provides an easy handle to carry the cell

Consider the table of standard reduction potentials shown below. Half-reaction $$E^{\circ}$$ $$Cl_{2} + 2e^{-}\rightarrow 2Cl^{-}$$ $$1.36\ V$$ $$O_{2} + 4H^{+} + 4e^{-} \rightarrow 2H_{2}O$$ $$1.23\ V$$ $$2H_{2}O + 2e^{-} \rightarrow H_{2} + 2OH^{-}$$ $$-0.83\ V$$ $$Rb^{+} + e^{-} \rightarrow Rb$$ $$-2.93\ V$$ Use the information from the table and your knowledge of electrochemistry to predict the CORRECT net ionic equation for the reaction that will occur when an aqueous solution of rubidium chloride undergoes electrolysis.

$$H_{2} + 2OH^{-} + 2Rb^{+} \rightarrow 2Rb + 2H_{2}O$$

$$4Cl^{-} + O_{2} + 4H^{+} \rightarrow 2H_{2}O + 2Cl_{2}$$

MCQ Questions for Class 12 Engineering Chemistry Electrochemistry Quiz 12

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.

0%
3, 4, 2, 1
0%
4, 3, 2, 1
0%
2, 3, 4, 1
0%
1, 4, 3, 2

Explanation

(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.

$Zn\, |\, Zn^{2+}(c_1)\, ||\, Zn^{2+}(c_2)\, |\, Zn$ . For this cell

$\Delta$ G is negative if:

0%
$c_1\, =\, c_2$
0%
$c_1\, >\, c_2$
0%
$c_2\, >\, c_1$
0%
None

Explanation

${ E }_{ cell }={ E }_{ cell }^{ o }-\dfrac { 0.0591 }{ 2 } log\dfrac { [{ Zn }^{ 2+ }]anode }{ [{ Zn }^{ 2+ }]cathode }$

At anode

$Zn\rightarrow { Zn }^{ 2+ }+2{ e }^{ - }$

At cathode

${ Zn }^{ 2+ }+2{ e }^{ - }\rightarrow Zn$

${ E }_{ cell }=0-\dfrac { 0.0591 }{ 2 } log\dfrac { { C }_{ 1 } }{ { C }_{ 2 } } \\ \triangle { C }_{ 1 }=-nF{ E }_{ cell }$

for

$\triangle { C }_{ 1 }=-ve$

${ E }_{ cell }=+ve\\ \Rightarrow { C }_{ 2 }>{ C }_{ 1 }$

The tarnishing of silver ornaments in atmosphere is due to:

0%
$Ag_{2}O$
0%
$Ag_{2}S$
0%
$Ag_{2}CO_{3}$
0%
$Ag_{2}SO_{4}$

Explanation

The tarnishing of silver ornaments in atmosphere is due to formation of silver oxide

$Ag_{2}O$ and silver sulphide

$Ag_{2}S$ .

$2Ag\, +\, \displaystyle \frac{1}{2}{O_{2}}\, \rightarrow\, Ag_{2}O$

$2Ag\, +\, H_{2}S\, \rightarrow\, Ag_{2}S\, +\, H_{2}$

Hence, option A and B are correct.

The electrical conductivity of a solution serves as a means of determining the end point in a chemical reaction, involved in the titration of acids, bases, or precipitation. Which of the following conductometric titration represent the curve of HCl vs

$NaOH$ :
x-axis

$\Rightarrow$ Volume of alkali added
y-axis

$\Rightarrow$ Conductivity

Explanation

Strong Acid with a Strong Base, example for.

$HCl$ with

$NaOH$ :
Before

$NaOH$ is added, a conductance is high because of the presence of highly mobile hydrogen ions. While the base is added, the conductance falls due to the replacement of hydrogen ions through the added cation as

$H^+$ ions react along with

$OH^ -$ ions to form undissociated water. That decrease in the conductance continues till the equivalence point. At the equivalence point, a solution holds just

$NaCl$ . After the equivalence point, the conductance increases because of the huge conductivity of

$OH^-$ ions.

