 2 }{ \left( BC \right) }_{ 2 }$$

$${ A }_{ 2 }{ \left( BC \right) }_{ 3 }$$

$${ A }_{ 3 }{ \left( { BC }_{ 4 } \right) }_{ 2 }$$

$${ A }_{ 2 }{ \left( { BC }_{ 4 } \right) }_{ 3 }$$

Solution

Question 4

$$+2,\: 0,\: 0,\: +6$$

$$+1,\: 0,\: +1,\: +6$$

$$+1,\: 0,\: 0,\: +4$$

None of the above

Solution

Question 5

Electrode of half-cell potential $$-0.44\ V$$ will act as anode

Electrode of half-cell potential $$-0.44\ V$$ will act as cathode

Electrode of half-cell potential $$0.799\ V$$ will act as anode

Electrode of half-cell potential $$-0.44\ V$$ will act as a positive terminal

Solution

Question 6

Solution

Question 7

$$\dfrac {(10.0)(15)(96,500)}{(59)(60)} g$$

$$\dfrac {(10.0)(15)(59)}{(60)(96,500)} g$$

$$\dfrac {(10.0)(15)(60)(59)}{(96,500)(2)} g$$

$$\dfrac {(96,500)(59)}{(10)(15)(60)(2)} g$$

Solution

Question 8

$$2Rb^{+} + 2Cl^{-} \rightarrow 2Rb + Cl_{2}$$

$$2H_{2}O \rightarrow 2H_{2} + O_{2}$$

$$H_{2} + 2OH^{-} + 2Rb^{+} \rightarrow 2Rb + 2H_{2}O$$

$$4Cl^{-} + O_{2} + 4H^{+} \rightarrow 2H_{2}O + 2Cl_{2}$$

Solution

Question 9

$$-0.78 V$$

$$0.10 V$$

$$0.78 V$$

$$-0.10 V$$

Solution

Question 10

$$+0.5$$

$$2.5$$

$$+4$$

$$+6$$

Solution

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