accepts electrons from hydrogen ions

donates electrons to sulphate ions

accepts electrons from sulphate ions

should able to form positive and negative ions

should be able to conduct charge without dissociating

positive ions move towards the negative electrode and negative ions towards the positive electrode.

positive ions move towards the positive electrode and negative ions towards the negative electrode.

both ions move towards both the electrodes in equal amounts until they are balanced.

$$E_{cell} > 0; \triangle G < 0$$

$$E_{cell} > 0; \triangle G^{\circ} < 0$$

MCQ Questions for Class 12 Engineering Chemistry Electrochemistry Quiz 8

How many electrons constitute current of one micro ampere in one second?

0%
$6.25\times10^8$
0%
$6.25\times10^{12}$
0%
$6.25\times10^9$
0%
$6.25\times10^{15}$

Explanation

Given: Current ,

$I = 1\mu A = 10^{-6}A$
We know that

$Q = ne, t = 1s$
Where

$n$ is the number of electrons and e is the charge on

$1$ electron and

$e = 1.6\times10^{-19}C$

$n = \displaystyle\frac{I}{e} = \displaystyle\frac{10^{-6}}{1.6\times10^{-19}} = 6.25\times10^{12} Electrons$

The weights of various ions liberated by passing the same current through different electrolytes are in the ratio of their:

0%
Atomic weights
0%
Equivalent weights
0%
Molecular weights
0%
Atomic numbers

Lead pipes are readily corroded by:

0%
water
0%
acetic acid
0%
conc. $H_2SO_4$
0%
dil. $H_2SO_4$

If copper is kept open in air, it slowly loses its shining brown surface and gains a green coating. It is due to the formation of:

0%
$CuCO_3$
0%
$CuSO_4$
0%
$CuCO_3.Cu(OH)_2$
0%
$CuO$

Explanation

Copper reacts with

$CO_2$ present in the air and forms a green coating on its surface due to the formation of basic copper carbonate.

$2Cu+O_2+CO_2+H_2O\rightarrow CuCO_3.Cu(OH)_2$

So, the correct option is

$C$

$1$ faraday

$= \ ?$

0%
$10000\ C$
0%
$95000\ C$
0%
$96.5 \ C$
0%
$96500\ C$

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 accepted value $96485.34$ coulombs in round figure $96500$ coulombs.

The hydrogen formed by the action of zinc on sulphuric acid forms a layer of gas on the surface of copper plate in a simple voltaic cell, because of which

0%
Internal resistance of the cell increases
0%
Hydrogen ions are unable to reach the anode
0%
Polarisation takes place at the anode
0%
All of the above

Explanation

The hydrogen formed by the action of zinc on sulphuric acid forms a layer of gas on the surface of a copper plate in a simple voltaic cell, because of which:-

$\rightarrow$ Internal resistance of cell increases

$\rightarrow$ Hydrogen ions are unable to reach the anode

$\rightarrow$ polarisation takes place at an anode.

For purification of impure metals by the process of electrolysis

0%
Salt solution of the impure metal is taken as an electrolyte
0%
Salt solution of that metal in pure form is taken as an electrolyte
0%
That metal in pure form is made as anode
0%
Both $(1)$ and $(3)$

Which of the following methods is employed to prevent rusting?

0%
Refining
0%
Alloying
0%
Greasing
0%
both (b) and (c)

Explanation

The rusting of iron can be prevented by greasing or alloying but refining is method to obtain pure metal from it's ore, thus, option

$D$ is correct.

During electroplating by electrolysis, the article to be electroplated is:

0%
Kept immersed in the electrolyte at the bottom.
0%
Made the cathode of an electrolytic cell.
0%
Made the anode of an electrolytic cell.
0%
Made the electrolyte of the electrolytic cell.

$m$ is the mass of silver deposited at the cathode by

$2 A$ current flowing for

$25 min$ through a silver voltameter, then the mass deposited by

$1.5 A$ current flowing for

$600 s$ is _________.

0%
$m$
0%
$3m$
0%
$0.3m$
0%
None

Explanation

$\displaystyle \frac {{m}'}{m} = \frac {{I}'{t}'}{It} = \frac {1.5 \times 600}{2 \times 25 \times 60} = \frac {900}{3000} = \frac {3}{10} = 0.3$

$\Rightarrow {m}'=0.3m$

The liberation of

$1$ mole of chlorine requires how many Faradays of electricity?