$\Delta G^{\ominus}$ for the following reaction is:

$4A1\, +\, 3O_2\, +\, 6H_2O\, +\, 4 \overset {\ominus} OH\, \rightarrow\, 4A1(OH)_4\, ^{\ominus}$

$E^{\ominus}_{cell}\, =\, 2.73\, V$

$\Delta _fG^{\ominus}_{(\overset {\ominus} OH)}\, =\, -157\, kJ\, mol^{-1}$

$\Delta _fG^{\ominus}_{(H_2O)}\, =\, -237\, kJ\, mol^{-1}$

0%
$-3.16\, \times\, 10^3\, kJ\, mol^{-1}$
0%
$-0.79\, \times\, 10^3\, kJ\, mol^{-1}$
0%
$-0.263\, \times\, 10^3\, kJ\, mol^{-1}$
0%
$+0.263\, \times\, 10^3\, kJ\, mol^{-1}$

Explanation

As we know,

$\Delta_rG^{\ominus}_{cell}\, =\, -n_{cell}FE^{\ominus}_{cell}\, =\, -12\, \times\, 96500\, \times\, 2.73$

$=\, -3.16\, \times\, 10^3\, kJ\, mol^{-1}$
For

$n_{cell},$ write the half cell reaction as:
Cathode :

$2H_2O(l)\, +\, O_2(g)\, +\, 4e^-\, \rightarrow\, 4 \overset{\ominus}O H (aq)$
Anode :

$Al(s)\, +\, 4 \overset{\ominus}O H\, (aq)\, \rightarrow\, Al(OH)_4^{\ominus}(aq)\, +\, 3e^-$

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

$NaClO_4$ ?

0%
$1$
0%
$2$
0%
$3$
0%
$1.5$

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

$dm^{3}$ )

0%
$-1$
0%
$0$
0%
$+1$
0%
$+2$

Explanation

$O_{2}\, +\, 2H_{2}O\, +\, 4e^{-}\, \Rightarrow\, 1\, mol\, \equiv\, 4\, F$

$\Rightarrow\, \displaystyle \frac{0.6}{24}mol \, \equiv\, 4\, \times\, \displaystyle \frac{0.6}{24}\, =\, 0.1\, F$

$Cr^{3+}\, +\, xe^{-}\, \rightarrow\, Cr^{(3\, -\, x)\oplus}$

$\Rightarrow\, 1\, mol\, \equiv\, x\, F\, \Rightarrow\, \displaystyle \frac{2.6}{52} mol\, \equiv\, \frac{x}{20}F\, \Rightarrow\, x\, -\, 2$

$\Rightarrow$ Oxidation state of

$Cr^{(3\, -\, 2)}\, =\, Cr^{+1}\, =\, +1$ .

The

$E^{\ominus}\, for\, Cu^{2+}/Cu^{\oplus},\, Cu^{\oplus} / Cu,\, Cu^{2+}/Cu,$ are 0.15 V, 0.50 V, and 0.325 V, respectively. The redox cell showing redox reaction

$2Cu^+, \rightarrow\, Cu^{2+}\, +\, Cu$ is made.

$E^{\ominus}$ of this cell reaction and

$\Delta G^{\ominus}$ may be :

0%
$E^{\ominus}\, =\, 0.175\, V\, or\, E^{\ominus}\, =\, 0.350\, V$
0%
$n = 1$ or $2$
0%
$\Delta G^{\ominus}\, =\, -33.775\, kJ$
0%
All of these

Explanation

$2Cu^{\oplus}\, \rightarrow\, Cu^{2+}\, +\, Cu;$
Cell I:

$Cu^{\oplus}\, \rightarrow\, Cu^{2+}\, +\, e$

$Cu^{\oplus}\, +\, e\, \rightarrow\, Cu$

$2Cu^{\oplus}\, \rightarrow\, Cu^{2+}\, +\, Cu$

$E^{\ominus}\, =\, -0.15\, +\, 0.50$
= +

$0.35 V$
n = 1
Cell II:

$Cu\, \rightarrow\, Cu^{2+}\, +\, 2e$

$2Cu^{\oplus}\, +\, 2e\, \rightarrow\, 2Cu$

$2Cu^{\oplus}\, \rightarrow\, Cu^{2+}\, +\, Cu$

$E^{\ominus}\, =\, -0.325\, +\, 0.50$
= +

$0.175 V$
n = 2

The passage of

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

0%
$6\times {10}^{23}$
0%
$9\times {10}^{23}$
0%
$12\times {10}^{23}$
0%
$3\times {10}^{23}$

Explanation

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

$\Rightarrow$

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

$\Rightarrow$

$n(e^-)= {1.5 \ moles}$ =

$1.5 \times {6.022} \times {10^{23}}$

$\Rightarrow$

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

What happens when lead storage battery is discharged?

