Electrochemistry Test

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Electrochemistry Test - 51

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Weekly Quiz Competition

Question 1
1 / -0

When molten lithium chloride $$(LiCl)$$ is electrolyzed, lithium metal is formed at the cathode. If current efficiency is $$75\%$$ then how many grams of lithium are liberated when $$1930\ C$$ of charge pass through the cell? (Atomic weight : $$Li=7$$)

A
$$0.105$$

B
$$0.120$$

C
$$0.28$$

D
$$0.240$$

Solution

Weight of metal deposited $$= \cfrac {ECt}{96500}$$; Applied charge= $$1930C$$
Efficiency= $$75$$%
Effective charge= $$\cfrac {75}{100}(1930)=1447.5C$$
$$96500C$$ liberate $$1$$ mole.
$$1447.5C$$ liberates $$\cfrac {1447.5}{96500}$$ moles $$=0.105$$ moles .
Question 2
1 / -0

How many Faraday are required to reduce one mol of $${ MnO }_{ 4 }^{ - }$$ to $${ Mn }^{ 2+ }$$?

A
1

B
2

C
3

D
5

Solution

$$Mn{O^{ - 4}} = M{n^{ + 2}} + 5{e^ - }$$
$$Mn$$ in $$Mn{O_4{^{ -}}}$$ has an oxidation state of +7.
Going from +7 to +2 requires an additional 5 electrons
1 mol of $$e=1\ F$$
5 mol of $$e = 5\ F$$
Question 3
1 / -0

A schematic representation of enthalpy changes for the reaction, $${ C }_{ graphite }$$ + $$\frac { 1 }{ 2 } { O }_{ 2 }$$ (g) $$\rightarrow $$ CO (g) is given below. The missing value is

A
+ 10.5 kJ

B
- 11.05 kJ

C
- 110.5 kJ

D
- 10.5 kJ

Solution

Since enthalpy is a state function therefore it does not depend on the path . Thats y the enthalpy of direct method and indirect method of formation of product ( i . e here CO2 ) will be same so thats y.....

-393.5 = X + (-283 KJ)

X = -110.5 KJ

Hence, the correct option is $$\text{C}$$
Question 4
1 / -0

Amount of charge is required to convert 17 gm $$H_2O_2$$ into $$O_2$$:-

A
1 F

B
2 F

C
6 F

D
None of these

Solution

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

mole of $$ H_{2}O_{2} = \dfrac{17}{34} = 0.5\ mol $$

For 0.5 mole, amount of change required
$$= 0.5\times 2\times 6.023\times 10^{23}\times 1.6\times 10^{-19} C$$
$$ = 6.023\times 10^{4}\times 1.6 C$$
$$ = 9.6\times 10^{4}C $$
$$ \Rightarrow 96500C = 1F $$
Question 5
1 / -0

Identify the correct half cell reaction of the following:
$$Zn + CuSO_4 \rightarrow ZnSO_4 + Cu$$

A
$$SO_4^{2-} \rightarrow SO_4 + 2e^-$$

B
$$Zn^{2+}+2e^-\rightarrow Zn$$

C
$$Cu \rightarrow Cu^{2+} +2e^-$$

D
$$Cu^{2+} +2e^- \rightarrow Cu$$
Question 6
1 / -0

If a pieces of iron gains $$10\%$$ of its mass due to partial rusting into $$Fe_{2}O_{3}$$ the percentage of total iron that has rusted is

A
$$3$$

B
$$13$$

C
$$23.3$$

D
$$25.67$$

Solution
Question 7
1 / -0

Electrolysis can be used to determine atomic masses. A current of $$0.550$$ A deposits $$0.55 $$ g of a certain metal in $$100$$ minutes. Calculate the atomic mass of the metal is $$ n = 3$$ :

A
$$100$$

B
$$45.0$$

C
$$48.25$$

D
$$144.75$$

Solution

$$I=0.55$$A
$$W=0.55$$g
$$t=100$$min$$=6000$$sec
$$n=3$$
We know that,
According to Faraday's $$1$$st law,
$$W=Z\times I\times t$$
$$\Rightarrow 0.55=\dfrac{Atomic mass}{n\times 96500}\times 0.55\times 6000$$
$$\Rightarrow 1=\dfrac{At. mass}{3\times 965}\times 60$$
$$\Rightarrow$$ At mass$$=104.2$$u
$$\cong 100u$$.
Question 8
1 / -0

A magnet is pulled away from a conducting coil in a circuit. The direction of the induced emf in the coil can be found from...........

