Electrochemistry Test

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Electrochemistry Test - 50

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Weekly Quiz Competition

Question 1
1 / -0

For the half reaction
$$B(s) \rightarrow B^{2+} + 2e^{-}$$ $$E_{1}^{\circ} = - 0.44 V$$
and $$B_{2}+\rightarrow B^{3+} + e^{-}$$ $$E_{2}^{\circ} = 1.3 V$$
What is $$E^{\circ}$$ for the reaction,
$$3e^{-} + B^{+3} \rightarrow B(s)$$

A
$$0.86$$ V

B
$$0.14$$ V

C
$$-0.14$$ V

D
$$-0.28$$ V

Solution

$$B(s)\rightarrow { B }^{ 2 }+{ 2e }^{ - }\quad \quad \quad { E }_{ 1 }^{ 0 }=-0.44V$$ $$- i$$
$${ B }^{ 2+ }\rightarrow { B }^{ 3+ }+{ e }^{ - }\quad \quad \quad { E }_{ 1 }^{ 0 }=1.3V$$ $$- ii$$
$${ B }^{ 3+ }+{ 3e }^{ - }\rightarrow B(s)\quad \quad \quad \quad { E }_{ 3 }^{ 0 }=?$$ $$- iii$$
iii $$=- (II+I)$$
Or, $$-3\times { E }_{ 3 }^{ 0 }\times F=-\left\{ -1\times 1.3\times F+0.44\times 2\times F \right\} $$
$$\therefore { E }_{ 3 }^{ 0 }=-0.14V$$
Question 2
1 / -0

The weight in grams of $$O_2$$ formed at $$Pt$$ anode during the electrolysis of aq. $$K_2SO_4$$ solution during the passage of one coulomb of electricity is:

A
$$\dfrac{16}{96800}$$

B
$$\dfrac{8}{96500}$$

C
$$\dfrac{32}{96500}$$

D
$$\dfrac{64}{96500}$$

Solution

Weight of oxygen $$=\dfrac{16}{96500}$$
Question 3
1 / -0

The electrolysis of aqueous solution of $$Cu{Br}_{2}$$ using platinum electrode would leads to:

A
$$Cu$$ at anode

B
$${Br}_{2}$$ gas at anode and $${O}_{2}$$ gas at cathode

C
$$Cu$$ at cathode

D
$${H}_{2}$$ gas at cathode

Solution

'Copper $$(Cu)$$ at the cathode' is the answer.

Detailed explanation:

Below is the chemical reaction behind the electrolysis of aqueous solution of $$CuBr_2$$ ( Copper Bromide) with use of Platinum electrodes (inert).

$$CuBr_2 \longrightarrow Cu + 2{Br}^-$$

Reaction that happens at Anode:

$$2{Br}^- \longrightarrow { Br}_2 + 2e^-$$

Reaction that happens at Cathode:

$$Cu + 2e^- \longrightarrow Cu$$

Option C is the correct answer.
Question 4
1 / -0

For the reaction:
$$F{ e }^{ 2+ }(aq)+A{ g }^{ + }(aq)\longrightarrow Ag(s)+F{ e }^{ +3 }(aq)$$

Predict which change will decrease the cell voltage?

A
Increase the amount of Ag

B
Decrease the concentration of $$A{ g }^{ + }$$

C
Increase the concentration of $$F{ e }^{ +3 }$$

D
Both $$B$$ and $$C$$

Solution

$$E_{cell}=E^o_{cell}-\cfrac {2.303RT}{nF}\log \cfrac {[Ag][Fe^{+3}]}{[Fe^{2+}][Ag^+]}$$

Increasing the concentration of $$Fe^{+3}$$ will decrease cell voltage $$(E_{cell})$$ .
Question 5
1 / -0

The number of coulombs required to deposit $$5.4 g$$ of Aluminium when the given electrode reaction is represented as $$Al^{3+}+3e^- \rightarrow Al$$.

A
$$1.83 \times 10^5C$$

B
$$57900C$$

C
$$5.86 \times 10^5 C$$

D
$$3F$$

Solution

According to Faraday's law of electrolysis

$$W=\cfrac{Z\times Q}{F}$$

$$W=$$ mass deposited $$=5.4g$$

$$Z=$$ Equivalent mass $$=\cfrac{Molar\quad mass}{e^-\quad accepted}$$

$$F=$$ Faraday constant $$=96500C$$

$$Q=$$ Charge

$$5.4g=(\cfrac{27g}{3})\times\cfrac{Q}{96500C}$$

$$Q=57900C$$

Thus, $$57900C$$ is required to deposit $$5.4g$$ of $$Al$$
Question 6
1 / -0

How many faraday charge is required to convert one mole of $${C}_{6}{H}_{5}NO_{2}$$ in to $${C}_{6}H_{5}NH_{2}$$?

A
$$8$$

B
$$12$$

C
$$6$$

D
$$32$$

Solution
Question 7
1 / -0

Given that $$ { Cl }_{ 2\left( g \right) }+{ 2e }^{ - }\longrightarrow { 2Cl }^{ - }\left( aq \right)$$ is the reduction half-reaction for the overall reaction $$ 2Ag\left( s \right) +{ Cl }_{ 2 }\left( g \right) \longrightarrow 2AgCl\left( s \right)$$.
What is the oxidation half reaction?

