Electrochemistry Test

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Electrochemistry Test - 60

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Weekly Quiz Competition

Question 1
1 / -0

Which one of the following species has maximum conductance in their aqueous solutions?

A
$$K_2PtCl_6$$

B
$$PtCl_4 \cdot 2NH_3$$

C
$$PtCl_4 \cdot 3NH_3$$

D
$$PtCl_4 \cdot 5NH_3$$

Solution

Greater the number of ions, greater the conductance.

(a) $$ K_2[Pt\, Cl_6]\rightleftharpoons 2K^+ +[Pt\, Cl_a]^{2-}$$ (3 ions)

(b) $$ [Pt (NH_3)_2Cl_4]$$ (No ions)

(c) $$ [Pt(NH_3)_3Cl_3]Cl \rightleftharpoons [Pt(NH_3)_3Cl_3]+Cl^-$$ (two ions)

(d) $$[Pt(NH_3)_5Cl]Cl_3\rightleftharpoons [Pt(NH_3)_5Cl]+3Cl^-$$ (4 ions), it has maximum electrolytic conductance
Question 2
1 / -0

For the reduction of silver ions will copper metal, the standard cell potential was found to be $$+0.46\ V$$ at $$25^{\circ}C$$. The value of standard gibbs energy, $$\Delta G^{\circ}$$ will be:
$$(F = 96500\ C\ mol^{-1})$$

A
$$89.0\ kJ$$

B
$$-89.0\ kJ$$

C
$$-44.5\ kJ$$

D
$$-98.0\ kJ$$

Solution

Cell reaction:-
$$Cu_{(s)}+2Ag_{(aq)}^{+}\rightarrow\,Cu_{(aq)}^{2+}+2Ag_{(s)}$$----------1
Standard Gibb's energy $$\Rightarrow\,\Delta ^{\circ}G=-nFE^{\circ}$$--------2
Reaction 1 involves a 2 electron change.
$$\therefore\,n=2$$.
$$E^{\circ}=0.46V\,\,,F=96500\,C\,mol^{-1}$$
Putting all values in Equation 2,
$$\Delta ^{\circ}G=-2\,mol\times96500\,C\,mol^{-1}\times0.46V$$
$$\therefore\Delta ^{\circ}G=-88700\,J$$. [$$\because\,CV=J$$]
$$\therefore\Delta =-88.7\,kJ\approx-89.0\,kJ$$.
Question 3
1 / -0

The approximate time duration in hours to electroplate 30 g of calcium from molten calcium chloride using a current of 5 amp is :
[Atomic mass of $$Ca = 40$$]

A
8

B
80

C
10

D
16

Solution

30 g of $$Ca$$ (atomic weight 40 g/mol) corresponds to $$\displaystyle \dfrac {30}{40} = 0.75$$ moles.

1 mole of $$Ca$$ corresponds to 2 moles of electrons. 0.75 moles of $$Ca$$ will correspond to $$\displaystyle 2 \times 0.75 = 1.5$$ moles of electron.

1 mole of electron corresponds to $$96500$$ coulombs of electricity. 1.5 moles of electrons will correspond to $$\displaystyle 96500 \times 1.5 = 144750$$ coloumbs of electricity.

The current is 5 amp.
The time required to electroplate calcium is $$\displaystyle =\dfrac{Q}{i}=\dfrac {144750}{5} = 28950 $$ seconds. Convert the unit of time from seconds to hours.

Time $$\displaystyle = \dfrac {28950}{3600} = 8$$ hours.
Question 4
1 / -0

An electrolysis of a oxytungsten complex ion using $$1.10\ A$$ for $$40$$ min produces $$0.838\ g$$ of tungsten. What is the charge of tungsten in the material? (Atomic weight: $$W = 184$$)

A
$$6$$

B
$$3$$

C
$$5$$

D
$$1$$

Solution

Ans :A
w = $$\dfrac{(mol.wt)it}{nF}$$
n =$$ \dfrac{184*1.10*40*60}{0.838*96500}$$
n = 6
o.s. of w = +6

Question 5

1 / -0

Consider a galvanic cell using solid $$Cu$$ and $$Fe$$ metals with their corresponding solutions.
What is the $$E^{\circ}_{cell}$$?

