Electrochemistry Test

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Electrochemistry Test - 46

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Weekly Quiz Competition

Question 1
1 / -0

For an electrochemical reaction occuring in a galvanic cell.
$$2{ Fe }_{ \left( aq \right) }^{ +3 }+{ Zn }_{ \left( s \right) }\rightarrow { Zn }_{ \left( aq \right) }^{ +2 }+2{ Fe }_{ \left( aq \right) }^{ +2 }$$
if concentration of $${ Fe }^{ +2 }$$ is increased.

A
$${E}_{cell}$$ will remain unchanged

B
$${E}_{cell}$$ will decrease

C
$$pH$$ of the solution will change

D
less non P-V work will be obtained

Solution

In given cell reaction we have,
at anode $$Zn \rightarrow Zn^{2+} + 2e^-$$
at cathode $$ 2Fe^{3+} + 2e^- \rightarrow 2Fe^{+2}$$

$$E_{cell} = E^0_{cell}$$ -$$\dfrac{0.0591}{2}$$ $$log$$$$\dfrac{Zn^{2+} (Fe^{2+})^2}{(Fe^{3+})^2}$$

so, on increasing the concentration of $$Fe^{2+}$$ions
$$E_{cell}$$ will decreases.
Question 2
1 / -0

The approximate time duration in hours to electroplate $$30\space g$$ of calcium from molten calcium chloride using a current of $$5 \space A$$ is (At. mass of $$Ca = 40g/mol$$)

A
80

B
10

C
16

D
8

Solution

$$ \begin{array}{l}\text{Equivalent weight of a metal is defined as }\\ \text { the mass of the metal that react with 1} \\ \text { mole of } \mathrm{H}^{+} \text {lons. } \\ \qquad 1 \mathrm{~F}=96500 \text { coulomb } \end{array}$$

$$ \text { Eq.wt }=\dfrac{\text { Molecular mass }}{\text { valency factor }} $$
$$ \begin{array}{l} =\dfrac{40}{2} = 20 \\\text{20g of Ca to 1F }\\ 1 g - \dfrac{1}{20}\\ 30g-\dfrac{1}{20} \times 30=1.5 \mathrm{~F} \end{array} $$

$$ \begin{array}{c} q=i t \\ 1.5 \times 96500=5 \times t \\ t=8.041 \\ \approx. 8sec \end{array} $$
Question 3
1 / -0

Consider the following equations for a cell reaction :
$$A + B \rightarrow C + D \,; E^o = x \,volt, \Delta G = \Delta G_1$$
$$2A +2 B \rightarrow 2C +2D \,; E^o = x \,volt, \Delta G = \Delta G_2$$

A
$$x = y, \Delta G_1 = \Delta G_2$$

B
$$x > y, \Delta G_1 > \Delta G_2$$

C
$$x = y, \Delta G_2 = 2\Delta G_1$$

D
$$x < y, \Delta G_2 = 2\Delta G_1$$

Solution

$$x=y, \Delta G_{2}=2\Delta G_{1}$$ because $$\Delta G$$ is an expensive property and eletrode potential is an intensive property i.e, electrote potential doesnt denpends upon the coefficient of the reactant and product in the equation of reaction
$$\because \Delta G=-n FE_{cell}$$ where $$n=no$$, of electrons used in the reaction which depends upon the coefficient of the reactant
Question 4
1 / -0

In which of the following conditions is the potential for the following half-cell reaction maximum?
$$2H^+ + 2e^{-} \longrightarrow H_2$$

A
$$1.0 \,M \,HCl$$

B
A solution having $$pH = 4$$

C
Pure water

D
$$1.0 \,M \,NaOH$$

Solution

In the given options,

(A). Concentration of $${ H }^{ + }$$ in 1.0 M HCl = $$1M$$

(B). Since, $$pH$$ of solution is 4.
So, concentration of $${ H }^{ + }$$ = $${ 10 }^{ -4 }M$$

(C). $$pH$$ of pure water = 7
So, concetration of $${ H }^{ + }$$ = $${ 10 }^{ -7 }M$$

(D). Concentration of $${ H }^{ + }$$ in 1.0 M $${ H }^{ + }$$ is much less than the all the above options.

Since, concentration of $${ H }^{ + }$$ is highest in option A. So, potential of option A is maximum .
So, correct answer is option A.
Question 5
1 / -0

Given that $$E^o_{Ag^+/Ag} = + 0.8V$$ and $$ E^o_{Ni^{+2}/Ni} = 0.25 V$$. Which of the following statement is true with regard to this?