0%
$0.5$ Faraday
0%
$0.25$ Faraday
0%
$2$ Faradays
0%
$4$ Faradays

Explanation

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

$Q={n(e^-)}\times {F}$
2mol electrons are liberated in production of 1mol of chlorine.
therefore electricity required for the process

$\Rightarrow$

$= {2}\times {F}$ =

$2$ Faradays

In an electrolyte,

$3.2 \times 10^{18}$ bivalent positive ions drift to the right per second while

$3.6 \times 10^{18}$ monovalent negative ions drift to the left per second. The current through the electrolyte is _______.

0%
$0.45 A$ to the left
0%
$0.45 A$ to the right
0%
$1.6 A$ to the right
0%
$1.6 A$ to the left

Explanation

Current

$I =$ net charge transported/second

$= 3.2 \times 10^{18} \times 2e + 3.6 \times 10^{18} \times e$

$= (6.u + 3.6) \times 10^{18} \times 1.6 \times 10^{-19}$

$= 10 \times 1.6 \times 10^{-1}$

$= 1.6 A$ to the right.

The process of lead action in a simple voltaic cell

0%
Catalyses the process of electrolysis
0%
Increases the efficiency of the cell
0%
Decreases the efficiency of the cell
0%
Increases the charge carriers

The U shaped vessel used in Nelson cell is lined inside with asbestos which:

0%
helps in the separation of cathode and anode compartments.
0%
acts as an insulator.
0%
helps $H^{+}$ ions to reduced to $H_{2}$ gas.
0%
allows the dissociation of steam.

What is the voltage of the voltaic cell

$Zn|Zn^{2+}||Cu^{2+}|Cu$ at

$298\ K$ if

$[Zn^{2+}] = 0.2\ M$ and

$[Cu^{2+}] = 4.0\ M$ ?

$Cu^{2+} + 2e^{-}\rightarrow Cu\ E^{\circ} = +0.34V$

$Zn^{2+} + 2e^{-}\rightarrow Zn\ E^{\circ} = -0.76\ V$
[Note:

$E = E^{\circ} - (0.0591/n)(\log Q)$

0%
$+1.10\ V$
0%
$-1.10\ V$
0%
$+1.07\ V$
0%
$+1.14\ V$
0%
$-1.07\ V$

Explanation

The representation of the voltaic cell shows that

$Zn$ is oxidised to

$Zn^{2+}$ , whereas,

$Cu^{2+}$ is reduced to

$Cu$ .

These two phenomenons can be written by two equations-

$Cu^{2+}$

$+$

$2e^-$

$\rightarrow$

$Cu$ .....

$E^o$

$=$

$+0.34V$ , (since

$Cu^{2+}$ is reduced).....

$eq(1)$

and

$Zn^{2+}$

$+$

$2e^-$

$\rightarrow$

$Zn$ .....

$E^o$

$=$

$-0.76V$

But the voltaic cell represents that

$Zn$ is oxidised to

$Zn^{2+}$ , hence the equation must be written as-

$Zn$

$\rightarrow$

$Zn^{2+}$

$+$

$2e^-$ .....

$E^o$

$=$

$+0.76V$ , (since

$Zn$ is oxidised)......

$eq(2)$

Adding

$eq(1)$ and

$eq(2)$ , we get,

$Cu^{2+}$

$+$

$Zn$

$+$

$2e^-$

$\rightarrow$

$Cu$

$+$

$Zn^{2+}$

$+$

$2e^-$ ......

$E^o$

$=$

$0.34+0.76$

$=$

$1.10V$

Or,

$Cu^{2+}$

$+$

$Zn$

$\rightarrow$

$Cu$

$+$

$Zn^{2+}$ .....

$E^o$

$=$

$+1.10V$

We also know the formula-

$E$

$=$

$E^o$

$-$

$\dfrac{0.0591}{n}$

$(logQ)$

Q can be written as

$Q$

$=$

$\dfrac{[Zn^{2+}]}{[Cu^{2+}]}$ and

$n$ is the number of electrons, that is

$2$

Putting the values of all the known quantities in the given equation we get,

$E$

$=$

$1.10$

$-$

$\dfrac{0.0591}{2}$

$log$

$\dfrac{[0.2]}{[4.0]}$

Thus,

$E$

$=$

$1.10-(-0.038)$

$E$

$=$

$1.138V$ , that is

$1.14V$

Option D is the correct answer.

