Redox Reactions Test

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Redox Reactions Test - 47

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Weekly Quiz Competition

Question 1
1 / -0

For the redox reaction, $$Zn+Cu^{2+}_{(0.1M)}\rightarrow Zn^{2+}_{(1M)}+Cu$$ taking place in a cell $$E^{\ominus}_{cell}=1.10$$V. $$E^{\ominus}_{Cell}$$ for the galvanic cell will be?

A
$$2.14$$V

B
$$1.80$$V

C
$$1.07$$V

D
$$0.82$$V

Solution

BY Nernst's equation, $$_{c \in 11}=E_{\text {cell }}^{0}-\dfrac{0.0591}{n} \log \theta_{1}$$ (at $$5^{\circ} \mathrm{C}$$

Here, $$theta=\dfrac{\left[z n^{+2}\right]}{\left[\left(\mu^{+2}\right]\right.}=\dfrac{1}{0 \cdot 1}=10$$

So, $$_{\text {cell }}=1.10-\dfrac{0.0591}{2} \log 10 .=1.10-0.0295$$
$$=1.0705 \mathrm{~V}$$
Thus, option (C) is correct.
Question 2
1 / -0

In the reaction $$4Fe+3O_2\rightarrow 4Fe^{3+}+6O^{2-}$$ which of the following statements is incorrect?

A
It is a redox reaction

B
Metallic iron is a reducing agent

C
$$Fe^{3+}$$ is an oxidising agent

D
Metallic iron is reduced to $$Fe^{3+}$$

Solution

$$ \text { Given, } 4 \mathrm{Fe}+3 \mathrm{O}_{2} \longrightarrow 4 \mathrm{Fe}^{+3}+6 \mathrm{O}^{2} \text { . }$$

clearly, given is a redox reaction in Which, Fe gets oxidised to $$\mathrm{Fe}^{+3}$$, ithive $$0_{2}$$ gets reduced to $$0^{2-}$$.

Thus motallic iron (Fe) is a reducing agent and $$\mathrm{Fe}^{+3}$$ is an oxidising agent.
So, option (D) is correct.
Question 3
1 / -0

When zinc reacts with very dilute $$HNO_3$$, the oxidation state of nitrogen changes from:

A
$$+5$$ to $$+1$$

B
$$+5$$ to $$-3$$

C
$$+5$$ to $$+4$$

D
$$+5$$ to $$+3$$

Solution

Generally, when zinc reacts with nitric acid, it forms zinc nitrate. But when we take very dilute solution of nitric acid, if forms ammonium nitrate as well.

So, here Nitrogen is reducing from $$+5$$ to $$-3$$.

$$4Zn(s)+\overset {+5}{10HNO_3}(aq)\longrightarrow 4Zn(NO_3)_2(aq)+\overset {-3}{N}H_4NO_3(aq)+3H_2O(l)$$

Option B is correct.
Question 4
1 / -0

Given:
$$X Na_2 HAsO_3+Y NaBrO_3+ZHCl\rightarrow NaBr+H_3AsO_4+NaCl$$
The values of $$X, \ Y$$ and $$Z$$ in the given redox reaction are respectively.

A
$$2, 1, 2$$

B
$$2, 1, 3$$

C
$$3, 1, 6$$

D
$$3, 1, 4$$

Solution

The balanced chemical equation can be represented as:

$$3Na_2HAsO_3+NaBrO_3+ 6HCl \rightarrow NaBr+3H_3AsO_4+6NaCl$$

Clearly, $$X=3$$, $$Y=1$$, and $$Z=6$$
Question 5
1 / -0

Calculate the equilibrium constant for the redox reaction at $$25^o$$C, $$Sr_{(s)}+Mg^{2+}_{(aq)}\rightarrow Sr^{2+}_{(aq)}+Mg_{(s)}$$, that occurs in a galvanic cell. $$E^o_{{Mg^{2+}}/{Mg}}=-2.37$$V and $$E^o_{{Sr}^2+/Sr}=-2.89$$V.

