Electrochemistry Test

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Electrochemistry Test - 57

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Weekly Quiz Competition

Question 1
1 / -0

The standard reduction potential data at 25$$^o$$C is given below:
$$E^o (Fe^{3+} , Fe^{2+}) = +0.77 V$$;
$$E^o (Fe^{2+} , Fe) = -0.44 V$$
$$E^o(Cu^{2+}, Cu) = +0.34 V$$;
$$E^o(Cu^+ , Cu) = +0.52 V$$
$$E^o (O_2(g) + 4H^+ + 4e^- \rightarrow 2H_2O ] = + 1.23 V$$;
$$E^o (O_2 (g) + 2H_2 O + 4e^- \rightarrow 4OH^-) = + 0.40 V$$
$$E^o (Cr^{3+} , Cr) = - 0.74 V$$;
$$E^o (Cr^{2+} , Cr) = - 0.91 V$$;
Match $$E^o$$ of the redox pair in List I with the values given in List II and select the correct answer using the code given below the lists.
List I List II
(P) $$E^o (Fe^{3+}, Fe)$$ (1) -0.18 V
(Q) $$E^o (4H_2O \rightleftharpoons 4H^+ + 4OH^-)$$ (2) -0.4 V
(R) $$E^o (Cu^{2+} + Cu \rightarrow 2 Cu^+)$$ (3) 0.04 V
(S) $$E^o(Cr^{3+}, Cr^{2+})$$ (4) -0.83 V

A
$$P-4, Q-1, R-2, S-3$$

B
$$P-2, Q-3, R-4, S-1$$

C
$$P-1, Q-2, R-3, S-4$$

D
$$P-3, Q-4, R-1, S-2$$

Solution

(P) The electrode reaction is $$3e + Fe^{3+}   \rightarrow Fe ;              \Delta G_1^o = - 3 \times E_1^o \times F$$
$$e + Fe^{3+}   \rightarrow Fe^{2+} ;              0\Delta G_2^o = - 1 \times 0.77 \times F$$
$$e + Fe^{3+}   \rightarrow Fe^{2+} ;              0\Delta G_2^o = - 1 \times 0.77 \times F$$
Add above reactions.
$$Fe^{3+} + 3e \rightarrow Fe;    \Delta G_1^o = \Delta G_2^o + \Delta G_3^o$$
$$\therefore \Delta G_1^o = -0.77F + 0.88 F = + 0.11 F$$
$$\therefore -3 E_1^o \times F = + 0.11 F$$
$$\therefore E_1^o = - 0.04$$
Thus P is (3)

(Q) The electrode reaction is $$4H_2O \rightleftharpoons 4H^+ + 4OH^-                       \Delta G_1^o = - 1 \times E_1^o \times F$$
$$ 2H_2O   \rightarrow O_2(g) + 4H^+ + 4e^- = - 1.23 V               \Delta G_2^o = - 1 \times -1.23 \times F$$.....(1)
$$O_2 (g) + 2H_2 O + 4e^- \rightarrow 4OH^- = + 0.40 V                \Delta G_3^o = - 1 \times 0.40 \times F$$......(2)
Add reactions (i) and (ii)
$$4H_2O \rightleftharpoons 4H^+ + 4OH^-                              \Delta G_1^o = \Delta G_2^o + \Delta G_3^o$$
$$\Delta G_1^o = -1.23 F +0.40 F$$
$$\therefore -1 \times E_1^o F =-0.83 F$$
$$E_1^o = -0.83 V$$
Thus Q is (2)

(R) The electrode reaction is
$$Cu^{2+} + Cu \rightarrow 2 Cu^+         \Delta G_1^o = - 1 \times E_1^o \times F$$
$$Cu^{2+} + e \rightarrow Cu^+         \Delta G_2^o = - 1 \times 0.34 \times F$$......(i)
$$ Cu \rightarrow Cu^++e         \Delta G_3^o = - 1 \times (-0.52) \times F$$......(ii)
Add equation (i) and (ii)
$$Cu^{2+} + Cu \rightarrow 2 Cu^+          \Delta G_1^o = \Delta G_2^o + \Delta G_3^o$$
$$\Delta G_1^o = + 0.34 F - 0.52 F$$
$$\therefore -1 \times E_1^o F = 0.18 F$$
$$E_1^o = -0.18 V$$
Thus R is (1)