A
$$Ag^+$$ is an oxidizing agent while $$Ni^{2+}$$ is a reducing agent

B
$$Ag^{2+}$$ can be reduced by silver metal

C
$$Ag^+$$ is a better oxidizing agent than $$Ni^{+2}$$ and $$Ni$$ is a better reducing agent that $$Ag$$

D
$$Ni^{2+}$$ is a better oxidizing agent than $$Ag^+$$ and $$Ag$$ is better reducing agent that $$Ni$$

Solution

In the given problem, $${ E }_{ { Ag }^{ + }/Ag }^{ 0 }\quad =\quad +0.8\quad V\\ { E }_{ { Ni }^{ 2+ }/Ni }^{ 0 }\quad =\quad +0.25\quad V$$
Here, $${ E }_{ { Ag }^{ + }/Ag }^{ 0 }>{ E }_{ { Ni }^{ 2+ }/Ni }^{ 0 }$$.
So, $${ Ag }^{ + }$$ can be easily reduced than $${ Ni }^{ 2+ }$$.
So, $${ Ag }^{ + }$$ acts as an oxidizing agent and $${ Ni }^{ 2+ }$$ acts as a reducing agent.
So, correct answer is option A.
Question 6
1 / -0

Which of the following statements about the spontaneous reaction occurring in a galvanic cell is always true?

A
$$E^o_{cell} > 0, \Delta G^o < 0$$, and $$Q < K$$

B
$$E^o_{cell} > 0, \Delta G^o < 0$$, and $$Q > K$$

C
$$E^o_{cell} > 0, \Delta G^o > 0$$, and $$Q > K$$

D
$$E_{cell} > 0, \Delta G^o < 0$$, and $$Q < K$$

Solution

For a reaction to be spontaneous we have to know that always $$\Delta G^0<0$$ always.
We know for a galvanic cell,
$$\Delta G=-nFE^0_{cell}$$
So for $$\Delta G^0<0$$ then $$E^{0}_{cell}>0$$.
We know that reaction will be spontaneous $$Q<K$$
And hence option D is correct answer.
Question 7
1 / -0

What is the amount of chlorine evolved, when 2A of current is passed for 30 minutes in an aqueous solution of NaCl?

A
1.32gm

B
4.56gm

C
9.81gm

D
12.6gm

Solution

Total Charge Supplied(Q) $$= I\times t$$
= 2 x 30 x60
= 3600 C
$$NaCl\rightarrow{Na}^{+} + {Cl}^{-}$$
96500 C liberate $${Cl}_{2}$$ = Eq.Wt of $${Cl}_{2}$$ = 35.5 g
3600 C liberates $${Cl}_2$$ = $$\dfrac{35.5}{96500} \times 3600$$ = 1.32 gm
Question 8
1 / -0

Which of the following does not evolve oxygen at anodes when electrolysis is carried?

A
dil, $${ H }_{ 2 }{ SO }_{ 4 }$$ with Pt electrode

B
Fused NaOH with Pt electrode

C
Acidic water with Pt electrode

D
dil, $${ H }_{ 2 }{ SO }_{ 4 }$$ with Cu electrode

Solution

In the electrolysis of dil. $$ H_2SO_4$$ using Cu electrodes, the reaction at anode occurs as:
Cu $$\rightarrow Cu^2$$$$^+$$ + $$e^-$$
The electrode is not inert an so participates in the reaction. In all other reaction, inert electrode i.e. Pt is used. Thus oxygen is evolved in these reactions.
Therefore D is the correct solution.
Question 9
1 / -0

Standard electrode potential data are useful for understanding the suitability of an oxidant in a redox titration. Some half cell reaction and their standard potential are given below:

A
$${ { MnO }_{ 4 } }^{ - }\left( aq. \right) +{ 8H }^{ + }\left( aq \right) +{ 5e }^{ - }\rightarrow { Mn }^{ 2 }\left( aq. \right) +{ 4H }_{ 2 }O\left( l \right) \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad { E }^{ 0 }=1.51V$$

B
$${ Cr }_{ 2 }{ { O }_{ 7 } }^{ 2 }\left( aq. \right) +{ 14H }^{ + }\left( aq. \right) +6e\rightarrow 2{ Cr }^{ 3 }\left( aq. \right) +{ 7H }_{ 2 }O\left( l \right) \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad { E }^{ 0 }=1.38V$$

C
$${ Fe }^{ 3+ }\left( aq. \right) +{ e }^{ - }\rightarrow { Fe }^{ 2+ }\left( aq. \right) \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad { E }^{ 0 }=0.77V$$

D
$${ C| }_{ 2 }\left( g \right) +{ 2e }^{ - }\rightarrow { 2C| }^{ - }\left( aq. \right) \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad { E }^{ 0 }=1.40V$$

Solution

$${ { MnO }_{ 4 } }^{ - }$$ will oxidise $$Cl^-$$ ion according to the equation:
$${ { MnO }_{ 4 } }^{ - } +16H^+ +10Cl^-$$ $$\rightarrow 2M_n^2$$$$^+$$ + $$8H_2O + 5Cl_2$$
The cell corresponding to this reaction is as follows:
Pt, $$Cl_2 (1 atm) | Cl^- ||$$ $${ { MnO }_{ 4 } }^{ - }$$, $$M_n^2$$$$^+$$, $$H^+ | Pt$$
$$E^0$$$$_c$$$$_e$$$$_l$$$$_l$$ being +ve, $$\Delta G^0$$ will be -ve and so the above reaction is feasible.
$${ { MnO }_{ 4 } }^{ - }$$ will not oxidise $$F_e$$$$^2$$$$^+$$ ion but also $$Cl^-$$ ion simultaneously.
Question 10
1 / -0

If $$9$$ gm $$H_2O$$ is electrolysed completely with the current of $$50\%$$ efficiency then?