A
$$2.69\times 10^{15}$$

B
$$2.69\times 10^{17}$$

C
$$3.69\times 10^{17}$$

D
$$3.69\times 10^{15}$$

Solution

(i) $${Mg^{2+}}_{(aq)}+2e^- \longrightarrow Mg_{(s)}; E^o_{Mg^{2+}/Mg}=-2.37V$$
$${Sr^{2+}}_{(aq)}+2e^- \longrightarrow Sr_{(s)}; E^o_{Sr^{2+}/Sr}=-2.89V$$
(ii) $$Sr_{(s)} \longrightarrow S{r^{2+}}_{(aq)}+2e^-;E^o_{Sr/Sr^{2+}}=2.89V$$

Adding (i) & (ii), we get,
$${Sr}_{(s)}+Mg^{2+}_{(aq)} \longrightarrow Sr^{2+}_{(aq)}+Mg_{(s)}; E^o= -2.37+2.89=0.52V$$
Now, $$\Delta G^o=-RTlnK_{eq}$$
$$\therefore -nFE^o= -RT lnK_{eq}$$
$$\therefore ln K_{eq}=\cfrac {nFE^o}{RT}=\cfrac {2 \times 96500 \times 0.52}{8.314 \times 298}$$
$$\therefore lnK_{eq}=40.507$$
$$\therefore K_{eq}= 3.69 \times 10^{17}$$
Question 6
1 / -0

Two oxidation states for chlorine are found in the compound:

A
$${ CaOCl }_{ 2 }$$

B
$$KCL$$

C
$${ KClO }_{ 3 }$$

D
$${ Cl }_{ 2 }{ O }_{ 7 }$$

Solution

Solution:- (A) $$CaO{Cl}_{2}$$
$$CaO{Cl}_{2}$$ consist of hypochlorite $$\left( {OCl}^{-} \right)$$ ion and chloride ion $$\left( {Cl}^{-} \right)$$
The oxidation number of oxygen in hypochlorite ion is -2.
As we know that the sum of the oxidation states of all the atoms in an ion is equal to the charge on the ion.
Let the oxidation number of chlorine in hypochlorite ion be x.
$$\therefore$$ Oxidation no. of chlorine in $${OCl}^{-}$$ ion-
$$x + \left( -2 \right) = -1$$
$$\Rightarrow \; x = -1 + 2 = +1$$
Therefore, oxidation number of chlorine in $${OCl}^{-}$$ is +1.
Oxidation number of chlorine in $${Cl}^{-}$$ is -1.
Hence in $$CaO{Cl}_{2}$$, the two $$Cl$$ atoms are in different oxidation states i.e., one as $${Cl}^{–}$$ having oxidation number of $$-1$$ and the other as $${OCl}^{–}$$ having oxidation number of $$+1$$.
Question 7
1 / -0

Oxidation state of cobalt in $$[{ Co }(NH_{ 3 }{ ) }_{ 4 } ({ H }_{ 2 }O)Cl]{ SO }_{ 4 }$$ ?

A
$$0$$

B
$$+4$$

C
$$-2$$

D
$$+3$$

Solution

$$\left[ Co {\left( N{H}_{3} \right)}_{4} \left( {H}_{2}O \right) Cl \right] S{O}_{4}$$
In the above compound,
Oxidation state of ammonia, $$\left( N{H}_{3} \right) = 0$$
Oxidation state of water, $$\left( {H}_{2}O \right) = 0$$
Oidation state of chlorine, $$\left( Cl \right) = -1$$
Oxidation state of sulphate ion, $$\left( S{O}_{4} \right) = -2$$
Let the oxidation state of cobalt, $$\left( Co \right)$$ be x.
As we know that sum of oxidation states of all atoms is equal to the overall charge on the compound.
$$\therefore x + \left( 4 \times 0 \right) + \left( 1 \times 0 \right) + \left( -1 \right) + \left( -2 \right) = 0$$
$$\Rightarrow \; x + 0 + 0 -1 -2 = 0$$
$$\Rightarrow \; x - 3 = 0$$
$$\Rightarrow \; x = +3$$
Hence, the oxidation state of cobalt, $$\left( Co \right)$$ in the compound $$\left[ Co {\left( N{H}_{3} \right)}_{4} \left( {H}_{2}O \right) Cl \right] S{O}_{4}$$ is +3.
Question 8
1 / -0