0%
${SO}_{2}$ is evolved
0%
Lead sulphate is consumed
0%
Lead is formed
0%
${H}_{2}{SO}_{4}$ is consumed

Explanation

The electrodes of the cells in a lead storage battery consist of lead grids. Dilute sulfuric acid

$H_2SO_4$ serves as the electrolyte. When the battery is delivering a current, i.e., discharging, the lead at the anode is oxidized. When the cell is under load, there is an electric field in the electrolyte that causes negative ions (

$SO_4^{2-}$ ) to drift toward the anode plate. During the discharge operation, acid is consumed and water is produced.

A process in which a thin film of a metal like gold, silver, etc., is deposited on another metallic article with the help of electricity.

0%
Electrorefining
0%
Electroplating
0%
Electrochemical process
0%
Galvanization

Select the correct statement.

0%
Faraday represents 96500 coulombs per second.
0%
Coulomb represents one ampere for 1/2 second.
0%
Coulomb represents 1/2 ampere for 1 second.
0%
Faraday represents charge of one mole electron.

Explanation

Faraday represents charge of one mole electron since

$1F=1C/mole$

The rusting of iron takes place as follows:

$\displaystyle 2H^{\oplus}\, +\, 2e^-\, +\, \frac{1}{2} O_2\, \rightarrow\, H_2O(l);\, E^{\ominus}\, =\, +1.23\, V$

$Fe^{2+}\, (aq)\, +\, 2e^-\, \rightarrow\, Fe(s);\, E^{\ominus}\, =\, -0.44\, V$
Calculate

$\Delta G^{\ominus}$ for the net process.

0%
$-322 kJ$ $mol^{-1}$
0%
$-152 kJ$ $mol^{-1}$
0%
$-76 kJ$ $mol^{-1}$
0%
$-161 kJ$ $mol^{-1}$

Explanation

Cathode :

$\displaystyle 2H^{\oplus}(aq)\, +\, 2e^-\, +\, \dfrac{1}{2} O_2(g)\, \rightarrow\, H_2O(l)$

$\Delta G^{\ominus}_1\, =\, -2\, \times\, FE^{\ominus}\, =\, -2\, \times\, F(1.23)$
Anode:

$Fe(s)\, \rightarrow\, Fe^{2+} (aq)\, +\, 2e^-$

$\Delta G^{\ominus}_2\, =\, -2\, \times\, FE^{\ominus}\, =\, -2\, \times\, F(0.44)$

$\displaystyle 2H^{\oplus}(aq)\, +\, \dfrac{1}{2} O_2(g)\, +\, Fe(s)\, \rightarrow\, H_2O(l)\, +\, Fe^{2+} (aq)$

$\Delta G^{\ominus}_{net}\, =\, [-2F(1.23)]\, +\, [-2F(0.44)]\, =\, -322.3\, kJ\, mol^{-1}$

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

0%
electrification
0%
electrolysis
0%
electrolyte
0%
electromagnetisation

Explanation

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.

1 Faraday can be defined as :

0%
the magnitude of the charge of 1 mole of electrons
0%
the magnitude of the electric dipole
0%
a fundamental constant of nature equal to $\displaystyle 6.63\times { 10 }^{ -34 }\ J \ s/photon$
0%
a constant that accounts for the existence of ions in solution
0%
the assignment of charges to individual atoms

Explanation

The charge of 1 mole of electrons is called the Faraday Constant, F.
It is calculated by multiplying the charge of one electron by the Avogadro constant.
1 Faraday =

$9.648 70 \times 10^{4}$ coulombs/faraday

What is the charge on an individual electron, if one

$F$ is equivalent to

$96,487$

${ C }/{ mol }{ e }$ ?

0%
$5.76\times { 10 }^{ 28 }{ C }/{ { e }^{ - } }$
0%
$6.022\times { 10 }^{ 23 }{ C }/{ { e }^{ - } }$
0%
$1.6\times { 10 }^{ 19 }{ C }/{ { e }^{ - } }$
0%
$1.6\times { 10 }^{ -19 }{ C }/{ { e }^{ - } }$

Explanation

The Faraday constant, denoted by the symbol F and named after Michael Faraday, is the magnitude of electric charge per mole of electrons.
It has the currently accepted value =

$96487 C/mole$ .