Standard Potential (V)	Reduction Half-Reaction
$$2.87$$	$$F_{2}(g) + 2e^{-} \rightarrow 2F^{-}(aq)$$
$$1.51$$	$$MnO_{4}^{-}(aq) + 8H^{+}(aq) + 5e^{-}\rightarrow Mn^{2+}(aq) + 4H_{2}O(l)$$
$$1.36$$	$$Cl_{2}(aq) + 3e^{-} \rightarrow 2Cl^{-}(aq)$$
$$1.33$$	$$Cr_{2}O_{7}^{2-} (aq) + 14H^{+}(aq) + 6e^{-} \rightarrow 2Cr^{3+}(aq) + 7H_{2}O(l)$$
$$1.23$$	$$O_{2}(g) + 4H^{+}(aq) + 4e^{-}\rightarrow 2H_{2}O(l)$$
$$1.06$$	$$Br_{2}(l) + 2e^{-} \rightarrow 2Br^{-}(aq)$$
$$0.96$$	$$NO_{3}^{-}(aq) + 4H^{+}(aq) + 3e^{-}\rightarrow NO(g) + H_{2}O(l)$$
$$0.80$$	$$Ag^{+}(aq) + e^{-} \rightarrow Ag(s)$$$
$$0.77$$	$$Fe^{3+} (aq) + e^{-} \rightarrow Fe^{2+}(aq)$$
$$0.68$$	$$O_{2}(g) + 2H^{+}(aq) + 2e^{-}\rightarrow H_{2}O_{2}(aq)$$
$$0.59$$	$$MnO_{4}^{-}(aq) + 2H_{2}O(l) + 3e^{-}\rightarrow MnO_{2}(s) + 4OH^{-}(aq)$$
$$0.54$$	$$I_{2}(s) + 2e^{-}\rightarrow 2I^{-}(aq)$$
$$0.40$$	$$O_{2}(g) + 2H_{2}O(l) + 4e^{-} \rightarrow 4OH^{-}(aq)$$
$$0.34$$	$$Cu^{2+}(aq) + 2e^{-} \rightarrow Cu(s)$$
$$0$$	$$2H^{+}(aq) + 2e^{-}\rightarrow H_{2}(g)$$
$$-0.28$$	$$Ni^{2+}(aq) + 2e^{-}\rightarrow Ni(s)$$
$$-0.44$$	$$Fe^{2+}(aq) + 2e^{-}\rightarrow Fe(s)$$
$$-0.76$$	$$Zn^{2+}(aq) + 2e^{-}\rightarrow Zn(s)$$
$$-0.83$$	$$2H_{2}O(l) + 2e^{-}\rightarrow H_{2}(g) + 2OH^{-}(aq)$$
$$1.66$$	$$Al^{3+}(aq) + 3e^{-}\rightarrow Al(s)$$
$$-2.71$$	$$Na^{+}(aq) + e^{-} \rightarrow Na(s)$$
$$-3.05$$	$$Li^{+}(aq) + e^{-}\rightarrow Li(s)$$

$$-0.78 V$$

$$0.10 V$$

$$0.78 V$$

$$-0.10 V$$

Solution

$$E^0$$ for Cu = 0.34 V

$$E^0$$ for Fe = -0.44V

so, $$E^0_{cell} = 0.34 - (-0.44) = 0.78 V$$

Question 6
1 / -0

How many coulombs of electricity are required for the oxidation of one mole of water to dioxygen?

A
$$9.65 \times { 10 }^{ 4 }C$$

B
$$1.93 \times { 10 }^{ 4 }C$$

C
$$1.93 \times { 10 }^{ 5 }C$$

D
$$19.3 \times { 10 }^{ 5 }C$$

Solution

$$H_2O\rightarrow \dfrac{1}{2}O_2$$ or $$H_2O\longrightarrow 2H^+ + O+ 2e^-$$

$$n_f$$ for $$O=2$$

Therefore 2 moles of electricity is required for oxidation of 1 mol of water.

Hence, charge required $$=2\times 96500\ C=1.93 \times10^5\ C$$
Question 7
1 / -0

Electrolysis of dilute aqueous NaCl solution was carried out by passing 10 mA current. The time required to liberate 0.01 moles of $$H_2$$ gas at the cathode is

A
$$9.65 \times 10^4 s$$

B
$$19.3 \times 10^4 s$$

C
$$28.95 \times 10^4 s$$

D
$$38.6 \times 10^4 s$$

Solution

0.01 moles of $$\displaystyle H_2$$ will require $$\displaystyle 2 \times 0.01 = 0.02$$ moles of electrons or $$0.02$$ faradays.

$$\displaystyle \dfrac {I(A) \times t(s)}{96500}= 0.02 $$

$$\displaystyle \dfrac { \dfrac {10 \: mA}{1000 \: mA/A}\times t(s)}{96500}=0.02 $$

$$\displaystyle t = 19.3 \times 10^4$$ s

Hence, the corret option is $$B$$
Question 8
1 / -0

The reaction taking place in the cell $$Pt|\underset {1\ atm}{H_{2}(g)}|HCl(1.0\ M)|AgCl|Ag$$ is:

