Redox Reactions Test

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Redox Reactions Test - 69

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Weekly Quiz Competition

Question 1
1 / -0

The stoichiometric constants for the reaction $$pCu + qHNO_3 \rightarrow rCu(NO_3)_2 + sNO + tH_2O$$ p, q, r ,s and t respectively are:

A
3, 3, 3, 2, 3

B
3, 2, 3, 2, 4

C
3, 8, 3, 2, 4

D
2, 3, 3, 3, 2

Solution

Given reaction:
$$pCu+qHNO_3 \rightarrow rCu(NO_3)_2+sNO+tH_2O$$

Assign the oxidation states as:

$$\overset {0}{Cu}+H\overset {+5}{N}O_3 \rightarrow \overset {+2}{Cu}(\overset{+5}{N}O_3)_2+\overset {+2}{N}O+H_2O$$

$$\therefore$$ for a total change of 6 electrons in reduction of N and oxidation of Cu multiply by 2 and 3 respectively

$$3\overset {0}{Cu}+2H\overset {+5}{N}O_3 \rightarrow 3\overset {+2}{Cu}(\overset{+5}{N}O_3)_2+2\overset {+2}{N}O+H_2O$$

-Balance N
$$3\overset {0}{Cu}+8H\overset {+5}{N}O_3 \rightarrow 3\overset {+2}{Cu}(\overset{+5}{N}O_3)_2+2\overset {+2}{N}O$$

-Balance O and H by adding $$H_2O$$
$$3\overset {0}{Cu}+8H\overset {+5}{N}O_3 \rightarrow 3\overset {+2}{Cu}(\overset{+5}{N}O_3)_2+2\overset {+2}{N}O+4H_2O$$

comparing the above balanced reaction with given reaction we get:
p=3,q=8,r=3,s=2,t=4
Question 2
1 / -0

For the redox reaction:

$$Zn+N{ O }^-_3\rightarrow Zn^{2+}+NH^+_4$$

in a basic medium, coefficients of $$Zn, \ NO^-_3$$ and $$OH^-$$ in the balanced equation respectively are:

A
$$4, 1, 7$$

B
$$7, 4, 1$$

C
$$4, 1, 10$$

D
$$1, 4, 10$$

Solution

$$ 4 Zn+1 \mathrm{NO}_{3}+10 \mathrm{H}^{+} \longrightarrow 4Zn^{+2}+\mathrm{NH}_{4}^{+}+3 \mathrm{H}_{2} \mathrm{O} $$
$$ \mathrm{Zn}, \mathrm{NO}_{3}^- \& \mathrm{^-OH}=4,1,10 $$
Question 3
1 / -0

Half cell reactions for some electrodes are given below:
I. $$A+{ e }^{ - }\longrightarrow { A }^{ - }\ ;\quad \quad \ { E }^{ 0 }=0.96V$$
II. $${ B }^{ - }+{ e }^{ - }\longrightarrow { B }^{ 2- }\ ;\quad { E }^{ o }=-0.12V$$
III. $${ C }^{ + }+{ e }^{ - }\longrightarrow C\ ;\quad \ { E }^{ o }=+0.18V$$
IV. $${ D }^{ 2+ }+{ 2e }^{ - }\longrightarrow D\ ;\quad { E }^{ o }=-1.12V$$

The largest potential will be generated in which cell?

A
$${ A }^{ - }|A\parallel { B }^{ - }|{ B }^{ 2 }$$

B
$$D|{ D }^{ 2+ }\parallel A|{ A }^{ - }$$

C
$${ B }^{ 2 }|{ B }^{ - }\parallel { C }^{ + }|C$$

D
$$D|{ D }^{ 2+ }\parallel { C }^{ + }|C$$

Solution
Question 4
1 / -0

When $${{\text{K}}_{\text{2}}}{\text{C}}{{\text{r}}_{\text{2}}}{{\text{O}}_{\text{7}}}$$ is converted into $${{\text{K}}_{\text{2}}}{\text{Cr}}{{\text{O}}_4}$$, the change in oxidation number of Cr is

