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CBSE Questions for Class 12 Engineering Chemistry Electrochemistry Quiz 13 - MCQExams.com
CBSE
Class 12 Engineering Chemistry
Electrochemistry
Quiz 13
$$E_{Fe^{+3}/ Fe^{+2}}^{0} = +0.77\ V; E_{Fe^{+3}/ Fe}^{0} = -0.036\ V$$. What is the value of $$E_{Fe/ Fe^{+2}}^{0}$$ ?
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$$+0.44\ V$$
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$$-0.44\ V$$
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$$+0.48\ V$$
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$$-0.48\ V$$
Given the standard half-cell potentials $$(E^o)$$ of the following as
$$Zn=Zn^{2+}+2e^-$$; $$E^o=+0.76$$V
$$Fe=Fe^{2+}+2e^{-}$$; $$E^o=+0.41$$V
Then the standard e.m.f. of the cell with the reaction $$Fe^{2+}+Zn\rightarrow Zn^{2+}+Fe$$ is?
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$$-0.35$$V
0%
$$+0.35$$V
0%
$$+1.17$$V
0%
$$-1.17$$V
Explanation
$$Zn^{+2}+2e^-\rightarrow Zn$$ ; $$E^o=-0.76\ V$$ (SRP) (Anode)
$$Fe^{+2}+2e^-\rightarrow Fe$$ ; $$E^o=-0.41\ V$$ (SRP) (Cathode)
$$E^o_{cell}=E^+_{Fe^{+2}/Fe}-E^o_{Zn^{+2}/Zn}$$
$$=-0.41+0.76=0.35V$$.
Hence, the correct option is $$\text{B}$$
What will be the reduction potential for the following half-cell reaction at $$298$$K?
(Given: $$[Ag^+]=0.1$$M and $$E^o_{cell}=+0.80$$V)
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$$0.741$$V
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$$0.80$$V
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$$-0.80$$V
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$$-0.741$$V
Explanation
A galvanic cell or simple battery is made of two electrodes. Each of the electrodes of a galvanic cell is known as a half cell. In a battery, the two half cells form an oxidizing-reducing couple. When two half cells are connected via an electric conductor and salt bridge, an electrochemical reaction is started.
Nernst equation is,
$$E_{cell}=E^{0}_{cell}-\dfrac{0.0591}{n}logQ$$
Here the reaction is,
$$Ag^{+}+e^{-}\rightarrow Ag$$
$$n=1$$
$$E^{0}_{cell}=0.80V$$
Hence,
$$E_{cell}=E^{0}_{cell}-\dfrac{0.0591}{n}log\dfrac{1}{[Ag^{+}]}$$
$$E_{cell}=0.80-\dfrac{0.0591}{1}log\dfrac{1}{0.1}$$
$$=0.741V$$
The time required for a current of $$3\ amp$$. to decompose electrolytically $$18\ g$$ of $$H_{2}O$$ is:
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$$18\ hour$$
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$$36\ hour$$
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$$9\ hour$$
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$$18\ seconds$$
Explanation
$$H_{ 2 }O\quad \rightarrow { O }_{ 2 }+\quad 2{ e }^{ - }$$
2 far a days of charge required for 1 mole
$$ \\ ie:\quad 2\quad \times \quad 96500\quad C\\ $$
$$\quad \quad \quad 2\quad \times \quad 96500\quad =\quad 3\times t\\$$
$$ \\ \dfrac { 2\times 96500 }{ 3 } =t\quad =\quad 2\times 32166.67\\\\\quad \quad \quad \quad \quad =\ 64333.34\ sec\\ $$
$$\\ \quad \quad \quad \quad \quad = \dfrac { 64333.34 }{ 60\times 60 } hours\\$$
$$ \\ \quad \quad \quad \quad \quad =17.87\quad hours$$
The time required for a current of $$3$$ amp. to decompose electrolytically $$18$$g of $$H_2O$$ is:
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$$18$$ hour
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$$36$$ hour
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$$9$$ hour
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$$18$$ seconds
Explanation
According to Faraday first law $$W \propto Q$$
and $$W = \dfrac{Ewt\ } {F}$$
$$Ewt.=$$ Equivalent weight
$$F =$$ Faraday constant
$$W=$$ weight
$$Q =$$ $$current(i) \times time(t)$$
$$\dfrac {W} {Ewt} = \dfrac {it} {F}$$
$$\dfrac {18\times2}{18} =\dfrac {3\times t} {96500} (Ewt = \dfrac{M.w.}{V.F.})$$
$$\dfrac{2\times 96500}{3\times 3600} = t$$
$$ t =64333s \approx 18hr$$
The correct option is [A].