A
$$AgCl + (1/2)H_{2} \rightarrow Ag + H^{+} + Cl^{-}$$

B
$$Ag + H^{+} + Cl^{-} \rightarrow AgCl + (1/2) H_{2}$$

C
$$2Ag^{+} + H_{2} \rightarrow 2Ag + 2H^{+}$$

D
$$2Ag + 2H^{+} \rightarrow 2Ag^{+} + H_{2}$$

Solution

In the given cell, $$H_{2}$$ is getting oxidised to $$H^{+}$$ and $$Ag^{+}$$ is reduced to $$Ag$$.
So, reaction taking place is $$AgCl+\cfrac{1}{2}H_{2}\rightarrow Ag+H^{+}+Cl^{-}$$
Question 9
1 / -0

The heat of combustion of ethanol in a bomb calorimeter is $$-670.48Kcal\ { mol }^{ -1 }$$ at $${25}^{o}C$$. What is $$\Delta E$$ at $${25}^{o}C$$ for the reaction?

A
-269.24 kcal

B
-469.28 kcal

C
-671.07 kcal

D
+770.48 kcal

Solution
Question 10
1 / -0

Consider an electrochemical cell in which the following reaction occurs and predict which changes will decreases the cell voltage:

$$Fe^{2+} (aq) + Ag^{+}(aq)\rightarrow Ag(s) + Fe^{3+}(aq)$$

(I) decreases the $$[Ag^{+}]$$ (II) increases in $$[Fe^{3+}]$$ (III) increase the amount of $$Ag$$

A
I only

B
II and III only

C
II only

D
I and II only

Solution

$$Fe_{(aq)}^{2+} + Ag^+_{(aq)} \rightarrow Ag_{(s)} + Fe^{3+} _{(aq)}$$

$$E_{cell} =E^o_{cell} - \cfrac{0.0591}{1} log \cfrac {[Fe^{3+}]}{[Fe^{2+}] [Ag^+]}$$

(I) On increasing $$[Fe^{3+}]$$
$$-log \cfrac{[Fe^{3+}]}{[Fe^{2+}][Ag^+]}$$ decreases

$$\because E^o_{Cell}$$ is constant.

So, $$E _{cell} $$ will decrease

(II) On decreasing $$[Ag^+]$$
$$-log \cfrac {[Fe^{3+}]}{[Fe^{2+}] [Ag^+]}$$ decreases

So, $$E_{cell} $$ will decrease

(III) Changing amount of $$Ag(s)$$ will not affect $$E_{cell}$$

Option D is correct.

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Electrochemistry Test - 60

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Electrochemistry Test - 60

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SHARING IS CARING

Question 1

$$K_2PtCl_6$$

$$PtCl_4 \cdot 2NH_3$$

$$PtCl_4 \cdot 3NH_3$$

$$PtCl_4 \cdot 5NH_3$$

Solution

Question 2

$$89.0\ kJ$$

$$-89.0\ kJ$$

$$-44.5\ kJ$$

$$-98.0\ kJ$$

Solution

Question 3

8

80

10

16

Solution

Question 4

$$6$$

$$3$$

$$5$$

$$1$$

Solution

Question 5

$$-0.78 V$$

$$0.10 V$$

$$0.78 V$$

$$-0.10 V$$

Solution

Question 6

$$9.65 \times { 10 }^{ 4 }C$$

$$1.93 \times { 10 }^{ 4 }C$$

$$1.93 \times { 10 }^{ 5 }C$$

$$19.3 \times { 10 }^{ 5 }C$$

Solution

Question 7

$$9.65 \times 10^4 s$$

$$19.3 \times 10^4 s$$

$$28.95 \times 10^4 s$$

$$38.6 \times 10^4 s$$

Solution

Question 8

$$AgCl + (1/2)H_{2} \rightarrow Ag + H^{+} + Cl^{-}$$

$$Ag + H^{+} + Cl^{-} \rightarrow AgCl + (1/2) H_{2}$$

$$2Ag^{+} + H_{2} \rightarrow 2Ag + 2H^{+}$$

$$2Ag + 2H^{+} \rightarrow 2Ag^{+} + H_{2}$$

Solution

Question 9

-269.24 kcal

-469.28 kcal

-671.07 kcal

+770.48 kcal

Solution

Question 10

I only

II and III only

II only

I and II only

Solution

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