A
0

B
6

C
4

D
3

Solution

For $${{\text{K}}_{\text{2}}}{\text{C}}{{\text{r}}_{\text{2}}}{{\text{O}}_{\text{7}}}$$
we consider oxidation no. of Cr as x , we will have
2(1) + 2(x) + 7 × (-2) = 0.….(as oxidation no. of alkali metals is 1 and of oxygen is -2 .Also we know charge on overall atom is 0)
Hence we get x = 6
Oxidation no. Of Cr is 6
for $${{\text{K}}_2}{\text{C}}{{\text{r}}_{\text{4}}}$$ similarly oxidation no. is 6
Hence change is 0
(A) is correct Option.
Question 5
1 / -0

Given below are the half-cell reactions:
$$Mn^{2+} + 2e^- \to Mn; E^o = -1.18V$$
$$2(Mn^{3+} + e^- \to Mn^{2+}); E^o = +1.51V$$
The $$E^o$$ for $$3Mn^{2+} \to Mn +2Mn^{3+}$$ will be:

A
$$-0.33V$$; the reaction will not occur

B
$$-0.33V$$; the reaction will occur

C
$$-2.69V$$; the reaction will not occur

D
$$-2.69V$$; the reaction will occur

Solution

Standard electrode potential of reaction will not change due to multiply the half-cell reactions with some numbers,
To get the main eq we have to reverse $$2nd$$ equation and add them
So $$E_3 = E_2+E_1$$
$$E_3 = -1.18+(-1.51)$$
$$E_3 = -2.69V$$
The reaction is not possible as the $$\Delta G$$ will come +ve for this case and that indicates reaction is non-spontaneous.
Question 6
1 / -0

In the electrolysis of aqueous sodium chloride solution which of the half-cell reaction will occur at the anode?

A
$$Na^+_{(aq)}+e^-\rightarrow Na_{(s)}; E^o_{cell}=-2.71$$V

B
$$2H_2O_{(l)}\rightarrow O_{2(g)}+4H^+_{(aq)}+4e^-; E^o_{cell}=1.23$$V

C
$$H^+_{aq}+e^-\rightarrow \dfrac{1}{2}H_{2(g)}; E^o_{cell}=0.00$$V

D
$$Cl^-_{(aq)}\rightarrow \dfrac{1}{2}Cl_{2(g)}+e^-; E^o_{cell}=1.36$$V

Solution

At anode, oxidation takes place.
According to preferential discharge theory, more is the $$\varepsilon_{cell}$$ of anode potential more is the tendency for reaction to take place.
hence, $$Cl_2$$ gas is evolved.
$$Cl^{\circleddash}\longrightarrow \cfrac{1}{2}Cl_2+e^-\quad\quad \varepsilon^o_{cell}=1.36$$ $$V$$
Question 7
1 / -0

$$KMnO_4$$ acts as an oxidizing agent in acidic medium. The number of moles of $$KMnO_4$$ that will be needed to react with one mole of sulphide ions in acidic solution is:

A
$$\dfrac{2}{5}$$

B
$$\dfrac{3}{5}$$

C
$$\dfrac{4}{5}$$

D
$$\dfrac{1}{5}$$

Solution

$$5\overline { e } +Mn{ O }_{ 4 }^{ -2 }\longrightarrow M{ n }^{ +2 }\times 2\\ +\quad \quad \quad { 5 }^{ -2 }+2\overline { e } \longrightarrow S\times 2\\ \_ \_ \_ \_ \_ \_ \_ \_ \_ \_ \_ \_ \_ \_ \_ \_ \\ 2Mn{ O }_{ 4 }^{ - }+5{ S }^{ -2 }\longrightarrow 2Mn^{ +2 }+5S$$
Divide reaction by $$5.$$
$$\cfrac { 2 }{ 5 } Mn{ O }_{ 4 }^{ - }+{ 5 }^{ -2 }\longrightarrow 2Mn^{ +2 }+5S$$
Hence $$1$$ mole of sulphide ions will require $$\cfrac { 2 }{ 5 } $$ moles of $$Mn{ O }_{ 4 }^{ - }$$ ions.
Question 8
1 / -0

The oxidation number and coordination number of $$M$$ in the compound, $$[M(SO_4)(NH_3)_5]$$ will be:

A
10 and 3

B
1 and 6

C
6 and 4

D
2 and 6

Solution

Since $$M$$ has 6 ligands attached to it, the co-ordination no. of the compound is 6. Here, the anionic species inside the co-ordination sphere is $$(SO_{4})^{2-}$$. Since the compound as a whole is neutral, the oxidation no. of $$M$$ should be 2.