When current is passed through molten sodium chloride

0%
sodium is deposited at the positive electrode and chlorine gas is formed at the negative electrode
0%
sodium is evaporated and chloride ions and formed at the negative electrode
0%
sodium is deposited at the positive electrode and chlorine is deposited at the negative electrode
0%
sodium is deposited at the negative electrode and the chlorine gas is formed at the positive electrode

Explanation

The electrolysis of sodium chloride

$(NaCl)$ produces the positive ion,

$Na^+$ and negative ions,

$Cl^-$ . Thus, the positive ion of sodium will be deposited at the negative electrode (cathode) and the chlorine ion will be deposited at the positive electrode (anode).

State True or False.

Pure water can be used as an electrolyte in the process of electrolysis.

0%
True
0%
False

When copper rod donates electrons to hydrogen ions, it gains ................. charge.

0%
positie
0%
negative
0%
no charge
0%
can't say

Copper electrode

0%
donates electrons to hydrogen ions
0%
accepts electrons from hydrogen ions
0%
donates electrons to sulphate ions
0%
accepts electrons from sulphate ions

In electrolytic solutions, carrier of charge is

0%
proton
0%
electron
0%
neutron
0%
ion

An electrolyte

0%
has positive charge
0%
has negative charge
0%
should be able to conduct charge without dissociating
0%
should able to form positive and negative ions

In electrolysis

0%
positive ions move towards the positive electrode and negative ions towards the negative electrode.
0%
positive ions move towards the negative electrode and negative ions towards the positive electrode.
0%
both ions move towards both the electrodes in equal amounts until they are balanced.
0%
none of the above

State True or False.

The byproducts of electrolysis of impure water are hydrogen and oxygen.

0%
True
0%
False

Explanation

Cell reactions during electrolysis of water are:

Anode:

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

Cathode:

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

Reaction products are thus hydrogen and oxygen gas.

option A is correct .

On electrolysis, water splits into

0%
positively charged hydrogen ions and negatively charged oxygen ions.
0%
negatively charged hydrogen ions and positively charged oxygen ions.
0%
hydrogen and oxygen atoms having positive and negative charges respectively.
0%
hydrogen and oxygen atoms having negative and positive charges respectively.

Explanation

The electrolysis of water

$(H_2O)$ produces the positive ion of hydrogen atom

$(H^+)$ and negative ion of oxygen

$(O^-)$ .

In the diagram above, what is the half reaction that occurs at the cathode?

0%
$Al \rightarrow Al^{3+} + 3e^-$
0%
$Ni^{2+} + 2e^- \rightarrow Ni$
0%
$Ni \rightarrow Ni^{2+} + 2e^-$
0%
$2Al^{3+} + 6e^- \rightarrow 2Al$
0%
None of these

Explanation

The half-reaction that occurs at the cathode is the reduction of Ni(II) ions to Ni atoms.

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

The gain of electrons is called reduction. Cathode is the electrode where reduction occurs. In the electrochemical series, Al (III) has more negative standard reduction potential than Ni (II).

Hence, Al is oxidized and Ni(II) is reduced.

Sulphate ions move towards

0%
copper electrode
0%
battery
0%
electrolyte
0%
zinc electrode

Electrolysis of a dilute solution of aqueous sodium chloride results in the cathode product:

0%
Sodium
0%
Hydrogen
0%
Chlorine
0%
Oxygen
0%
Peroxide

Explanation

In electrolysis of aqueous sodium chloride, both

$H^+$ and

$Na^+$ will compete with each other for reduction and as the reduction potential of

$H^+$ is more compared to

$Na^+$ , so

$H^+$ will get reduced first at cathode and hydrogen gas will be liberated at cathode.

For

$Cu(s) + NO_3^- (aq) + H^+(aq) \rightarrow Cu^{2+}(aq) + NO_2(g) + H_2O(l)$ , which of the following takes place?

0%
$Cu(s)$ is oxidized
0%
$H^+ (aq)$ is oxidized
0%
$Cu(s)$ is reduced
0%
$H^+ (aq)$ is reduced
0%
$NO_3^-$ is oxidized.

Explanation

Oxidation state of

$Cu$ changes from

$0$ to

$+2$

Oxidation state of

$N$ changes from

$+5$ to

$+4$

Oxidation state of

$H^+$ remains same

therefore,

$Cu$ is getting oxidised and

$NO_3^-$ is getting reduced.