The constant F has a simple relation to two other physical constants:

$F = e. N_A$
where

$e ≈ 1.6021766×10^{−19} C$

$N_A = 6.022 \times 10^{23} mol^{-1}$

$N_A$ is the Avogadro constant (the ratio of the number of particles, N, which is unitless, to the amount of substance, n, in units of moles), and e is the elementary charge or the magnitude of the charge of an electron. This relation is true because the amount of charge of a mole of electrons is equal to the amount of charge in one electron multiplied by the number of electrons in a mole.

About an electrolyte cell, which of the following is true?

0%
An electric current causes an otherwise non-spontaneous chemical reaction to occur
0%
Reduction occurs at the anode
0%
A spontaneous electrochemical reaction produces an electric current
0%
The electrode to which electrons flow is where oxidation occurs
0%
None of the above

For a reaction, where the

$\Delta { G }^{ o }=-553.91kJ$ and

$2$ electrons are transferred, what will be the

${ E }_{ cell }^{ o }$ of the reaction?

0%
$-2.87 V$
0%
$-0.00287 V$
0%
$0.00287 V$
0%
$2.87 V$

Explanation

$\Delta G = -n F E_{cell}^o$

$-553.91 = -2\times 96500\times E_{cell}^o$

$E_{cell}^o= \dfrac {553910}{96500\times 2} = 2.87V$

Assertion : A voltaic cell spontaneously converts chemical energy into electrical energy.

Reason : A voltaic cell needs an externally applied current to work.

0%
Both Assertion and Reason are true and Reason is the correct explanation of Assertion
0%
Both Assertion and Reason are true but Reason is not the correct explanation of Assertion
0%
Assertion is true but Reason is false
0%
Assertion is false but Reason is true
0%
Both Assertion and Reason are false

The below reaction is the gold-plating process reaction.

${ Au }^{ 3+ }\left( aq \right) +3{ e }^{ - }\rightarrow Au\left( s \right)$
If

$0.600 g$ of

$Au$ is plated onto a metal, how many coulombs are used?

0%
$299 C$
0%
$868 C$
0%
$2,990 C$
0%
$8,680 C$

Explanation

$Au^{3+} + 3e^- \rightarrow Au(s)$
1 mole of

$Au^{3+}$ requires 3 mol of electrons for reduction to

$Au(s)$ .
Moles of

$Au$ in 0.600 g =

$\dfrac {0.600}{197}= 0.003$
1 mol of

$Au^{3+}$ requires 3 mol , so 0.003 mol of

$Au^{3+}$ requires

$= 3 \times 0.003 = 0.009$ moles of electrons.
1 mol of electron has = 96500 C charge
0.009 moles of electrons has =

$96500 \times 0.009 = 868 C$

In an electrochemical cell, reduction takes place at anode BECAUSE oxidation always takes place at the cathode.

0%
F, F
0%
F,T
0%
T,T
0%
T,F

How many Faraday (

$F$ ) are required for the reduction of

$1$ mole of

${ Ni }^{ 2+ }$ to

$Ni\left( s \right)$ ?

0%
$1 F$
0%
$2 F$
0%
$96, 487 F$
0%
$6.022\times { 10 }^{ 23 }F$

Explanation

$Ni^{2+} + 2e^- \rightarrow Ni$

1 mole of

$Ni^{2+}$ requires 2 moles of electrons to be reduced to

$Ni$ .

1 mole of electrons has

$1 F$ charge, hence 2 moles of electrons have

$2F$ charge.

Energy is required for the electrolysis of water because

${H}_{2}O$ has less chemical potential energy than the products formed,

${H}_{2}$ and

${O}_{2}$ .

0%
True
0%
False

Explanation

Energy is required for the electrolysis of water because

$H_2O$ has less chemical potential energy than the products,

$H_2$ and

$O_2$ .
Water reacts at the anode to form oxygen and positively charged hydrogen ions (protons).

Which of these processes could be associated with the following reaction.