Solution A,B and C of the same strong electrolyte offered resistances of $$50\ \Omega,\ 100\ \Omega,\ and\ 150\ \Omega$$ in a given conductivity cell. The resistance observed if they are mixed in a volume proportion which is reciprocal of their resistances and tested in the same conductivity cell would be:
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$$67.3\ \Omega$$
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$$81.8\ \Omega$$
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$$100\ \Omega$$
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$$300\ \Omega$$
Explanation
Let value be $$v_1,\ v_2$$ and $$v_3$$
$$\dfrac {v_1}{R_1}+\dfrac {v_2}{R_2}+\dfrac {v_3}{R_3}=\dfrac {v_1 +v_2 +v_3}{Req}$$
$$R_1=52\ \Omega\ R_2=100\ \Omega \ R_3=150\ \Omega$$
$$v_1=\dfrac {1}{50}\ v_2=\dfrac {1}{100}\ v_3=\dfrac {1}{150}$$
$$\dfrac {v_1}{R_1}+\dfrac {v_2}{R_2}+\dfrac {v_3}{R_3}=\dfrac {v_1 +v_2 +v_3}{Req}$$
$$\dfrac {1}{50\times 50}+\dfrac {1}{100\times 100}+\dfrac {1}{150\times 150}=\dfrac {\dfrac {1}{50}+\dfrac {1}{100}+\dfrac {1}{150}}{Req}$$
$$\dfrac {36+9+4}{150\times 50\times 4}=\dfrac {\dfrac {6+3+2}{900}}{Req}$$
$$Req\ \dfrac {49}{150\times 150\times 4}=\dfrac {11}{300}$$
$$Req\ =\dfrac {3300}{49}=67.3\ \Omega$$
Which of the following is the cell reaction that occurs when the following half-cells are combined?
$$I_2+2e^-\rightarrow 2I^-(1M)$$; $$E^o=+0.54$$V
$$Br_2+2e^-\rightarrow 2Br^-(1M); E^o=+1.09$$V
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$$2Br^-+I_2\rightarrow Br_2+2I^-$$
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$$I_2+Br_2\rightarrow 2I^-+2Br^-$$
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$$2I^-+Br_2\rightarrow I_2+2Br^-$$
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$$2I^-+2Br^-\rightarrow I_2Br_2$$
Explanation
A galvanic cell or simple battery is made of two electrodes. Each of the electrodes of a galvanic cell is known as a half cell. In a battery, the two half cells form an oxidizing-reducing couple. When two half cells are connected via an electric conductor and salt bridge, an electrochemical reaction is started.
Reduction potential of both $$I_2$$ and $$Br_2$$ is given. We have to note that more the value of reduction potential then less the element has the tendency to get reduced and hence $$I_2$$ having more reduction potential acts as anode i.e. undergoes oxidation. And $$Br_2$$ having less reduction potential than $$I_2$$ undergoes reduction and acts as cathode.
