Electrochemistry Test

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Electrochemistry Test - 56

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Weekly Quiz Competition

Question 1
1 / -0

The temperature coefficient of a standard Cd-cell is $$-5.0 \times 10^{-5} Vk^{-1}$$ whose emf at 25$$^o$$C is 1.018 V. During the cell operation, the temperature will:

A
Increase

B
Decreases

C
Either

D
Remains constant

Solution

$$\because \Delta H = \Delta G + T \Delta S$$
$$= -n FE_{cell} + nFT \left ( \frac{\partial E}{\partial T} \right)_{\rho}$$ [n = 2 for Cd$$^{2+} + 2e^- \rightarrow$$ Cd & putting other values]
$$= -199.35 kJ/mol$$
Now, $$\Delta H < O \Rightarrow$$ Exothermic reaction
$$\Rightarrow $$ Heat will be released during cell reaction & increasing the temperature of system.
Question 2
1 / -0

Battery charging equipment is generally installed:

A
in well ventilated location.

B
in clean and dry place.

C
as near as practical to the battery being charged.

D
in location having all above features.

Solution

The ventilation is required to maintain the temperature level in battery. The clean and dry place maintains the battery from the short circuit by wet condition. The charging source as possible as be near with the battery. Otherwise, more power loss happens in the electrical wire. Hence, we need the above all features to install the battery charging equipment.
Question 3
1 / -0

Directions For Questions

The cell potential $$(E_{cell})$$ of a reaction is related as $$\triangle G = -nF\space E_{cell}$$, where $$\triangle G$$ represents max useful electrical work
$$n$$ = no. of moles of electrons exchanged during the reaction
for reversible cell reaction $$d(\triangle G) = \triangle_rV)dp - (\triangle_rS).dT$$
at constant pressure $$\quad d(\triangle G) = -(\triangle_rS).dT$$
$$\because\quad$$ At constant pressure $$\quad \triangle G = \triangle H - T.\triangle S\quad ...(1)$$
$$ \triangle G = \triangle H + T\left(\displaystyle\frac{d(\triangle g)}{dT}\right)_p \quad ...(2)$$
$$\left(\displaystyle\frac{dE_{cell}}{dT}\right)_p$$ is known as temperature coefficient of the e.m.f. of the cell

...view full instructions

The temperature coefficient of the e.m.f. of cell, $$\left(\displaystyle\frac{dE}{dT}\right)_p$$ is given by:

A
$$\displaystyle\frac{nF}{\triangle S}$$

B
$$\displaystyle\frac{\triangle S}{nF}$$

C
$$\displaystyle\frac{\triangle S}{nFT}$$

D
$$-nFE$$

Solution

$$E_{cell} = -\dfrac { \Delta G}{nF}$$

$$\Delta G=\dfrac {\Delta S}{dT}$$

The temperature coefficient $$(\dfrac {dE}{dT})= \dfrac {\Delta S/ dT}{dT. nF}= \dfrac {\Delta S}{nF}$$
Question 4
1 / -0

A resistance of 50 $$\Omega$$ is registered when two electrodes are suspended into a beaker containing a dilute solution of a strong electrolyte such that exactly half of the them are submerged into solution. If the solution is diluted by adding pure water (negligible conductivity) so as to just completely submerge the electrodes, the new resistance offered by the solution would be:

A
50 $$\Omega$$

B
100 $$\Omega$$

C
25 $$\Omega$$

D
200 $$\Omega$$

Solution

The expression for the resistance of the solution is:

$$R = \dfrac{1}{k} \dfrac{l}{A}$$

Initially, only half of the electrodes were submerged but after dilution, they are completely submerged. Hence, the value of the area of electrode A in the above equation is doubled.

Due to the dilution, the volume of the solution is doubled and the molar concentration is reduced to half. Due to this, the value of specific conductance k is reduced to half.

Thus k is halved while the A is doubled. Hence R remains the same, i.e., 50 $$\Omega$$.
Question 5
1 / -0

The indication of the state of charge of a battery is best given by:

A
specific gravity of electrolyte

B
temperature of electrolyte

C
colour of electrolyte.

D
level of electrolyte.

Solution

During the charging and discharging time, the value of specific gravity of electrolyte varies. For example, during the charging time, the value of specific gravity increases from the electric current level and during the discharging time, the value of specific gravity decreases from the electric current level. But, the rest of all factors which shown in above are not changed. Hence, the state of charge of a battery is given by the specific gravity of electrolytic.
Question 6
1 / -0

Internal resistance of a battery cell decreases with :

A
an increase in area of the plates inside the electrolyte

B
an increase in distance between the two electrodes

C
a decrease in size of the electrodes

D
an increase of age of the battery

Solution

Internal resistance of a battery cell:
1. increases with increase in distance between two electrodes.
2. decreases with increase in concentration of electrolyte.
3. decreases with increase in area of the plates inside the electrolyte.
4. decreases with increase in size of the electrodes.
Question 7
1 / -0

The E$$^o$$ in the given diagram is:

A
0.523

B
0.613

C
0.715

D
0.825

Solution

$$CiO_3    + 2H_2 O +4e \rightarrow CIO^- + 4OH^- ; \Delta G_1^o$$
$$CIO^- + H_2O + e \rightarrow \frac{1}{2}Cl_2 + 2OH^-         ; \Delta G_2^o$$
$$\frac{1}{2} Cl_2 + e \rightarrow Cl^-$$ $$;\Delta G_3^o$$
---------------------------------------------------------------
$$ClO_3^- + 3H_2O + 6e \rightarrow Cl^- + 6OH^-$$ $$;\Delta G_3^o$$
---------------------------------------------------------------
$$\therefore    \Delta G^o = \Delta G_1^o + \Delta G_2^o + \Delta G_3^o$$
     $$-6FE^o = - 4F \times 0.54 - 1 F \times 0.45 - 1F \times 1.07$$
$$\therefore   E^o = + \frac{3.68}{6} = + 0.613 V$$
Question 8
1 / -0

The $$\Delta $$ H for the reaction at 0$$^o$$C,
$$Zn + 2AgCl \rightarrow ZnCl_2 + 2Ag$$
is - 52.05 kcal. If the e.m.f. of the cell is 1.015 volt, the temperature coefficient of the cell at 0$$^o$$C in $$V/^oC$$ is:

A
$$-4.311 \times 10^{-4}$$

B
$$2.388 \times 10^{-4}$$

C
$$4.311 \times 10^{-4}$$

D
$$4.311 \times 10^{-5}$$

Solution

The temperature coefficient is $$\frac {2.303R}{nF}(T-T')=\frac {2.303 \times 2}{2 \times 96500}\times 10 = 2.388 \times 10^{-4} V/^{o}C$$.
Question 9
1 / -0

The temperature coefficient $$\displaystyle \left ( \frac{\partial E}{\partial T} \right )_P $$ for a given cell is $$1.5 \times 10^{-4} JK^{-1}$$ at 300 K. The change in entropy during cell reaction is:
$$Pb(s) + HgCl_2(aq) \rightarrow PbCl_2(aq) + Hg(l)$$

A
28.95 JK$$^{-1}$$

B
14.47 JK$$^{-1}$$

C
57.9 JK$$^{-1}$$

D
21.70 JK$$^{-1}$$

Solution

As we know,

$$\displaystyle \left ( \frac{\partial E}{\partial T} \right )_P = \frac{\Delta S}{n F}$$

$$\therefore \Delta S = \displaystyle \left ( \frac{\partial E}{\partial T} \right )_P \times n F$$

$$=1.5 \times 10^{-4} \times 2 \times 96500 =28.95 JK^{-1}$$ (as n is 2)
Question 10
1 / -0

The Gibbs energy change accompanying a given process is -85.77 kJ mol$$^{-1}$$ at 25$$^o$$C and -83.68 kJ mol$$^{-1}$$ at 35$$^o$$C. The heat of reaction of process at 30$$^o$$C is:

A
-148.05 kJ mol$$^{-1}$$

B
-138.15 kJ mol$$^{-1}$$

C
-168.25 kJ mol$$^{-1}$$

D
none of these

Solution

Assume that the enthalpy change and the entropy change are not affected by the temperature change of about $$10^oC$$.

$$\Delta G =\Delta H-T \Delta S$$

$$-85770=\Delta H-298 \times \Delta S$$...... (1)

$$-83680=\Delta H-308\Delta S$$......(2)

From (1) $$\Delta S =\dfrac {\Delta H+85770}{298}$$......(3)

Substitute (3) into (2)

$$-83860=\Delta H- \dfrac {308}{298}(\Delta H+85770)$$

$$\Delta H= -148052 \: kJ/mol = -148.05 J/mol$$

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Electrochemistry Test - 56

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Electrochemistry Test - 56

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Question 1

Increase

Decreases

Either

Remains constant

Solution

Question 2

in well ventilated location.

in clean and dry place.

as near as practical to the battery being charged.

in location having all above features.

Solution

Question 3

$$\displaystyle\frac{nF}{\triangle S}$$

$$\displaystyle\frac{\triangle S}{nF}$$

$$\displaystyle\frac{\triangle S}{nFT}$$

$$-nFE$$

Solution

Question 4

50 $$\Omega$$

100 $$\Omega$$

25 $$\Omega$$

200 $$\Omega$$

Solution

Question 5

specific gravity of electrolyte

temperature of electrolyte

colour of electrolyte.

level of electrolyte.

Solution

Question 6

an increase in area of the plates inside the electrolyte

an increase in distance between the two electrodes

a decrease in size of the electrodes

an increase of age of the battery

Solution

Question 7

0.523

0.613

0.715

0.825

Solution

Question 8

$$-4.311 \times 10^{-4}$$

$$2.388 \times 10^{-4}$$

$$4.311 \times 10^{-4}$$

$$4.311 \times 10^{-5}$$

Solution

Question 9

28.95 JK$$^{-1}$$

14.47 JK$$^{-1}$$

57.9 JK$$^{-1}$$

21.70 JK$$^{-1}$$

Solution

Question 10

-148.05 kJ mol$$^{-1}$$

-138.15 kJ mol$$^{-1}$$

-168.25 kJ mol$$^{-1}$$

none of these

Solution

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