$\displaystyle 2{ H }_{ 2 }O\rightarrow 2{ H }_{ 2 }+{ O }_{ 2 }$ .
i. Electrolysis
ii. Neutralization
iii. Decomposition

0%
i only
0%
iii only
0%
i and iii only
0%
i and ii only
0%
ii and iii only

Explanation

Electrolysis of water is its decomposition to give hydrogen and oxygen gases due to the passage of an electric current.

$2H_2O + energy \rightarrow O_2 + 2H_2$

A salt bridge :

0%
is not used in a galvanic cell
0%
completes the circuit
0%
increases the rate of reaction
0%
provides an easy handle to carry the cell
0%
None of these

The standard Gibbs free energy change

$(\triangle G^{\circ}$ in

$kJ\ mol^{-1})$ , in a Daniel cell

$(E_{cell}^{\circ} = 1.1V)$ , when

$2$ moles of

$Zn(s)$ is oxidized at

$298\ K$ , is closest to:

0%
$-212.3$
0%
$-106.2$
0%
$-424.6$
0%
$-53.1$

Explanation

$\Delta G=-nFE^o_{cell}$

$=-2\times 96500\times1.1=-212.3\ KJ$

Hence, option C is correct.

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

0%
$K_2PtCl_6$
0%
$PtCl_4 \cdot 2NH_3$
0%
$PtCl_4 \cdot 3NH_3$
0%
$PtCl_4 \cdot 5NH_3$

Explanation

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

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

$Ca = 40$ ]

0%
8
0%
80
0%
10
0%
16

Explanation

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.

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%
$-0.78 V$
0%
$0.10 V$
0%
$0.78 V$
0%
$-0.10 V$

Explanation

$E^0$ for Cu = 0.34 V

$E^0$ for Fe = -0.44V

so,

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

A positive cell potential tells you ________.

0%
the reaction is not spontaneous
0%
the reaction is spontaneous
0%
the reaction will not occur without intervention
0%
$\Delta G$ is negative
0%
both the reaction is spontaneous and $\Delta G$ is negative

Explanation

$\Delta G=- nFE_{cell}^o$
A positive cell potential tells you that both the reaction is spontaneous and

$\Delta G$ is negative.

Consider the table of standard reduction potentials shown below.

Half-reaction	$E^{\circ}$
$Cl_{2} + 2e^{-}\rightarrow 2Cl^{-}$	$1.36\ V$
$O_{2} + 4H^{+} + 4e^{-} \rightarrow 2H_{2}O$	$1.23\ V$
$2H_{2}O + 2e^{-} \rightarrow H_{2} + 2OH^{-}$	$-0.83\ V$
$Rb^{+} + e^{-} \rightarrow Rb$	$-2.93\ V$

Use the information from the table and your knowledge of electrochemistry to predict the CORRECT net ionic equation for the reaction that will occur when an aqueous solution of rubidium chloride undergoes electrolysis.

0%
$2Rb^{+} + 2Cl^{-} \rightarrow 2Rb + Cl_{2}$
0%
$2H_{2}O \rightarrow 2H_{2} + O_{2}$
0%
$H_{2} + 2OH^{-} + 2Rb^{+} \rightarrow 2Rb + 2H_{2}O$
0%
$4Cl^{-} + O_{2} + 4H^{+} \rightarrow 2H_{2}O + 2Cl_{2}$

Explanation

From the given table, we have;

Rb as the lowest reduction potential therefore it undergo oxidation

and

$Cl_2$ have the highest oxidation potential therefore it undergo reduction.

so, the net reaction becomes

$2Rb^+ + 2Cl^- \rightarrow 2Rb + Cl_2$

A steady current of

$10.0$ Amps is passed through a nickel production cell of

$15$ minutes.
Which of the following is the correct expression for calculating the number of grams of nickel produced?
Note:

$\bullet 1\ faraday = 96,500\ Coulombs$

$\bullet$ The electroytic cell involves the following half-reaction:

$Ni_{(aq)}^{2+} + 2e^{-} \rightarrow Ni_{(s)}$

0%
$\dfrac {(10.0)(15)(96,500)}{(59)(60)} g$
0%
$\dfrac {(10.0)(15)(59)}{(60)(96,500)} g$
0%
$\dfrac {(10.0)(15)(60)(59)}{(96,500)(2)} g$
0%
$\dfrac {(96,500)(59)}{(10)(15)(60)(2)} g$