Oxidation number of $$Fe$$ in $${ Fe }_{ 3 }{ O }_{ 4 }$$ are:

A
$$+2$$ and $$+3$$

B
$$+1$$ and $$+2$$

C
$$+1$$ and $$+3$$

D
None

Solution

$${Fe}_{3}{O}_{4}$$ contains $$Fe$$ atoms of both $$+2$$ and $$+3$$ oxidation number. It is a stoichiometric mixture of Ferrous $$\left( FeO \right)$$ and Ferric $$\left( {Fe}_{2}{O}_{3} \right)$$ oxides combined as $$FeO.{Fe}_{2}{O}_{3}$$.
Oxidation state of oxygen in $$FeO$$ and $${Fe}_{2}{O}_{3}$$ is -2.
Let the oxidation state of Fe in $$FeO$$ and $${Fe}_{2}{O}_{3}$$ be x and y respectively.
As we know that the sum of the oxidation states of all the atoms or ions in a neutral compound is zero.
Therefore,
Oxidation state of $$Fe$$ in $$FeO$$-
$$x + \left( -2 \right) = 0$$
$$\Rightarrow \; x = 2$$
Oxidation state of $$Fe$$ in $${Fe}_{2}{O}_{3}$$-
$$2 \left( y \right) + 3 \left( -2 \right) = 0$$
$$\Rightarrow \; 2y = 6$$
$$\Rightarrow \; y = 3$$
Hence the oxidation number of $$Fe$$ in $${Fe}_{3}{O}_{4}$$ are +2 and +3 respectively.
Question 9
1 / -0

In which of the following pair oxidation number of $$Fe$$ is same:

A
$${ K }_{ 3 }{ Fe(CN) }_{ 6 },{ Fe }_{ 2 }{ O }_{ 3 }$$

B
$${ Fe(CN) }_{ 5 },{ Fe }_{ 2 }{ O }_{ 3 }$$

C
$${ Fe }_{ 2 }{ O }_{ 3, }FeO$$

D
$${ Fe }_{ 2 }({ SO }_{ 4 })_{ 3 },{ K }_{ 4 }Fe({ C) }_{ 6 }$$

Solution

Let the oxidation state of Fe in $${K}_{3}Fe{\left( CN \right)}_{6}$$ and $${Fe}_{2}{O}_{3}$$ be x and y respectively.
As we know that the sum of the oxidation states of all the atoms or ions in a neutral compound is zero.
Therefore,
Oxidation state of $$Fe$$ in $${K}_{3}Fe{\left( CN \right)}_{6}$$-
$$\because \; CN$$ is a negative ligand.
$$\therefore$$ Oxidation number of $$CN$$ is -1.
Oxidation number of $$K$$ is +1.
$$\therefore \; 3 \left( +1 \right) + x + 6 \left( -1 \right) = 0$$
$$\Rightarrow \; x - 3 = 0$$
$$\Rightarrow \; x = 3$$
Oxidation state of $$Fe$$ in $${Fe}_{2}{O}_{3}$$-
Oxidation no. of $$O$$ is -2.
$$2 \left( y \right) + 3 \left( -2 \right) = 0$$
$$\Rightarrow \; 2y = 6$$
$$\Rightarrow \; y = 3$$
Hence the oxidation number of $$Fe$$ in $${K}_{3}Fe{\left( CN \right)}_{6}$$ and $${Fe}_{2}{O}_{3}$$ is same.
Question 10
1 / -0

The oxidation number of C in $${ CH }_{ 4 }, { CH }_{ 3 }Cl, { CH }_{ 2 }Cl_{ 2 }, { CH }Cl_{ 3 }$$ and $$CCl_{ 4 }$$ are respectively :

A
$$+4, +2, 0, -2, -4$$

B
$$+2, +4, 0, -4, -2$$

C
$$-4, -2, 0, +2, +4$$

D
$$-2, -4, 0, +4, +2$$

Solution

Electronegativity order in carbon, hydrogen, and chlorine is-
$$Cl > C > H$$
Hence, oxidation state of hydrogen and chlorine in all the given compounds will be +1 and -1 respectively.