Whats the potential of the reaction below given the half-reaction potentials:

$2Fe_2+ + Cl_2 \rightarrow 2Fe^{3+} + 2Cl^-$

$Fe^{3+} + e^- \rightarrow Fe^{2+}; E = 0.77 V$

$Cl_2 + 2e^- \rightarrow 2Cl^- ; E = 1.36 V$

0%
$0.18 V$
0%
$0.59 V$
0%
$1.05 V$
0%
$2.13 V$
0%
$2.90 V$

Explanation

$2Fe^{2+}+Cl_2\rightarrow 2Fe^{3+}+2Cl^-$

$E^0_{reaction}=E^0_{cathode}-E^0_{anode}$

$=1.36-0.77$

$=0.59V$

In a double bond connecting two atoms there is sharing of:

0%
2 electrons
0%
4 electrons
0%
1 electron
0%
none

Which of the following statements is/are not correct?

0%
$Zn-Cu$ cell is called Daniell cell
0%
Rust is $Fe_{2}O_{3}$
0%
Saline water slows down rusting
0%
Pure metals undergo corrosion faster than impure metals

Explanation

$(A)$

$Zn-Cu$ cell is Daniell cell

oxidation

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

${ Red }^{ n }$

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

$(B)$

$4Fe+3{ O }_{ 2 }+2n{ H }_{ 2 }O\longrightarrow 2{ Fe }_{ 2 }{ O }_{ 3 }.n{ H }_{ 2 }O$

rust

$(C)$ Saline water increase rusting as salt is strong electrolyte and ions present increases the reaction.

$(D)$ Impure metals undergo rusting faster.

I. Cathode is

$-ve$ terminal both in electrochemical and electrolytic cells.
II. Reduction occurs at cathode both in galvanic as well as electrolytic cells.
III. Chemical change in electrolytic cell is non-spontaneous.

Which of the above statements are correct?

0%
All the statements are correct
0%
II and III are correct
0%
I and III are correct
0%
Only II is correct

In electrolysis of very dilute

$NaOH$ solution using platinum electrodes:

0%
$H_{2}$ is evolved at cathode
0%
$H_{2}$ is produced at anode
0%
$Na$ is obtained at cathode
0%
$O_{2}$ is produced at anode

Explanation

In aqueous NaOH solution, on electrolysis, we have the reduction potential of

$H^+$ is much more than

$Na^+$ , so the reaction takes place as;

At anode:

$4OH^- \rightarrow 2H_2O + O_2 + 4e^-$

At cathode:

$4H^+ +4e^- \rightarrow 2H_2$

The main factors which affect corrosion are :

0%
Position of metal in electrochemical series
0%
Presence of $CO_{2}$ in water
0%
Presence of impurities coating
0%
All are correct

Explanation

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,

What is the sign of

$\triangle G^{\circ}$ and the value of

$K$ for a electrochemical cell for which

$E^{\circ}_{cell} = 0.80\ V$ ?

0%
$\triangle G^{\circ} = -, K = > 1$
0%
$\triangle G^{\circ} = +, K = > 1$
0%
$\triangle G^{\circ} = +, K = < 1$
0%
$\triangle G^{\circ} = -, K = < 1$

Explanation

$\rightarrow$ 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

$K>1$ .

During the electrolysis of

$AgNO_{3}$ (using

$Pt$ electrodes) concentration around cathode as well as anode falls from

$4M$ to

$3M$ . What will happen if this happened with

$Ag$ electrodes?

0%
Result will remain same
0%
Concentration around cathode will fall from $4M$ to $3M$ but around anode will increase from $4M$ to $5M$
0%
Reverse of statement (b)
0%
Concentration increases from $4M$ to $5M$ on both the electrodes

Which of the following is/are correct about 1 mole electrons?