Explanation

According to Faraday's Law

$W= \cfrac {Z \times I \times t}{F}$

$W$ = mass of compound

$Z$ = Equivalent mass

$F$ = 96500

$I$ = Current in Ampere

$t$ = time in seconds

$\Rightarrow W= \left(\cfrac {59}{2}\right)\cfrac {\times 15\times 60 sec\times 10A}{96500}$

$\Rightarrow W= \cfrac {(10.0)(15)(60)(59)}{(96500)(2)}g$

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

0%
$9.65 \times { 10 }^{ 4 }C$
0%
$1.93 \times { 10 }^{ 4 }C$
0%
$1.93 \times { 10 }^{ 5 }C$
0%
$19.3 \times { 10 }^{ 5 }C$

Explanation

$H_2O\rightarrow \dfrac{1}{2}O_2$ 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

$=2\times 96500\ C=1.93 \times10^5\ C$

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

0%
$9.65 \times 10^4 s$
0%
$19.3 \times 10^4 s$
0%
$28.95 \times 10^4 s$
0%
$38.6 \times 10^4 s$

Explanation

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

$B$

$250.0\ ml$ sample of a

$0.20\ M\ Cr^{3+}$ is electrolysed with a current of

$96.5\ A$ . If the remaining

$[Cr^{3+}]$ is

$0.1\ M$ the duration of process is ?

0%
$25\ sec$
0%
$50\ sec$
0%
$100\ sec$
0%
$75\ sec$

Two half-cells have potentials

$-0.44$ and

$0.799$ volt respectively. These two are coupled to make a galvanic cell. Which of the following will be true?

0%
Electrode of half-cell potential $-0.44\ V$ will act as anode
0%
Electrode of half-cell potential $-0.44\ V$ will act as cathode
0%
Electrode of half-cell potential $0.799\ V$ will act as anode
0%
Electrode of half-cell potential $-0.44\ V$ will act as a positive terminal

Explanation

Galvanic Cell:-

$\rightarrow$ In Galvanic cell,left hand side (anode), oxidation occurs. At right hand side electrode(cathode) , reduction occurs.

$\rightarrow$ The sum of the oxidation half reaction at anode and reduction half reaction at cathode is the overall cell reaction.

$\rightarrow$ Electrode with positive value of electrode potential suggest forward reaction i.e. reduction at cathode. So, electrode with 0.799 volt will act as cathode.

$\rightarrow$ Electrode with negative value of electrode potential suggest reverse reaction i.e. oxidation at anode. Hence, electrode with -0.44 volt will act as anode.

$W$

$g$ of

$Ag$ is deposited at the cathode of one electrolytic cell due to the passage of

$1A$ of current for

$1h$ . The time required for passage of current to deposit

$W$

$g$ of

$Mg$ by the same value of current is:

0%
$3.0h$
0%
$9h$
0%
$2.5h$
0%
$1h$

The reaction taking place in the cell

$Pt|\underset {1\ atm}{H_{2}(g)}|HCl(1.0\ M)|AgCl|Ag$ is:

0%
$AgCl + (1/2)H_{2} \rightarrow Ag + H^{+} + Cl^{-}$
0%
$Ag + H^{+} + Cl^{-} \rightarrow AgCl + (1/2) H_{2}$
0%
$2Ag^{+} + H_{2} \rightarrow 2Ag + 2H^{+}$
0%
$2Ag + 2H^{+} \rightarrow 2Ag^{+} + H_{2}$

Explanation

In the given cell,

$H_{2}$ is getting oxidised to

$H^{+}$ and

$Ag^{+}$ is reduced to

$Ag$ .

So, reaction taking place is

$AgCl+\cfrac{1}{2}H_{2}\rightarrow Ag+H^{+}+Cl^{-}$

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:

$W = 184$ )

0%
$6$
0%
$3$
0%
$5$
0%
$1$

Explanation

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

$0.50\ L$ of a

$0.60\ M\ SnSO_{4}$ solution is electrolysed for a period of

$30.0$ min using a current of

$4.60\ A$ . If in electrodes are used, what is the final concentration of

$Sn^{2+}$ remaining in the solution?