(S) The electrode reaction is
$$Cr^{3+} + e \rightarrow Cr^{2+};      \Delta G_1^o = - 1 \times E_1^o \times F$$
$$Cr^{3+} + 3e \rightarrow Cr;      \Delta G_2^o = - 3 \times (-0.74) \times F$$......(i)
$$Cr^{2+} + 2e \rightarrow Cr;      \Delta G_3^o = - 2 \times (-0.91) \times F$$......(ii)
Substract equation (ii) from (i)
$$Cr^{3+} + e \rightarrow Cr^{2+};             \Delta G_1^o = \Delta G_2^o - \Delta G_3^o$$
$$\Delta G_1^o = + 2.22 F - 1.82 F$$
$$\therefore -1 \times E_1^o F = -0.4 F$$
$$E_1^o = - 0.4 V$$
Thus S is (2)
Question 2
1 / -0

The standard reduction potentials $$E^{\circleddash}$$ for the half reactions are follows :

$$Zn\, \rightarrow\, Zn^{2+}\, +\, 2e^{-};\, \quad\, E^{\circleddash}\, =\, +0.76\, V$$
$$Fe\, \rightarrow\, Fe^{2+}\, +\, 2e^{-};\, \quad\, E^{\circleddash}\, =\, 0.41\, V$$

The EMF for the cell reaction $$Fe^{2+}\, Zn\, \rightarrow\, Zn^{2+}\, +\, Fe$$ is:

A
$$- 0.35 V$$

B
$$+ 0.35 V$$

C
$$+ 1.17 V$$

D
$$- 1.17 V$$

Solution

$$Fe^{2+}\, +\, Zn\, \rightarrow\, Zn^{2+}\, +\, Fe$$
$$Zn\, \rightarrow\, Zn^{2+}\, +\, 2e^-;\, \quad\, E^{\circleddash}\, =\, +0.76\, V$$
$$Fe\, \rightarrow\, Fe^{2+}\, +\, 2e^-;\, \quad\, E^{\circleddash}\, =\, +0.41\, V$$
These are oxidation potentials.
Reduction potentials are equal and opposite.
Fe forms cathode and Zn forms anode.
$$E^{\circleddash}_{cell}\, =\, (E^{\circleddash}_{red})_c\, +\, (E^{\circleddash}_{oxid})_a$$
= (-0.41 + 0.76) V
= 0.35 V
Question 3
1 / -0

For the reaction :
$$4Al(s)\, +\, 3O_{2}(g)\, +\, 6H_{2}O\, +\, 4\, \overset{\ominus}{O}H\, \rightarrow\, 4[Al(OH)_{4}]^{\oplus}$$
$$E^{\ominus}_{cell}\, =\, 2.73\, V$$
If $$\Delta_{f}G^{\ominus}\, [\overset{\oplus}{O}H]\, =\, -\, 157\, kJ\, mol^{-1}$$;
$$\Delta_{f}G^{\ominus}\, (H_{2}O)\, =\, -\, 237.2\, kJ\, mol^{-1}$$.
The value of $$\Delta_{f}G^{\ominus}\, [Al(OH)_{4}]^{\ominus}$$ is:
[free energy of formation of $$(Al(OH)_{4}^{\ominus}]$$.

A
$$\Delta_{f}G^{\ominus}[Al(OH)_{4}]^{\ominus}\, =\, + 3.50\, \times\, 10^{3}\, kJ\, mol^{-1}$$

B
$$\Delta_{f}G^{\ominus}[Al(OH)_{4}]^{\ominus}\, =\, + 2.40\, \times\, 10^{3}\, kJ\, mol^{-1}$$

C
$$\Delta_{f}G^{\ominus}[Al(OH)_{4}]^{\ominus}\, =\, - 1.30\, \times\, 10^{3}\, kJ\, mol^{-1}$$

D
$$None \:\:of \:\:these $$

Solution

$$4A\, \overset{0}{l}(s)\, +\, 3O_{2}(g)\, +\, 6H_{2}O\, +\, 4\, \overset\, {\ominus}{O}H$$
$$\rightarrow\, 4[\overset{+3}{Al}(OH)_{4}]^{\ominus}$$
$$E^{\ominus}_{cell}\, =\, 2.73\, V$$
Here, $$n\, =\, 4\, \times\, 3\, =\, 12$$ (Change in oxidation number of Al for four moles)
$$\Delta G^{\ominus}\, =\, -\, 12\, \times\, 96500\, \times\, 2.73\, =\, - 3.16\, \times\, 10^{3}\, kJ$$
$$\Delta G^{\ominus}\, =\, 4\, \Delta_{f}G^{\ominus}[Al(OH)_{4}]^{\ominus}\, -\, [6\Delta_{f}G^{\ominus}(H_{2}O)\, +\, 4\Delta_{f}G^{\ominus}(\overset{\ominus}{O}H)]$$
$$\Rightarrow\, -\, 3.16\, \times\, 10^{3}\, =\, 4x\, -\, [6(- 237.2)\, +\, 4(-\, 157)]$$
$$\Rightarrow\, 4x\, \Rightarrow\, - 5212.54\, \Rightarrow\, xd\, =\, - 1.30 \, \times\, 10^{3}\, kJ\, mol^{-1}$$
$$\Delta_{f}G^{\ominus}[Al(OH)_{4}]^{\ominus}\, =\, - 1.30\, \times\, 10^{3}\, kJ\, mol^{-1}$$

