Electrochemistry Test

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Electrochemistry Test - 40

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Question 1
1 / -0

The half-cell reaction is the one that:

A
takes place at one electrode

B
consumes half a unit of electricity

C
involves half a mole of electrolyte

D
goes half way to completion

Solution

A half reaction is either the oxidation or reduction reaction component of a redox reaction. A half reaction is obtained by considering the change in oxidation states of individual substances involved in the redox reaction.
Half-reactions are often used as a method of balancing redox reactions.
Question 2
1 / -0

On passing one faraday of electricity through a dilute solution of an acid, the volume of hydrogen obtained at NTP is__________.

A
$$22400\ mL$$

B
$$1120\ mL$$

C
$$2240\ mL$$

D
$$11200\ mL$$

Solution

$$2H^++2e^-\longrightarrow H_2$$
$$2$$ mol of $$e^-\longrightarrow 1$$ mol of $$H_2$$
$$1$$ mole of $$e^- \longrightarrow \cfrac {1}{2}$$ mol of $$H_2$$
$$1F \longrightarrow \cfrac {1}{2}$$ mol of $$H_2$$
At NTP, $$1$$ mol $$H_2 \longrightarrow 22.4$$ litre
$$\cfrac {1}{2}$$ mol $$\longrightarrow 11.2$$ litre= $$11200ml$$
Question 3
1 / -0

The half-cell reaction for the corrosion,
$$2H^{+} + \dfrac {1}{2}O_{2} + 2e^{-} \rightarrow H_{2}O; E^{\circ} = 1.23\ V$$
$$Fe^{2+} + 2e^{-} \rightarrow Fe(s); E^{\circ} = -0.44\ V$$
Find the $$\triangle G^{\circ}$$ (in kJ) for the overall reaction.

A
$$-76\ kJ$$

B
$$-322\ kJ$$

C
$$-161\ kJ$$

D
$$-152\ kJ$$

Solution

$$Fe(s)\rightarrow Fe^{2+} + 2e^{-}; \triangle G_{1}^{\circ}$$
$$\underline {2H^{+} + 2e^{-} + \dfrac {1}{2}O_{2}\rightarrow H_{2}O(l); \triangle G_{2}^{\circ}}$$
$$\underline {Fe(s) + 2H^{+} + \dfrac {1}{2}O_{2}\rightarrow Fe^{2+} + H_{2}O; \triangle G_{3}^{\circ}}$$
Applying $$\triangle G^{\circ}_{1} + \triangle G^{\circ}_{2} = \triangle G^{\circ}_{3}$$
$$\triangle G^{\circ}_{3} = (-2F \times 0.44) + (-2F\times 1.23)$$
$$=(-2 \times 96500\times 0.44) + (-2\times 96500\times 1.23)$$$
$$= -322310\ J $$
$$= -322\ kJ$$.
Question 4
1 / -0

What is the sign of $$\triangle G^{\circ}$$ and the value of $$K$$ for a electrochemical cell for which $$E^{\circ}_{cell} = 0.80\ V$$?

A
$$\triangle G^{\circ} = -, K = > 1$$

B
$$\triangle G^{\circ} = +, K = > 1$$

C
$$\triangle G^{\circ} = +, K = < 1$$

D
$$\triangle G^{\circ} = -, K = < 1$$

Solution

$$\rightarrow$$ The standard free energy change of a cell reaction and emf of cell is given by,
$$\triangle G^{\circ}=-nFE_{cell}^{\circ}$$------------1
$$\rightarrow$$ Emf of cell and equilibrium constant $$K$$ is given by ,
$$E^{\circ}=\cfrac{0.0591}{n}\log K$$------------2
$$\rightarrow$$ The value of $$K$$ gives the extent of the cell reaction. If the value of $$K$$ is large, the reaction proceeds to larger extent.
$$\rightarrow$$ For spontaneous reaction, $$\triangle G^{\circ} $$ is negative and $$K>1$$.
Question 5
1 / -0

The number of electrons involved when one faraday of electricity is passed through an electrolytic solution is:

A
$$96500$$

B
$$8\times 10^{6}$$

C
$$12\times 10^{16}$$

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

Solution

$$1$$ Faraday= $$96500$$ coulombs
$$1F$$ is charge carried by $$1$$ mole of electrons so $$6\times 10^{23}$$ electrons are involved.
Question 6
1 / -0

For the cell reaction,
$$\underset {(C_{1})}{Cu^{2+}}(aq.) + Zn(s) \rightarrow \underset {(C_{2})}{Zn^{2+}}(aq.) + Cu(s)$$
the change in free energy $$(\triangle G)$$ at a given temperature is a function of____________.