A
$$ Ag\left( s \right) \longrightarrow Ag^{ + }\left( aq \right) +e^{ - }$$

B
$$ Ag\left( s \right) +{ Cl }^{ - }\left( aq \right) \longrightarrow Ag\left( s \right) +e^{ - }$$

C
$$ Ag\left( s \right) +{ Cl }_{ 2 }\left( g \right) +e^{ - }\longrightarrow AgCl\left( s \right) +Cl^{ - }\left( aq \right)$$

D
$$ { 2Cl }^{ - }\left( aq \right) \longrightarrow { Cl }_{ 2 }\left( g \right) +2e^{ - }$$

Solution
Question 8
1 / -0

From the following half-cell reaction and their potential, What is the smallest possible standard e.m.f spontaneous reaction?
$${ PO }_{ 4 }^{ 3- } (aq) + 2H_2O(I) + 2e^- \longrightarrow { HPO }_{3}^{2-} + 3OH^-(aq); E^o = -1.05 V$$
$$ PbO _2(s) + H_2O(I) + 2e^- \longrightarrow PbO(S) + 2OH^-(aq); E^o = +0.28 V$$
$$IO{-}{3}(aq) + 2H_2O(I) + 4e^- \longrightarrow IO^-(aq) + 4OH^-(aq); E^o = +0.56 V$$

A
$$+0.00$$

B
$$+0.74$$

C
$$+0.56$$

D
$$+0.28$$

Solution
Question 9
1 / -0

The time taken by the galvanic cell which operates almost ideally under reversible conditions at a current of $${ 10 }^{ -16 }$$ A to deliver 1 mole of electron is:

A
$$19.30 \times{ 10 }^{ 20 }s$$

B
$$4.825 \times{ 10 }^{ 20 }s$$

C
$$9.65 \times{ 10 }^{ 20 }s$$

D
$$3.4 \times{ 10 }^{ 11 }s$$

Solution

According to the problem:
Given that:
Current $$ = {10^{ - 16}}A$$
Charge $$=96548\,C$$ (Charge on $$1$$ mole of electrons)

Using the formula,
Charge $$(Q)=$$ current$$(A) \times $$Time $$(T)$$

Therefore,
$$T = \left( {\frac{{96548}}{{{{10}^{ - 16}}}}} \right)$$

Implies that
$$T = 9.65 \times {10^{20}}sec$$
Question 10
1 / -0

Which of the following is a strong electrolyte?

A
$$ \mathrm { Ca } \left( \mathrm { NO } _ { 3 } \right) _ { 2 } $$

B
HCN

C
$$ \mathrm { H } _ { 2 } \mathrm { SO } _ { 3 } $$

D
$$ \mathrm { NH } _ { 4 } \mathrm { OH } $$

Solution

A solute/ solution that completely ionizes or dissociates in a solution is called as strong electrolyte. Here $$Ca(NO_3)_2$$ is a strong electrolyte, as it completely dissociates into $$Ca^{2+}$$ and $$NO_3^-$$ ions in an aqueous solution.

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Electrochemistry Test - 50

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Electrochemistry Test - 50

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Question 1

$$0.86$$ V

$$0.14$$ V

$$-0.14$$ V

$$-0.28$$ V

Solution

Question 2

$$\dfrac{16}{96800}$$

$$\dfrac{8}{96500}$$

$$\dfrac{32}{96500}$$

$$\dfrac{64}{96500}$$

Solution

Question 3

$$Cu$$ at anode

$${Br}_{2}$$ gas at anode and $${O}_{2}$$ gas at cathode

$$Cu$$ at cathode

$${H}_{2}$$ gas at cathode

Solution

Question 4

Increase the amount of Ag

Decrease the concentration of $$A{ g }^{ + }$$

Increase the concentration of $$F{ e }^{ +3 }$$

Both $$B$$ and $$C$$

Solution

Question 5

$$1.83 \times 10^5C$$

$$57900C$$

$$5.86 \times 10^5 C$$

$$3F$$

Solution

Question 6

$$8$$

$$12$$

$$6$$

$$32$$

Solution

Question 7

$$ Ag\left( s \right) \longrightarrow Ag^{ + }\left( aq \right) +e^{ - }$$

$$ Ag\left( s \right) +{ Cl }^{ - }\left( aq \right) \longrightarrow Ag\left( s \right) +e^{ - }$$

$$ Ag\left( s \right) +{ Cl }_{ 2 }\left( g \right) +e^{ - }\longrightarrow AgCl\left( s \right) +Cl^{ - }\left( aq \right)$$

$$ { 2Cl }^{ - }\left( aq \right) \longrightarrow { Cl }_{ 2 }\left( g \right) +2e^{ - }$$

Solution

Question 8

$$+0.00$$

$$+0.74$$

$$+0.56$$

$$+0.28$$

Solution

Question 9

$$19.30 \times{ 10 }^{ 20 }s$$

$$4.825 \times{ 10 }^{ 20 }s$$

$$9.65 \times{ 10 }^{ 20 }s$$

$$3.4 \times{ 10 }^{ 11 }s$$

Solution

Question 10

$$ \mathrm { Ca } \left( \mathrm { NO } _ { 3 } \right) _ { 2 } $$

HCN

$$ \mathrm { H } _ { 2 } \mathrm { SO } _ { 3 } $$

$$ \mathrm { NH } _ { 4 } \mathrm { OH } $$

Solution

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