$$2I^-\rightarrow I_2+2e^-$$ (Oxidation)
$$Br_2+2e^{-}\rightarrow 2Br^-$$ (Reduction)
$$\_ \_ \_ \_ \_ \_ \_ \_ \_ \_ \_ \_ \_ \_ \_ \_ \_ \_ \_ \_ \_ \_ \_ \_ \_ \_ \_ \_ $$
$$2I^-+Br_2\rightarrow I-2+2Br^-$$ is net cell reaction
$$\_ \_ \_ \_ \_ \_ \_ \_ \_ \_ \_ \_ \_ \_ \_ \_ \_ \_ \_ \_ \_ \_ \_ \_ \_ \_ \_ $$
Hence option (C) is correct.
$$E^o$$ values for the half cell reactions are given:
$$Cu^{2+}+e^-\rightarrow Cu^+$$; $$E^o=0.15$$V
$$Cu^{2+}+2e^-\rightarrow Cu$$; $$E^o=0.34$$V
What will be the $$E^o$$ of the half-cell: $$Cu^{+}+e^-\rightarrow Cu$$?
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$$+0.49$$V
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$$+0.19$$V
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$$+0.53$$V
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$$+0.30$$V
Explanation
Given,
$$Cu^{2+}+e^{-}\rightarrow Cu^{+}; E^{0}_{1}=0.15V,\Delta G^{0}_{1},n_1=1$$
$$Cu^{2+}+2e^{-}\rightarrow Cu; E^{0}_{2}=0.34V,\Delta G^{0}_{2},n_2=2$$
$$Cu^{+}+e^{-}\rightarrow Cu; E^{0}_{3}=?V,\Delta G^{0}_{3},n_3=1$$
As $$\Delta G$$ is an extensive property,
$$\Delta G^{0}_{3}=\Delta G^{0}_{2}-\Delta G^{0}_{1}$$
$$\implies$$ $$ -n_3FE^{0}_{3}=-n_2FE^{0}_{2}+ n_1FE^{0}_{1}$$
$$-E^{0}_{3}=-2\times0.34+1\times0.15$$
$$E^{0}_{3}=0.68-0.15=0.53V$$
Hence, option C is correct.
In the electrolysis of $$AgNO_3$$ solution 0.7 g of Ag is deposited after a certain period of time. Calculate the quantity of electricity required in coulomb. Molar mass of Ag is $$107.9 g mol^{-1}$$.
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$$426 ^oC$$
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$$536^oC$$
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$$626^o C$$
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$$736^o C$$
Explanation
$$Mass \ Deposited = 0.7 \ g$$
$$n=1$$
$$Molar \ mass = 107.9 \ gmol^{-1}$$
$$\boxed {m= \cfrac {Atomic \ Mass}{n \times F} \times i \times t}$$
$$0.7= \cfrac {107.9}{1 \times 96500} \times i \times t$$
$$\cfrac {0.7 \times 96500}{107.9}= i \times t$$
$$i \times t = 626.04 \ C$$
The quantity of electricity required in Coulomb is $$626 \ ^oC$$
An electrochemical cell can behave like an electrolytic cell when:
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$$E_{cell} = 0$$
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$$E_{cell} > E_{ext}$$
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$$E_{ext} > E_{cell}$$
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$$E_{cell} = E_{ext}$$
Explanation
We can study the effect of opposing potential applied to a galvanic cell as follows.
1) When $$E_{external}< 1.1V$$
In a galvanic cell, if an external opposite potential is
applied and increased slowly,
we find that the reaction continues to take
place till the opposing voltage reaches the
value 1.1 V.
2) When
$$E_{external}= 1.1V$$
When the external voltage is 1.1V the reaction
stops altogether and no current flows
through the cell.
3) When
$$E_{external}> 1.1V$$
Any further increase in
the external potential again starts the
reaction but in the opposite direction
. It now functions as an electrolytic
cell, a device for using electrical energy
to carry non-spontaneous chemical
reactions.