A
$$96500$$ charge is required

B
$$2\times 96500$$ C charge is required

C
$$5.6$$ L of $$O_2$$ at STP will be formed

D
$$11.2$$ L of $$O_2$$ at STP will be formed

Solution

Number of moles of water= $$\cfrac {9}{18}=0.5$$
Balanced equation for electrolysis of water
$$2H_2O\longrightarrow 2H_2+O_2$$
$$\therefore 2$$ moles of $$H_2O$$ dissociates to produce $$1$$ mole of $$O_2$$ at STP.
$$\therefore 0.5$$ moles of $$H_2O$$ produces $$0.25$$ moles of $$O_2$$ at STP.
$$\therefore 0.25$$ moles of $$O_2$$ will occupy= $$0.25 \times 22.4=5.6 litres$$ .

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Electrochemistry Test - 46

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Electrochemistry Test - 46

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Question 1

$${E}_{cell}$$ will remain unchanged

$${E}_{cell}$$ will decrease

$$pH$$ of the solution will change

less non P-V work will be obtained

Solution

Question 2

80

10

16

8

Solution

Question 3

$$x = y, \Delta G_1 = \Delta G_2$$

$$x > y, \Delta G_1 > \Delta G_2$$

$$x = y, \Delta G_2 = 2\Delta G_1$$

$$x < y, \Delta G_2 = 2\Delta G_1$$

Solution

Question 4

$$1.0 \,M \,HCl$$

A solution having $$pH = 4$$

Pure water

$$1.0 \,M \,NaOH$$

Solution

Question 5

$$Ag^+$$ is an oxidizing agent while $$Ni^{2+}$$ is a reducing agent

$$Ag^{2+}$$ can be reduced by silver metal

$$Ag^+$$ is a better oxidizing agent than $$Ni^{+2}$$ and $$Ni$$ is a better reducing agent that $$Ag$$

$$Ni^{2+}$$ is a better oxidizing agent than $$Ag^+$$ and $$Ag$$ is better reducing agent that $$Ni$$

Solution

Question 6

$$E^o_{cell} > 0, \Delta G^o < 0$$, and $$Q < K$$

$$E^o_{cell} > 0, \Delta G^o < 0$$, and $$Q > K$$

$$E^o_{cell} > 0, \Delta G^o > 0$$, and $$Q > K$$

$$E_{cell} > 0, \Delta G^o < 0$$, and $$Q < K$$

Solution

Question 7

1.32gm

4.56gm

9.81gm

12.6gm

Solution

Question 8

dil, $${ H }_{ 2 }{ SO }_{ 4 }$$ with Pt electrode

Fused NaOH with Pt electrode

Acidic water with Pt electrode

dil, $${ H }_{ 2 }{ SO }_{ 4 }$$ with Cu electrode

Solution

Question 9

$${ { MnO }_{ 4 } }^{ - }\left( aq. \right) +{ 8H }^{ + }\left( aq \right) +{ 5e }^{ - }\rightarrow { Mn }^{ 2 }\left( aq. \right) +{ 4H }_{ 2 }O\left( l \right) \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad { E }^{ 0 }=1.51V$$

$${ Cr }_{ 2 }{ { O }_{ 7 } }^{ 2 }\left( aq. \right) +{ 14H }^{ + }\left( aq. \right) +6e\rightarrow 2{ Cr }^{ 3 }\left( aq. \right) +{ 7H }_{ 2 }O\left( l \right) \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad { E }^{ 0 }=1.38V$$

$${ Fe }^{ 3+ }\left( aq. \right) +{ e }^{ - }\rightarrow { Fe }^{ 2+ }\left( aq. \right) \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad { E }^{ 0 }=0.77V$$

$${ C| }_{ 2 }\left( g \right) +{ 2e }^{ - }\rightarrow { 2C| }^{ - }\left( aq. \right) \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad { E }^{ 0 }=1.40V$$

Solution

Question 10

$$96500$$ charge is required

$$2\times 96500$$ C charge is required

$$5.6$$ L of $$O_2$$ at STP will be formed

$$11.2$$ L of $$O_2$$ at STP will be formed

Solution

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