Hence, the correct option is $$D$$
Question 9
1 / -0

A metal, M forms chlorides in its +2 and +4 oxidation states. Which of the following statements about these chlorides is correct?

A
$$MCl_2$$ is less ionic than $$MCl_4$$

B
$$MCl_2$$ is less easily hydrolysed than $$MCl_4$$

C
$$MCl_2$$ is more volatile than $$MCl_4$$

D
$$MCl_2$$ is more soluble in anhydrous ethanol than $$MCl_4$$
Question 10
1 / -0

The magnetic moment of $$M^{x+}$$ (atomic number $$= 25$$) is $$\sqrt{15}$$ . Then, the oxidation number $$x$$ of $$M$$ is:

A
4

B
3

C
2

D
none of these

Solution

$$\sqrt { n\left( n+2 \right) } =\sqrt { 15 } $$
Squaring both sides
$$\Rightarrow n\left( n+2 \right) =15$$
$$\Rightarrow { n }^{ v }+2n-15=0$$
$$\Rightarrow { n }^{ v }+5n-3n-15=0$$
$$\Rightarrow n\left( n+5 \right) -3\left( n+5 \right) =0$$
$$\Rightarrow \left( n+5 \right) \left( n-3 \right) =0$$
Solving we get,
$$n=3$$
Thus there are $$3$$ unpaired electrons in the element.
The configuration of an element with atomic number $$2S$$ is $$\rightarrow$$ $${ 1S }^{ 2 }{ 2S }^{ 2 }{ 2P }^{ 6 }{ 3S }^{ 2 }{ 3P }^{ 6 }{ 4S }^{ 2 }{ 3d }^{ 5 }$$. This configuration has electrons in the valence shell.
We have been given that the symbol of the element is $${ M }^{ x+ }$$.
So, oxidation number of $$M=5-3$$
$$=2$$

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Re-Attempt Weekly Quiz Competition

Redox Reactions Test - 69

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Redox Reactions Test - 69

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SHARING IS CARING

Question 1

3, 3, 3, 2, 3

3, 2, 3, 2, 4

3, 8, 3, 2, 4

2, 3, 3, 3, 2

Solution

Question 2

$$4, 1, 7$$

$$7, 4, 1$$

$$4, 1, 10$$

$$1, 4, 10$$

Solution

Question 3

$${ A }^{ - }|A\parallel { B }^{ - }|{ B }^{ 2 }$$

$$D|{ D }^{ 2+ }\parallel A|{ A }^{ - }$$

$${ B }^{ 2 }|{ B }^{ - }\parallel { C }^{ + }|C$$

$$D|{ D }^{ 2+ }\parallel { C }^{ + }|C$$

Solution

Question 4

0

6

4

3

Solution

Question 5

$$-0.33V$$; the reaction will not occur

$$-0.33V$$; the reaction will occur

$$-2.69V$$; the reaction will not occur

$$-2.69V$$; the reaction will occur

Solution

Question 6

$$Na^+_{(aq)}+e^-\rightarrow Na_{(s)}; E^o_{cell}=-2.71$$V

$$2H_2O_{(l)}\rightarrow O_{2(g)}+4H^+_{(aq)}+4e^-; E^o_{cell}=1.23$$V

$$H^+_{aq}+e^-\rightarrow \dfrac{1}{2}H_{2(g)}; E^o_{cell}=0.00$$V

$$Cl^-_{(aq)}\rightarrow \dfrac{1}{2}Cl_{2(g)}+e^-; E^o_{cell}=1.36$$V

Solution

Question 7

$$\dfrac{2}{5}$$

$$\dfrac{3}{5}$$

$$\dfrac{4}{5}$$

$$\dfrac{1}{5}$$

Solution

Question 8

10 and 3

1 and 6

6 and 4

2 and 6

Solution

Question 9

$$MCl_2$$ is less ionic than $$MCl_4$$

$$MCl_2$$ is less easily hydrolysed than $$MCl_4$$

$$MCl_2$$ is more volatile than $$MCl_4$$

$$MCl_2$$ is more soluble in anhydrous ethanol than $$MCl_4$$

Question 10

4

3

2

none of these

Solution

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