0%
$6.023 \times 10^{23}$ electrons
0%
$5.48 \times 10^{-7}\, kg$
0%
96500 coulomb charge
0%
None of these

Explanation

1 mole is having

$N_A$ number of particles

so, 1 mole of electron have

$6.023 * 10^{23}$ electrons

mass of 1 electron =

$9.1 * 10^{-31}$ Kg

so, mass of 1 mole of electron=

$9.1 * 10^{-31}$ Kg

$*$

$6.023 * 10^{23}$ =

$5.48 * 10^{-7}$ kg

and 1 mole of electron = 96500 C

The amount of charge which the battery must have been used is:

0%
$9.88\ F$
0%
$8.98\ F$
0%
$8.89\ F$
0%
$7.88\ F$

Explanation

wt. of

$H_2SO_4$ at initial = 3.5*1000*.39*1.294 = 1766.31 gm

wt. of

$H_2SO_4$ after decay =3.5*1000*1.3*0.20 = 797.3 gm

amount deposite = 969 gm

n-fector of

$H_2SO_4$ = 1

w = zQ

Q = 969.0F

$\dfrac{1}{98}$

Q = 9.88F

For a spontaneous reaction

$\Delta {G}^{o}$ and

${E}^{o}$ cell will be respectively

0%
$-ve$ and $+ve$
0%
$+ve$ and $-ve$
0%
$+ve$ and $+ve$
0%
$-ve$ and $-ve$

Explanation

For a spontaneous reaction, the standard change in Gibbs free energy

$\displaystyle \Delta G^0$ and the standard cell potential

$\displaystyle E^0_{cell}$ will be negative and positive respectively.

$\displaystyle \Delta G^0 < 0$

$\displaystyle E^0_{cell} > 0$

Note:

$\displaystyle \Delta G^0 = -nFE^0_{cell}$

Thus,

$\displaystyle \Delta G^0$ and

$\displaystyle E^0_{cell}$ have opposite signs.

Pick out the pair in which the energy change of one is reverse of the energy change in the other :
(1) radio (2) fluorescent lamp (3) toaster (4) photoelectric cell

0%
1 and 2
0%
2 and 3
0%
2 and 4
0%
3 and 4

1 C electricity deposits:

0%
half of electrochemical equivalent of $Ag$
0%
electrochemical equivalent of $Ag$
0%
96500 g of $Ag$
0%
10.8 g of $Ag$

Explanation

1 C electricity deposits electrochemical equivalent of

$Ag$ .

Note:

Electrochemical equivalent is the mass of the ion deposited by passing a current of one ampere for one second (i.e, one coulomb).

According to Faraday's first law of electrolysis, the amount of substance that undergoes oxidation or reduction at each electrode during electrolysis is directly proportional to the amount of electricity that passes through the cell.

The mass and charge of 1 mole electrons will be:

0%
1 kg; 96500 C
0%
0.55 mg; 96500 C
0%
1.55 mg; 96500 C
0%
5.5 mg; 96500 C

In which of the following case, increase on concentration of ion cause increase in

$E_{cell}$ ?

0%
$Pt(H_2)H^+(aq)$
0%
$Pt|Quinhydrone|H^+(aq)$
0%
$Ag|Ag^+(aq)$
0%
$Ag, AgCl|Cl^-(aq)$

Explanation

(a)

$\displaystyle\frac{1}{2}H_2(g)\rightarrow H^+(aq)+e^-$

$E_{cell}=E^o_{cell}-0.0591$ log

$[H^+]$
(c)

$Ag(s)\rightarrow Ag^+(aq)$

$E_{cell}=E^o_{cell}-0.0591log[Ag^+]$
(d)

$Ag(s)+Cl^-(aq)\rightarrow AgCl(s)+e^-$

$E_{cell}=E^o_{cell}-0.0591log\displaystyle \frac{1}{[Cl^-]}$ .
Thus, in case(d), the

$E_{cell}$ increases with increase in the concentration of ions.

For the above cell.

0%
$E_{cell} < 0; \triangle G > 0$
0%
$E_{cell} > 0; \triangle G < 0$
0%
$E_{cell} < 0; \triangle G^{\circ} > 0$
0%
$E_{cell} > 0; \triangle G^{\circ} < 0$

Explanation

In the above electrolytic cell we have,

$E_{cell}$ =

$-0.0591$

$log$

$\dfrac{(oxi)}{(red)}$

$E_{cell}$ =

$-0.0591$

$log$

$\dfrac{0.05}{1}$

$E_{cell}$ = 0.13599 V

$\Delta$ G=

$-nFE_{cell}$

$\Delta$ G is negative

Use of electrolysis is not done in _____________.

0%
production of $Na$
0%
production of water
0%
purification of metal
0%
production of $KOH$

The standard reduction potential for

$Zn^{2+}/ Zn, Ni^{2+}/ Ni$ and

$Fe^{2+}/ Fe$ are

$-0.76, 0.23$ and

$-0.44\ V$ , respectively.

