Electrochemistry Test

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Electrochemistry Test - 65

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Weekly Quiz Competition

Question 1
1 / -0

The number of coulombs required to reduce $$12.3\ g$$ of nitrobenxene to aniline is :

A
$$115000\ C$$

B
$$5700\ C$$

C
$$57900\ C$$

D
$$None\ of\ these$$

Solution

Reaction for reduction :

$${ C }_{ 6 }{ H }_{ 5 }{ NO }_{ 2 }+{ 6H }^{ + }+6{ e }^{ - }\longrightarrow { C }_{ 6 }{ H }_{ 5 }{ NH }_{ 2 }+2{ H }_{ 2 }O$$

Given, molar mass of $${ C }_{ 6 }{ H }_{ 5 }{ NO }_{ 2 }=123$$

Here, $$12.3\ g$$ nitrobenzene mole $$=\dfrac{6\times 12.3}{123}=0.6$$

$$\therefore $$ Charge $$=0.6\times 96500$$ coloumbs
$$=57900$$
Question 2
1 / -0

Alizarin belongs to the class of

A
Vat dyes

B
Mordant dye

C
Basic dye

D
Reactive dye

Solution
Question 3
1 / -0

The standard electrode potential for the following reaction is $$+1.33V$$. What is the potential at $$pH=2.0$$?
$${Cr}_{2}{O}_{7}^{2-}(aq. $$M$$)+14{H}^{+}(aq)+6{e}^{-}\rightarrow 2{Cr}^{3+} (aq, $$1M$$)+7{H}_{2}O(l)$$

A
$$+1.820V$$

B
$$+1.990V$$

C
$$+1.608V$$

D
$$+1.0542V$$

Solution

$$E_{Cr_{2}O_{7}^{2-}\, |\, Cr^{3+}} = E^{\circ}_{Cr_{2}O_{7}^{2-}\, |\, Cr^{3+}} - \dfrac{0.0591}{6} \times \left ( log \dfrac{[Cr^{3+}]^{2}}{[Cr_{2}O_{7}^{2-}]} \times \dfrac{1}{[H^{+}]^{14}} \right ) $$

$$E_{Cr_{2}O_{7}^{2-}\, |\, Cr^{3+}} = 1.33 - \dfrac{0.0591}{6} log \dfrac{1}{(0.01)^{14}} $$

$$ \Rightarrow $$ $$1.0542\,V$$ {Option D}
Question 4
1 / -0

Arrange the following metals in the order of their decreasing reactivity?
$$Fe, Cu, Mg, Ca, Zn, Ag$$

A
$$Ca> Zn> Mg> Cu> Ag. Fe$$

B
$$Ca> Zn>Cu> Mg> Ag> Fe$$

C
$$Ca> Mg> Zn> Fe> Cu> Ag$$

D
$$Ca> Mg> Cu> Zn> Fe> Ag$$

Solution
Question 5
1 / -0

On the basis of the information available from the reaction$$\dfrac { 4 }{ 3 } { Al }+O_{ 2 }\rightarrow\dfrac { 2 }{ 3 } { Al }_{ 2 }{ O }_{ 3 },$$$$\triangle G=-827KJ{ mol }^{ -1 }of{ O }_{ 2 }$$ The minimum e.m.f. required to carry out an electrolysis of$$\left( F=96500C{ mol }^{ -1 } \right)$$

A
$$2.14\ V$$

B
$$4.28\ V$$

C
$$6.42\ V$$

D
$$8.56\ V$$

Solution
Question 6
1 / -0

In the galvanic cell, $$Ca|{ Ca }^{ 2+ }(1M)\parallel { Ag }^{ + }(1M)|Ag$$, the electrons will travel in the external circuit:

A
From $$Ag$$ to $$Ca$$

B
From $$Ca$$ to $$Ag$$

C
Electrons do not travel in the external circuit

D
In any direction

Solution
Question 7
1 / -0

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

A
$$+0.33V$$; the reaction will occur

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

C
$$+2.69V$$; the reaction will occur

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

Solution

Standard potential of reaction $$[E^0]$$ is given by,
$$E_0^{cell}=E_R-E_P$$
$$Mn^{2+}+2e^-\longrightarrow Mn\quad E^0=-1.18V\\2Mn^{2+}\longrightarrow 2Mn^{3+}+2e^-\quad E^0=-1.51V\\\underline{\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad}$$
$$3Mn^{2+}\longrightarrow Mn+2Mn^{3+}$$ $$E^0=-1.18+(-1.51)\\=-2.69V$$
The reaction is non spontaneous, $$E^0=-2.69V$$
Question 8
1 / -0

In the cell, $$Zn|Zn^{2+}||Cu^{2+}|Cu,$$ the negative electrode is

A
$$Cu$$

B
$$Cu^{2+}$$

C
$$Zn$$

D
$$Zn^{2+}$$
Question 9
1 / -0

Given: $$E_{I_{2}/I^{-}}^{0} (1M) = +0.54V$$ and $$E_{Br_{2}/Br^{-}}^{0} (1M) = +1.09 V$$

On this basis, the feasible reaction is:

A
$$2Br^{-} + I_{2} \rightarrow Br_{2} +2I^{-}$$

B
$$I_{2} + Br_{2} \rightarrow 2I^{-} + 2Br^{-}$$

C
$$2I^{-} +Br_{2} \rightarrow I_{2} + 2Br^{-}$$

D
$$2I^{-} + 2Br^{-} \rightarrow I_{2} +Br_{2}$$

Solution
Question 10
1 / -0

How many coulombs of electricity are required for the oxidation of $$1$$ mole of $$\mathrm { H } _ { 2 } \mathrm { O }$$ to $$\mathrm { O } _ { 2 }$$ ?

A
$$9.65 \times 10 ^ { 4 } C$$

B
$$4.825 \times 10 ^ { 5 } \mathrm { C }$$

C
$$1.93 \times 10 ^ { 5 } C$$

D
$$1.93 \times 10 ^ { 4 } C$$

Solution