Question 4

1 / -0

An aqueous solution of X is added slowly to an aqueous solution of Y as shown in List I. The variation in conductivity of theses reactions is given in List II. Match List I with List II and select the correct using the code given below the lists:

List-A	List-B
(P) $$\underset {X}{(C_2H_5)_3N}+\underset {Y}{CH_2COOH}$$	(1) Conductivity decreases and then increases
(Q) $$\underset {X}{KI(0.1M)}+\underset {Y}{AgNO_3(0.01M)}$$	(2) Conductivity decreases and then does not change much
(R) $$\underset {X}{CH_2COO}+\underset {Y}{KOH}$$	(3) Conductivity increases and then does not change much
(S) $$\underset {X}{NaOH}+\underset {Y}{HI}$$	(4) Conductivity does not change much and then increases.

3, 4, 2, 1

4, 3, 2, 1

2, 3, 4, 1

1, 4, 3, 2

Solution

(P) $$\underset {X}{(C_2H_5)_3N}+\underset {Y}{CH_2COOH}$$
(3) Since a weak base is added to weak acid, the conductivity
increases and then dos not change much.
(Q) $$\underset {X}{KI(0.1M)}+\underset {Y}{AgNO_3(0.01M)}$$
(4) Since an strong electrolyte is added to strong electrolyte, the conductivity dos not change much and then increases.
(R) $$\underset {X}{CH_2COO}+\underset {Y}{KOH}$$
(2) Since a weak acid is added to strong base, the conductivity decreases and then does not change much.
(S) $$\underset {X}{NaOH}+\underset {Y}{HI}$$
(1) Since a strong base is added to an acid, the conductivity decreases and then increases.

Question 5
1 / -0

Which of the following is used as electrodes in batteries?

A
Diamond

B
Graphite

C
Activated charcoal

D
Carbon black

Solution

Natural graphite has found uses in zinc-carbon batteries, in electric motor brushes, and various specialized applications. Graphite is also commonly used in the form of powders, and sticks for the purpose of writing or drawing. Graphite of various hardness or softness results in different qualities and tones when used as an artistic medium.
Question 6
1 / -0

Directions For Questions

The standard reduction potentials of some half reactions are given as below
$$Cu^{2+} + e \rightarrow Cu^+; E^o_1 = 0.15 V$$
$$Cu^{+} + e \rightarrow Cu; E^o_2 = 0.5 V$$
$$Ag^{+} + e \rightarrow Ag; E^o_3 = 0.799 V$$
The decrease in Gibbs energy $$(- \Delta G) $$ is responsible for the production of equivalent amount of electrical work. Thus, $$- \Delta G = n F \times E$$

...view full instructions

The $$E^o_{cell} $$ for $$Cu | Cu^{2+}$$ in volt is:

A
$$0.35$$

B
$$-0.35$$

C
$$+0.325$$

D
$$-0.325$$

Solution

Given,
$$Cu^{2+} + e \rightarrow Cu^+; -\Delta G^o_1 = E^o_1 F \times 1$$

$$Cu^{+} + e \rightarrow Cu; -\Delta G^o_2 = E^o_2 F \times 1$$

----------------------------
by adding these equations

$$Cu^{2+} + 2e \rightarrow Cu; -\Delta G^o_3 = -[\Delta G^o_1 + \Delta G^o_2] = - [0.15 F + 0.5 F]$$

or $$2 \times E^o_3 F = 0.65 F$$

(as $$\Delta G =-nFE^o$$)

$$\therefore E^o_3 = 0.325 V$$

So for reverse reaction.

$$\therefore Cu \rightarrow Cu^{2+} + 2e E^o = - 0.325 V$$
Question 7
1 / -0

Separating the components of a homogeneous mixture using electricity is called:

A
electrification

B
electrolysis

C
electrolyte

D
electromagnetisation

Solution

Separation of component of homogeneous mixture using electricity is called electrolysis.
Electrolysis is conversion of electrical energy $$\rightarrow $$ chemical energy.
Example: Electroplating, Electrolysis of $${ aq.CuSO }_{ 4 }\quad { sol }^{ n }$$ etc.
Question 8
1 / -0

For the electrolytic production of $$NaClO_4$$ from $$NaClO_3$$ according to the reaction $$NaClO_3 + H_2O \rightarrow NaClO_4 + H_2$$. How many faradays of electricity would be required to produce 0.5 mole of $$NaClO_4$$ ?