In $$C{H}_{4}$$-
Let the oxidation state of carbon in $$C{H}_{4}$$ be x.
$$\therefore \; x + \left(4 \times \left( +1 \right) \right) = 0$$
$$\Rightarrow \; x = -4$$

In $$C{H}_{3}Cl$$-
Let the oxidation state of carbon in $$C{H}_{3}Cl$$ be x.
$$\therefore \; x + \left(3 \times \left( +1 \right) \right) + \left( -1 \right) = 0$$
$$\Rightarrow \; x + 3 - 1 = 0$$
$$\Rightarrow \; x = -2$$

In $$C{H}_{2}{Cl}_{2}$$-
Let the oxidation state of carbon in $$C{H}_{3}{Cl}_{2}$$ be x.
$$\therefore \; x + \left(2 \times \left( + 1 \right) \right) + \left( 2 \times \left( -1 \right) \right) = 0$$
$$\Rightarrow \; x + 2 - 2 = 0$$
$$\Rightarrow \; x = 0$$

In $$CH{Cl}_{3}$$-
Let the oxidation state of carbon in $$CH{Cl}_{3}$$ be x.
$$\therefore \; x + 1 + \left(3 \times \left( -1 \right) \right) = 0$$
$$\Rightarrow \; x + 1 - 3 = 0$$
$$\Rightarrow \; x = +2$$

In $$C{Cl}_{4}$$-
Let the oxidation state of carbon in $$C{Cl}_{4}$$ be x.
$$\therefore \; x + \left(4 \times \left( -1 \right) \right) = 0$$
$$\Rightarrow \; x = +4$$

Hence, the correct option is $$C$$.

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Redox Reactions Test - 47

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Redox Reactions Test - 47

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Question 1

$$2.14$$V

$$1.80$$V

$$1.07$$V

$$0.82$$V

Solution

Question 2

It is a redox reaction

Metallic iron is a reducing agent

$$Fe^{3+}$$ is an oxidising agent

Metallic iron is reduced to $$Fe^{3+}$$

Solution

Question 3

$$+5$$ to $$+1$$

$$+5$$ to $$-3$$

$$+5$$ to $$+4$$

$$+5$$ to $$+3$$

Solution

Question 4

$$2, 1, 2$$

$$2, 1, 3$$

$$3, 1, 6$$

$$3, 1, 4$$

Solution

Question 5

$$2.69\times 10^{15}$$

$$2.69\times 10^{17}$$

$$3.69\times 10^{17}$$

$$3.69\times 10^{15}$$

Solution

Question 6

$${ CaOCl }_{ 2 }$$

$$KCL$$

$${ KClO }_{ 3 }$$

$${ Cl }_{ 2 }{ O }_{ 7 }$$

Solution

Question 7

$$0$$

$$+4$$

$$-2$$

$$+3$$

Solution

Question 8

$$+2$$ and $$+3$$

$$+1$$ and $$+2$$

$$+1$$ and $$+3$$

None

Solution

Question 9

$${ K }_{ 3 }{ Fe(CN) }_{ 6 },{ Fe }_{ 2 }{ O }_{ 3 }$$

$${ Fe(CN) }_{ 5 },{ Fe }_{ 2 }{ O }_{ 3 }$$

$${ Fe }_{ 2 }{ O }_{ 3, }FeO$$

$${ Fe }_{ 2 }({ SO }_{ 4 })_{ 3 },{ K }_{ 4 }Fe({ C) }_{ 6 }$$

Solution

Question 10

$$+4, +2, 0, -2, -4$$

$$+2, +4, 0, -4, -2$$

$$-4, -2, 0, +2, +4$$

$$-2, -4, 0, +4, +2$$

Solution

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