The reaction,

$X + Y^{2+} \rightarrow X^{2+} + Y$ will be spontaneous when

0%
$X = Ni, Y = Fe$
0%
$X = Ni, Y = Zn$
0%
$X = Fe, Y = Zn$
0%
$X = Zn, Y = Ni$

Explanation

The reaction is spontaneous when

$\triangle G < 0$ and

$E_{cell}^{\circ} > 0$ .

$\because \triangle G = -nFE^{\circ}$ and

$E^{\circ} > 0$

Also,

$E^{\circ}_{cell} = E^{\circ}_{oxi} + E^{\circ}_{red}$ (or)

$E^{\circ}_{cathode} - E^{\circ}_{anode}$

Here,

$E^{\circ}_{cell}$ for

(a)

$-0.67\ V$ , i.e.

$E^{\circ}_{cell} < 0$ (b)

$-0.99\ V$ , i.e.

$E^{\circ}_{cell} < 0$

(c)

$-0.32\ V$ , i.e.

$E^{\circ}_{cell} < 0$ (d)

$0.99\ V$ , i.e.

$E^{\circ} > 0$ .

Which one is/ are correct among the following?
Given, the half cell emf's

$E^{0}_{Cu^{+2}/ Cu} = 0.337, E^{0}_{Cu^{+1}/ Cu} = 0.521$ .

0%
$Cu^{+1}$ disproporationates
0%
$Cu$ and $Cu^{2+}$ comproportionates (reverse of disproportionation into $Cu^{+}$ ).
0%
$E^{0}_{Cu/ Cu^{+2}} + E^{0}_{Cu^{+2}/ Cu}$ is positive
0%
All of these

Explanation

$\begin{array}{l} Cu\left| { C{ U^{ 2+ } } } \right. \parallel C{ u^{ + } }\left| { { C_{ 4 } } } \right. \\ E_{ cell }^{ ^{ \circ } }=E_{ Cathode }^{ ^{ \circ } }-E_{ anode }^{ ^{ \circ } } \\ \Rightarrow 0.521-0.337\Rightarrow +ve\, \, value. \end{array}$

So, Reaction is

$\begin{array}{l} Cu\to C{ u^{ 2+ } }+2{ e^{ - } }\to \left( i \right) & \\ 2C{ u^{ + } }\to 2{ e^{ - } }\to 2Cu\to \left( { ii } \right) & \\ 2C{ u^{ + } }\to C{ u^{ 2+ } }+Cu & C{ u^{ + } }\, \, is\, \, disproportinates. \\ Cu+C{ u^{ 2+ } }\, \, comproportionates & \end{array}$

Option D is correct. All of these.

$1000\ mL\ of\ 1\ M\ CuSO_{4}(aq)$ is electrolysed by

$9.65$ ampere current for

$100$ sec using

$Pt-electrode$ . Which of the following is incorrect statement?

0%
Blue colour intensity decreases during electrolysis
0%
Blue colour intensity remains constant if $Cu-electrode$ used
0%
$pH$ of solution is $8$ after electrolysis
0%
$28\ mL$ of $CH_{4}$ (at $1\ atm$ and $273\ K)$ undergoes combustion by $O_{2}$ , liberated during electrolysis

In hydrogen-oxygen fuel cell the carbon rods are immersed in hot aqueous solution of.

0%
$KCl$
0%
$KOH$
0%
$H_2SO_4$
0%
$NH_4Cl$

Explanation

In the hydrogen-oxygen fuel cell, the carbon rods are immersed in hot aqueous solution of KOH (potassium hydroxide). KOH dissociates to provide hydroxide ions. Both anodic half reaction and cathodic half reaction involve hydroxide ions.

$\displaystyle 2H_2+4OH^- \rightarrow 4H_2O+4e^-$

$\displaystyle O_2+2H_2O +4e^- \rightarrow 4OH^-$

Voltaic cell can be recharged by:

0%
by addition of fresh solution
0%
by replacing external circuit with external source of electricity
0%
by removal of solution
0%
by heating it

CBSE Questions for Class 12 Engineering Chemistry Electrochemistry Quiz 8 - MCQExams.com

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Practice Class 12 Engineering Chemistry Quiz Questions and Answers

CBSE Questions for Class 12 Engineering Chemistry Electrochemistry Quiz 8 - MCQExams.com

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Practice Class 12 Engineering Chemistry Quiz Questions and Answers

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