A
$$1$$

B
$$2$$

C
$$3$$

D
$$1.5$$

Solution
Question 9
1 / -0

Chromium plating can involve the electrolysis of an electrolyte of an acidified mixture of chromic acid and chromium sulphate. If during electrolysis the article being plated increases in mass by 2.6 g and $$0.6\, dm^{3}$$ of oxygen are evolved at an inert anode, the oxidation state of chromiumions being discharged must be : (assuming atomic weight of Cr = 52 and 1 mole of gas at room temperature and pressure occupies a volume of 24 $$dm^{3}$$)

A
$$-1$$

B
$$0$$

C
$$+1$$

D
$$+2$$

Solution

$$O_{2}\, +\, 2H_{2}O\, +\, 4e^{-}\, \Rightarrow\, 1\, mol\, \equiv\, 4\, F$$
$$\Rightarrow\, \displaystyle \frac{0.6}{24}mol \, \equiv\, 4\, \times\, \displaystyle \frac{0.6}{24}\, =\, 0.1\, F$$
$$Cr^{3+}\, +\, xe^{-}\, \rightarrow\, Cr^{(3\, -\, x)\oplus}$$
$$\Rightarrow\, 1\, mol\, \equiv\, x\, F\, \Rightarrow\, \displaystyle \frac{2.6}{52} mol\, \equiv\, \frac{x}{20}F\, \Rightarrow\, x\, -\, 2$$
$$\Rightarrow$$ Oxidation state of $$Cr^{(3\, -\, 2)}\, =\, Cr^{+1}\, =\, +1$$.
Question 10
1 / -0

The passage of $$1.5$$ Faradays of electricity corresponds to the flow of how many electrons?

A
$$6\times {10}^{23}$$

B
$$9\times {10}^{23}$$

C
$$12\times {10}^{23}$$

D
$$3\times {10}^{23}$$

Solution

$$n(e^-)= \dfrac{Q}{F}$$

$$\Rightarrow$$ $$n(e^-)= \dfrac{1.5F}{F}$$

$$\Rightarrow$$ $$n(e^-)= {1.5 \ moles}$$ = $$1.5 \times {6.022} \times {10^{23}}$$

$$\Rightarrow$$ $${9} \times {10^{23}} \ electrons$$

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Re-Attempt Weekly Quiz Competition

Electrochemistry Test - 57

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Electrochemistry Test - 57

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SHARING IS CARING

Question 1

$$P-4, Q-1, R-2, S-3$$

$$P-2, Q-3, R-4, S-1$$

$$P-1, Q-2, R-3, S-4$$

$$P-3, Q-4, R-1, S-2$$

Solution

Question 2

$$- 0.35 V$$

$$+ 0.35 V$$

$$+ 1.17 V$$

$$- 1.17 V$$

Solution

Question 3

$$\Delta_{f}G^{\ominus}[Al(OH)_{4}]^{\ominus}\, =\, + 3.50\, \times\, 10^{3}\, kJ\, mol^{-1}$$

$$\Delta_{f}G^{\ominus}[Al(OH)_{4}]^{\ominus}\, =\, + 2.40\, \times\, 10^{3}\, kJ\, mol^{-1}$$

$$\Delta_{f}G^{\ominus}[Al(OH)_{4}]^{\ominus}\, =\, - 1.30\, \times\, 10^{3}\, kJ\, mol^{-1}$$

$$None \:\:of \:\:these $$

Solution

Question 4

3, 4, 2, 1

4, 3, 2, 1

2, 3, 4, 1

1, 4, 3, 2

Solution

Question 5

Diamond

Graphite

Activated charcoal

Carbon black

Solution

Question 6

$$0.35$$

$$-0.35$$

$$+0.325$$

$$-0.325$$

Solution

Question 7

electrification

electrolysis

electrolyte

electromagnetisation

Solution

Question 8

$$1$$

$$2$$

$$3$$

$$1.5$$

Solution

Question 9

$$-1$$

$$0$$

$$+1$$

$$+2$$

Solution

Question 10

$$6\times {10}^{23}$$

$$9\times {10}^{23}$$

$$12\times {10}^{23}$$

$$3\times {10}^{23}$$

Solution

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