$[at. wt.\ of\ Sn = 119]$

0%
$0.257\ M$
0%
$0.544\ M$
0%
$0.189\ M$
0%
$0.514\ M$

Explanation

Ans : A

Given : 0.6 mol in 1 L

then in 0.5 L = 0.3 mol of

$Sn^{2+}$

Mass deposited W = zit

$\dfrac{119}{2\times96500}\times4.6\times1800$

= 5.1 g

Rest mass in solution = 30.6 gm

Now molority of solution =

$\dfrac{30.6}{119\times1}$

= o.257 M

Calculate the potential of an indicator electrode, which originally contained

$0.1 M \ MnO_4^-$ and

$1.72 M \ H^+ ,$ and was treated with

$Fe^{2+} ,$ necessary to reduce 90% of

$KMnO_4$ to

$Mn^{2+}$ .

Given:

$[E^0_{MnO_4^-} /_{ Mn^{+2} } = 1.51 V]$

0%
$1.4 V$
0%
$1.5 V$
0%
$1.6 V$
0%
$1.3 V$

The density of copper is

$8 gm/cc.$ Number of coulumbs required to plate an area of

$10 cm \times 10 cm$ on both sides to a thickness of

$10^{-2} cm$ using

$CuSO_4$ solution as electrolyte is:

0%
$24,250$
0%
$48,124$
0%
$96,500$
0%
$10,000$

Consider the following half-cell reactions and associated standard half-cell potentials and determine the maximum voltage that can be obtained by combination resulting in spontaneous processes:

$AuBr_{4}^{-}(aq) + 3e^{-} \rightarrow Au(s) + 4Br^{-}(aq); E^{\circ} = -0.86\ V$

$Eu^{3+}(aq) + e^{-} \rightarrow Eu^{2+}(aq); E^{\circ} = -0.43\ V$

$Sn^{2+}(aq) + 2e^{-} \rightarrow Sn(s); E^{\circ} = -0.14\ V$

$IO^{-}(aq) + H_{2}O(l) + 2e^{-}\rightarrow I^{-}(aq) + 2OH^{-}(aq); E^{\circ} = + 0.49\ V$

0%
$+ 0.72$
0%
$+ 1.54$
0%
$+1.00$
0%
$+ 1.35$

Explanation

$\begin{array}{l} AuBr_{ 4 }^{ - }\left( { aq } \right) +3{ e^{ - } }\to Au\left( s \right) +4B{ r^{ - } }\left( { aq } \right) ;\, \, E^{ \circ }=-0.86V\to Anode \\ I{ O^{ - } }\left( { ar } \right) +{ H_{ 2 } }O\left( l \right) +2{ e^{ - } }\to { I^{ - } }\left( { aq } \right) +2O{ H^{ - } }\left( { aq } \right) ;E^{ \circ }=+0.49\to cathode \\ E{ ^{ \circ }_{ cell } }=E{ ^{ \circ }_{ cathode } }-E{ ^{ \circ }_{ anode } }\Rightarrow 0.49-\left( { -0.86 } \right) \Rightarrow +1.35V \end{array}$

Consider the half-cell reduction reactions :

${ Mn }^{ 2+ }+{ 2e }^{ - }\rightarrow Mn,{ E }^{ o }=-1.18\ V$

${ Mn }^{ 2+ }\rightarrow { 2n }^{ 3+ }+{ e }^{ - },{ E }^{ o }=-1.51\ V$
The

${ E }^{ o }$ for the reaction

${ 3Mn }^{ 2+ }\rightarrow { Mn }^{ 0 }+{ 2Mn }^{ 3+ }$ and possibility of the forward reaction are respectively:

0%
-4.18 V and yes
0%
+0.33 V and yes
0%
+2.69 V and no
0%
-2.69 V and no

Explanation

The standard electrode potential of reaction can be calculated as,

$E_{ 0 }=E_{ R }−E_{ P }$

Where,

$E_{ R }$ is the SRP of the reactant,

$E_{ P }$ is the SRP of the product.

${ Mn }^{ 3+ }→Mn^{ 2+ }$ this reaction takes place with the

$E_{ 0 }-1.51 V$

$Mn^{ 2+ }→Mn$ this takes place with

$E_{ 0 } -1.18 V$

$E_{ 0 }=−1.51 V−1.81 V$

$=-2.69V$

and the reaction is will not occurs due to spontaneity.