A
$$ln\ C_{1}$$

B
$$-RT ln \left (\dfrac {C_{2}}{C_{1}}\right )$$

C
$$ln\ (C_{1} + C_{2})$$

D
$$ln\ C_{2}$$

Solution

Gibbs free energy$$\triangle G^{\circ}=-RT\,ln\,K$$-----------1
For reaction:-
$$Cu_{(aq)}^{2+}+Zn_{(s)}\rightarrow\,Zn_{(aq)}^{2+}+Cu_{(s)}$$
$$K=\cfrac{Product}{Reactant}=\cfrac{[Zn_{(aq)}^{2+}][Cu_{(s)}]}{Cu_{(aq)}^{2+}][Zn_{(s)}]}=\cfrac{C_{2}}{C_{1}}$$---------------2
[In pure form, $$Cu_{s}$$ & $$Zn_{(s)}=1$$]
$$\therefore\,\triangle G^{\circ}=-RT\,ln(\cfrac{C_{2}}{C_{1}})$$------------From Equation 1 & 2.
Question 7
1 / -0

The standard emf of a galvanic cell involving cell reaction with $$n = 2$$ is found to be $$0.295\ V$$ at $$25^{\circ}C$$. The equilibrium constant of the reaction would be:
Given: $$F = 96500\ C\ mol^{-1}; R = 8.314\ JK^{-1} mol^{-1}$$.

A
$$2\times 10^{11}$$

B
$$4\times 10^{12}$$

C
$$1\times 10^{2}$$

D
$$1\times 10^{10}$$

Solution
Question 8
1 / -0

The measured potential for,
$$Mg^{2+} + 2e^{-}\rightleftharpoons Mg(s)$$
does not depend upon:

A
raising the temperature

B
increasing the concentration of $$Mg^{2+}$$ ions

C
making the magnesium plate bigger

D
purity of magnesium plate

Solution

Measured Potential (EMF) do not depend on size of electrode. So, on making the $$Mg$$ plate bigger there will be no change in measured potential.
Question 9
1 / -0

The main factors which affect corrosion are :

A
Position of metal in electrochemical series

B
Presence of $$CO_{2}$$ in water

C
Presence of impurities coating

D
All are correct

Solution

Corrosion is a process in which the metals react with the atmosphere and weakens the metal.
The main factors that effect corrosion are
(i) Presence of a protective coating.
(ii) Presence of $${ CO }_{ 2 }$$ in water.
(iii) Presence of impurities coating
Ans :- A,B,C,

Question 10

1 / -0

Match the thermodynamic properties (List I) with their relation (List II).

List I	List II
A. Free energy change ($$\Delta { G }^{ o }$$)	(i) $$RT\log _{ e }{ K } $$
B. Entropy change $$\Delta { S }^{ o }$$	(ii) $$-nFE$$
C. $$\Delta { H }^{ o }$$ enthalpy change of a reaction in standard state	(iii) $${ RT }^{ 2 }{ \left( \cfrac { d\ln { K } }{ dT } \right) }_{ P }$$
D. Standard free energy change $$(\Delta { G }^{ o })$$	(iv) $$-{ \left\{ \cfrac { d\Delta { G }^{ o } }{ dT } \right\} }_{ P }$$

A-i; B-ii; C-iii; D-iv

A-ii; B-iv; C-iii; D-i

A-iv; B-ii; C-iii; D-i

A-i; B-ii; C-iv; D-iii

Solution

A. We have, $$\Delta$$G= -nFE

B and C is explained in above figure

D. $$\Delta$$$$G^0$$= -RT$$log_e$$K

hence, the answer is B

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Electrochemistry Test - 40

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Electrochemistry Test - 40

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SHARING IS CARING

Question 1

takes place at one electrode

consumes half a unit of electricity

involves half a mole of electrolyte

goes half way to completion

Solution

Question 2

$$22400\ mL$$

$$1120\ mL$$

$$2240\ mL$$

$$11200\ mL$$

Solution

Question 3

$$-76\ kJ$$

$$-322\ kJ$$

$$-161\ kJ$$

$$-152\ kJ$$

Solution

Question 4

$$\triangle G^{\circ} = -, K = > 1$$

$$\triangle G^{\circ} = +, K = > 1$$

$$\triangle G^{\circ} = +, K = < 1$$

$$\triangle G^{\circ} = -, K = < 1$$

Solution

Question 5

$$96500$$

$$8\times 10^{6}$$

$$12\times 10^{16}$$

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

Solution

Question 6

$$ln\ C_{1}$$

$$-RT ln \left (\dfrac {C_{2}}{C_{1}}\right )$$

$$ln\ (C_{1} + C_{2})$$

$$ln\ C_{2}$$

Solution

Question 7

$$2\times 10^{11}$$

$$4\times 10^{12}$$

$$1\times 10^{2}$$

$$1\times 10^{10}$$

Solution

Question 8

raising the temperature

increasing the concentration of $$Mg^{2+}$$ ions

making the magnesium plate bigger

purity of magnesium plate

Solution

Question 9

Position of metal in electrochemical series

Presence of $$CO_{2}$$ in water

Presence of impurities coating

All are correct

Solution

Question 10

A-i; B-ii; C-iii; D-iv

A-ii; B-iv; C-iii; D-i

A-iv; B-ii; C-iii; D-i

A-i; B-ii; C-iv; D-iii

Solution

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