A
Faraday's law of electromagnetic induction

B
Joule's law

C
Lenz's law

D
Fleming's left hand rule

Solution

A magnet is pulled away from a conducting coil in a circuit. The direction of the induced emf in the coil can be found from Faraday's law of electromagnetic induction.
Question 9
1 / -0

On the basis of the electrochemical theory of aqueous corrosion, the reaction occurring at the cathode is :

A
$$O_{2(g)}+4H_{(aq)}^{+}+4e^{-}\rightarrow 2H_{2}O_{(l)}$$

B
$$Fe_{(s)}\rightarrow Fe_{(aq)}^{2+}+2e^{-}$$

C
$$Fe_{(aq)}^{2+}\rightarrow Fe_{(aq)}^{3+}+e^{-}$$

D
$$H_{2(g)}+2OH_{(aq)}^{-}\rightarrow 2H_{2}O_{(l)}+2e^{-}$$

Solution

Answer is Option $$A$$
At cathode : $$2{{H}^{+}}_{(aq)}+2{{e}^{-}}\to 2H$$
$$2H+\frac{1}{2}{{O}_{2}}\to {{H}_{2}}O$$
$$\overline{2{{H}^{+}}+\frac{1}{2}{{O}_{2}}+2{{e}^{-}}\to {{H}_{2}}O}$$
Question 10
1 / -0

Three faradays of electricity are passed through molten $$Al_2O_3$$, aqueous solution of $$CuSO_4$$ and molten $$NaCl$$ taken in different electrolytic cells. The amount of $$Al$$ $$Cu$$ and $$Na$$ deposited at the cathodes will be in the ratio of?

A
$$1$$ mole $$: 2$$ mole $$: 3$$ mole

B
$$3$$ mole $$: 2$$ mole $$: 1$$ mole

C
$$1$$ mole $$: 1.5$$ mole $$: 3$$ mole

D
$$1.5$$ mole $$: 2$$ mole $$: 3$$ mole

Solution

In $$Al_{2}O_{3}$$, $$Al$$ has charge $$3+$$
$$ Al^{3+} + 3e^{-} \rightarrow Al $$
Thus, 3 Faraday are required for $$1$$ mole. So, $$3F$$ will produce 1 mole of $$Al_{2}O_{3}$$
In $$CuSO_{4}$$, $$Cu$$ has charge $$2+$$
$$ Cu^{2+} + 2e^{-} \rightarrow Cu $$
Thus, $$2$$ Faraday are required for $$1$$ mole. So, $$3F$$ will produce $$1.5$$ moles of $$CuSO_{4}$$
In $$NaCl$$, $$Na$$ has charge $$1+$$
$$ Na^{+} + e^{-} \rightarrow Na $$
Thus, $$1$$ Faraday are required for $$1$$ mole, So, $$3F$$ will produce $$3$$ mole of $$NaCl$$
Hence, ratio will be $$1$$ mole : $$1.5$$ mole : $$3$$ mole

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Electrochemistry Test - 51

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Electrochemistry Test - 51

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Question 1

$$0.105$$

$$0.120$$

$$0.28$$

$$0.240$$

Solution

Question 2

1

2

3

5

Solution

Question 3

+ 10.5 kJ

- 11.05 kJ

- 110.5 kJ

- 10.5 kJ

Solution

Question 4

1 F

2 F

6 F

None of these

Solution

Question 5

$$SO_4^{2-} \rightarrow SO_4 + 2e^-$$

$$Zn^{2+}+2e^-\rightarrow Zn$$

$$Cu \rightarrow Cu^{2+} +2e^-$$

$$Cu^{2+} +2e^- \rightarrow Cu$$

Question 6

$$3$$

$$13$$

$$23.3$$

$$25.67$$

Solution

Question 7

$$100$$

$$45.0$$

$$48.25$$

$$144.75$$

Solution

Question 8

Faraday's law of electromagnetic induction

Joule's law

Lenz's law

Fleming's left hand rule

Solution

Question 9

$$O_{2(g)}+4H_{(aq)}^{+}+4e^{-}\rightarrow 2H_{2}O_{(l)}$$

$$Fe_{(s)}\rightarrow Fe_{(aq)}^{2+}+2e^{-}$$

$$Fe_{(aq)}^{2+}\rightarrow Fe_{(aq)}^{3+}+e^{-}$$

$$H_{2(g)}+2OH_{(aq)}^{-}\rightarrow 2H_{2}O_{(l)}+2e^{-}$$

Solution

Question 10

$$1$$ mole $$: 2$$ mole $$: 3$$ mole

$$3$$ mole $$: 2$$ mole $$: 1$$ mole

$$1$$ mole $$: 1.5$$ mole $$: 3$$ mole

$$1.5$$ mole $$: 2$$ mole $$: 3$$ mole

Solution

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