Electrochemistry Test

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Electrochemistry Test - 63

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Weekly Quiz Competition

Question 1
1 / -0

An electrochemical cell can behave like an electrolytic cell when:

A
$$E_{cell} = 0$$

B
$$E_{cell} > E_{ext}$$

C
$$E_{ext} > E_{cell}$$

D
$$E_{cell} = E_{ext}$$

Solution

We can study the effect of opposing potential applied to a galvanic cell as follows.
1) When $$E_{external}< 1.1V$$
In a galvanic cell, if an external opposite potential is applied and increased slowly, we find that the reaction continues to take place till the opposing voltage reaches the value 1.1 V.
2) When $$E_{external}= 1.1V$$
When the external voltage is 1.1V the reaction stops altogether and no current flows through the cell.
3) When $$E_{external}> 1.1V$$
Any further increase in the external potential again starts the reaction but in the opposite direction. It now functions as an electrolytic cell, a device for using electrical energy to carry non-spontaneous chemical reactions.
Question 2
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 3
1 / -0

The reaction, $$2B{r^ - }\left( {aq} \right) + S{n^{2 + }}\left( {aq} \right) \to B{r_2}\left( 1 \right) + Sn\left( s \right)$$ with the standard potentials,$$E{^\circ _{Sn}} = - 0.114V,\,E{^\circ _{B{r_2}}} = + 1.09V$$, is:

A
spontaneous in reverse direction

B
spontaneous in forward direction

C
at equilibrium

D
non-spontaneous in reverse direction

Solution

$$2Br^-+Sn^{2+}\longrightarrow Br_2+Sn(s)$$
$$E_{cell}=E^o_{Sn/Sn^{2+}}-E^o_{Br_2/Br^-}=-0.114-0.109=-0.223$$ $$V$$
Hence, the reaction will be spontaneous in reverse direction.
$$Br+Sn\longrightarrow Sn^{+2}+2Br^+$$
$$E_{cell}=+0.223$$ $$V$$
Question 4
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 5
1 / -0

Which of the following relation is not correct?

A
Equivalent conductivity - S $$c{m}e{q^{ - 1}}$$

B
Conductance -ohm$$^{ - 1}$$

C
Specific conductance - $$S cm^{-1}$$

D
Cell constant - cm$$^{ - 1}$$

Solution
Question 6
1 / -0

Chromium metal can be plated out from an acidic solution containing $$CrO_3$$ according to the following equation $$CrO_3(aq)+6H^+(aq)+6e^-\rightarrow Cr(s)+3H_2O$$
Calculate how many grams of chromium will be placed out by $$24000$$ coulombs (Atomic Weight of Cr$$=52$$).

A
$$2.15$$ gram

B
$$0.36$$ gram

C
$$21.5$$ gram

D
None

Solution

$$CrO_3 \ (aq) + 6H^+ \ (aq) + 6e^- \rightarrow Cr \ (s) + 3H_2O$$
$$n=6$$
Atomic weight of $$Cr=52$$
$$i \ times t = 24000 \ C$$
$$\boxed {m= \cfrac {Atomic \ Weight}{n \times f} \times i \times t}$$
$$= \cfrac {52}{6 \times 96500} \times 24000= 2.155 \ g$$
$$\therefore \ 2.15 \ g$$ of chromium will be placed out by $$24000 \ C$$
Question 7
1 / -0

What would you observe if you set up the following electrochemical cell
$${\text{Ag|Agn}}{{\text{o}}_3}\left( {0.001M} \right)||AgN{O_3}(1M)|Ag$$

A
Electrons will flow from left to right, causing a decrease in the $$\left[ {A{g^ + }} \right]$$ concentration in the right cell.

B
Electrons will flow from right to left, causing an increases in the $$\left[ {A{g^ + }} \right]$$ concentration in the left cell and a decrease in the $$\left[ {A{g^ + }} \right]$$ concentration in the right cell.

C
Electrons will flow from left to right, causing an increase in the $$\left[ {A{g^ + }} \right]$$ concentration in the left cell, and a decrease in the $$\left[ {A{g^ + }} \right]$$ concentration in the right cell.

D
Electrons will flow from right to left, causing a decrease in the $$\left[ {A{g^ + }} \right]$$ concentration in the right cell
Question 8
1 / -0

A constant electric current flows for 4 hours through two electrolytic cells connected in series. One contains $$AgNO_3$$ solution and second contains $$CuCl_2$$ solution. During this time, 4 grams of $$Ag$$ is deposited in the first cell. How many grams of Cu is deposited in the second cell?

A
2.176 g

B
0.176 g

C
1.176 g

D
None of the above

Solution

since $$ Cl $$ & $$ Ag $$ both are in series,
no. of eq. of $$ cl $$= no. of eq. of $$Ag$$
No./ eq of $$ Ag = \dfrac{4gm}{107}$$
& no. of eq.$$ cl =$$ no. of eq. of $$ Ag = \dfrac{4}{107}$$
eq. wt. of $$ cl = \dfrac{4}{107} \times \dfrac{1}{2} \times 63.5$$
$$ \boxed{eq. wt of $$ Cl = 1.176 gm}$$
Question 9
1 / -0

$$Cd$$ amalgam is prepared by electrolysis of a solution of $$CdCl_2$$ using a mercury cathode. Current of how much ampere must be passed for $$100$$ seconds in order to prepare $$20\%$$ $$Cd-Hg$$ amalgam on a cathode of $$2$$ g mercury? (At. wt. of Cd$$=112.40$$)

A
$$37.77$$ ampere

B
$$8.58$$ ampere

C
$$4.29$$ ampere

D
$$17.16$$ ampere

Solution
Question 10
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.

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Electrochemistry Test - 63

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Electrochemistry Test - 63

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Question 1

$$E_{cell} = 0$$

$$E_{cell} > E_{ext}$$

$$E_{ext} > E_{cell}$$

$$E_{cell} = E_{ext}$$

Solution

Question 2

$$Na^+_{(aq)}+e^-\rightarrow Na_{(s)}; E^o_{cell}=-2.71$$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 3

spontaneous in reverse direction

spontaneous in forward direction

at equilibrium

non-spontaneous in reverse direction

Solution

Question 4

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

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

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

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

Solution

Question 5

Equivalent conductivity - S $$c{m}e{q^{ - 1}}$$

Conductance -ohm$$^{ - 1}$$

Specific conductance - $$S cm^{-1}$$

Cell constant - cm$$^{ - 1}$$

Solution

Question 6

$$2.15$$ gram

$$0.36$$ gram

$$21.5$$ gram

None

Solution

Question 7

Electrons will flow from left to right, causing a decrease in the $$\left[ {A{g^ + }} \right]$$ concentration in the right cell.

Electrons will flow from right to left, causing an increases in the $$\left[ {A{g^ + }} \right]$$ concentration in the left cell and a decrease in the $$\left[ {A{g^ + }} \right]$$ concentration in the right cell.

Electrons will flow from left to right, causing an increase in the $$\left[ {A{g^ + }} \right]$$ concentration in the left cell, and a decrease in the $$\left[ {A{g^ + }} \right]$$ concentration in the right cell.

Electrons will flow from right to left, causing a decrease in the $$\left[ {A{g^ + }} \right]$$ concentration in the right cell

Question 8

2.176 g

0.176 g

1.176 g

None of the above

Solution

Question 9

$$37.77$$ ampere

$$8.58$$ ampere

$$4.29$$ ampere

$$17.16$$ ampere

Solution

Question 10

$$-0.33V$$; 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

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