A current of

$i$ ampere was passed for

$t$ sec, through three calls

$P,Q$ and

$T$ connected in series. These contain respectively silver nitrate, mercuric nitrate, and mercurous nitrate. At the cathode of the cell

$P,0.216g$ of

$Ag$ was deposited. The weights of mercury deposited in the cathode of

$Q$ and

$R$ respectively are: (at wt. of

$Hg=200.59$ )

0%
$0.4012$ and $0.8024g$
0%
$0.4012$ and $0.2006g$
0%
$0.2006$ and $0.4012g$
0%
$0.1003$ and $0.2006g$

Explanation

Cells are connected in series therefore same amount of current will Pass through them:

Faraday's second law:

$\dfrac{W_1}{W_2}=\dfrac{E_1}{E_2}$

(a) For Mercuric nitrate

$(Hg^{2+})$

$\dfrac{W_1}{(W_2)Ag}=\dfrac{E_1}{(E_2)Ag}$

$W_1=\dfrac{200.59}{2\times 108 }\times 0.216$

$=0.2006g$

(b) For mercurous nitrate

$(Hg^+)$

$\dfrac{W_1}{(W_2)Ag}=\dfrac{E_1}{(E_2)Ag}$

$(\therefore E_{Ag}=108)$

$W_1=\dfrac{200.59}{108}\times 0.216$

$=0.4012g$

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?

0%
-269.24 kcal
0%
-469.28 kcal
0%
-671.07 kcal
0%
+770.48 kcal

Electrolysis can be used to determine atomic masses. A current of

$0.550\ A$ deposits

$0.55$ . If a certain metal in

$100\ minutes$ . Calculate the atomic mass of the metal if eq. mass

$-$ mole mass

$/ 3$ .

0%
$100$
0%
$45.0$
0%
$48.25$
0%
$144.75$

Explanation

According to Faraday's law of Electrolysis

$W=(\cfrac { M }{ nf } ).\cfrac { I.t }{ F }$

$0.55ℊ=(\cfrac { M }{ 3 } )\times 0.550\times (100\times 60)s$

$M=48.25ℊ$

Thus atomic mass of metal is

$48.25$

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

$Ag$

0%
I only
0%
II and III only
0%
II only
0%
I and II only

Explanation

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

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

(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.

What will be

$\Delta H$ for the reaction;

$Ag(s)+\dfrac{1}{2}Hg_2Cl_2(s)\rightarrow AgCl(s)+Hg(l)$ at

$25^0c$ , if this reaction can be conducted in a cell for which the emf = 0.0455 volt at this temperature with temperature coefficient

$3.389\times10^{-4} volt\, deg^{-1}$ ?

0%
+1280 cal
0%
+640 cal
0%
-1280 cal
0%
-640 cal

Explanation

$Ag\left( s \right) +1/2{ Hg }_{ 2 }{ Cl }_{ 2 }\left( s \right) \longrightarrow AgCl\left( s \right) +Hg\left( l \right)$

$E=0.0455$ volt

$\dfrac { dE }{ dT } =3.389\times { 10 }^{ -4 }$ Volt

${ deg }^{ -1 }$

We know that,

$-nFE=\Delta H-nFT\left( \dfrac { dE }{ dT } \right)$

$\Rightarrow \quad E=-\dfrac { \Delta H }{ nF } +T\left( \dfrac { dE }{ dT } \right)$

$\Rightarrow \quad 0.0455=-\dfrac { \Delta H }{ 1\times 96485 } +298\times 3.389\times { 10 }^{ -4 }$

$\Rightarrow \quad \Delta H=+5354.16J=+1280Cal$

${ E }^{ 0 }$ value of

$Mg/{ Mg }^{ +2 }$ is

$+2.37$ V of

$Zn/{ Zn }^{ +2 }$ is

$+0.76$ V and that of

$Fe/{ Fe}^{ +2 }$ is

$+0.44$ V, which of the following statements is correct?

0%
Zinc will reduce ${ Mg }^{ +2 }$
0%
Zn will reduce ${ Fe }^{ +2 }$
0%
Mg oxidizes Fe
0%
Zn oxidizes Fe

CBSE Questions for Class 12 Engineering Chemistry Electrochemistry Quiz 12 - MCQExams.com

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Practice Class 12 Engineering Chemistry Quiz Questions and Answers

CBSE Questions for Class 12 Engineering Chemistry Electrochemistry Quiz 12 - MCQExams.com

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Practice Class 12 Engineering Chemistry Quiz Questions and Answers

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