Half cell reactions for some electrodes are given below:
I. $$A+{ e }^{ - }\longrightarrow { A }^{ - }\ ;\quad \quad \ { E }^{ 0 }=0.96V$$
II. $${ B }^{ - }+{ e }^{ - }\longrightarrow { B }^{ 2- }\ ;\quad { E }^{ o }=-0.12V$$
III. $${ C }^{ + }+{ e }^{ - }\longrightarrow C\ ;\quad \ { E }^{ o }=+0.18V$$
IV. $${ D }^{ 2+ }+{ 2e }^{ - }\longrightarrow D\ ;\quad { E }^{ o }=-1.12V$$
The largest potential will be generated in which cell?
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$${ A }^{ - }|A\parallel { B }^{ - }|{ B }^{ 2 }$$
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$$D|{ D }^{ 2+ }\parallel A|{ A }^{ - }$$
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$${ B }^{ 2 }|{ B }^{ - }\parallel { C }^{ + }|C$$
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$$D|{ D }^{ 2+ }\parallel { C }^{ + }|C$$
Which of the following relation is not correct?
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Equivalent conductivity - S $$c{m}e{q^{ - 1}}$$
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Conductance -ohm$$^{ - 1}$$
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Specific conductance - $$S cm^{-1}$$
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Cell constant - cm$$^{ - 1}$$
The quantity of charge required to obtain one mole of aluminium from $$Al_2O_3$$ is:
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$$1$$ F
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$$6$$ F
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$$3$$ F
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$$2$$ F
Explanation
From $$Al_2O_3\rightarrow 2Al$$ to obtain one mole of $$Al$$ we consider,
$$2Al^{3+}+6e^{-}\rightarrow2Al$$
So, 6F of charge is required to obtain 2 moles of $$Al$$ from $$Al_2O_3$$
Thus, 3F of charge is required to obatain 1 mole of $$Al$$ from $$Al_2O_3$$
What would you observe if you set up the following electrochemical cell
$${\text{Ag|Agn}}{{\text{o}}_3}\left( {0.001M} \right)||AgN{O_3}(1M)|Ag$$
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Electrons will flow from left to right, causing a decrease in the $$\left[ {A{g^ + }} \right]$$ concentration in the right cell.
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Electrons will flow from right to left, causing an increases in the $$\left[ {A{g^ + }} \right]$$ concentration in the left cell and a decrease in the $$\left[ {A{g^ + }} \right]$$ concentration in the right cell.
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Electrons will flow from left to right, causing an increase in the $$\left[ {A{g^ + }} \right]$$ concentration in the left cell, and a decrease in the $$\left[ {A{g^ + }} \right]$$ concentration in the right cell.
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Electrons will flow from right to left, causing a decrease in the $$\left[ {A{g^ + }} \right]$$ concentration in the right cell
The specific conductance of a saturated solution of silver bromide is $$\kappa \,{\text{S}}\,{\text{c}}{{\text{m}}^{ - 1}}$$. The limiting ionic conductivity of $${\text{A}}{{\text{g}}^ + }\,{\text{and}}\,{\text{B}}{{\text{r}}^ - }$$ ions are x and y, respectively. The solubility of silver bromide in $${\text{g}}{{\text{L}}^{{\text{ - 1}}}}$$ is:
[molar mass of AgBr = 188]
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$$\dfrac{{\kappa \times 1000}}{{{\text{x}} - {\text{y}}}}$$
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$$\dfrac{\kappa }{{{\text{x}} + {\text{y}}}} \times 188$$
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$$\dfrac{{\kappa \times 1000 \times 188}}{{{\text{x}} + {\text{y}}}}$$
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$$\dfrac{{{\text{x}} \times {\text{y}}}}{\kappa } \times \dfrac{{1000}}{{188}}$$
Explanation
Given Specific conductance of AgBr = K S $${\text{c}}{{\text{m}}^{ - 1}}$$
$$\Lambda {^\circ _{A{g^ + }}} = x$$ $$\Lambda {^\circ _{B{r^ - }}} = y$$
$$\Lambda {^\circ _{AgBr}} = \Lambda {^\circ _{A{g^ + }}} + \Lambda {^\circ _{B{r^ - }}} = x + y$$
$$\Lambda {^\circ _{AgBr}} = {\text{K}} \times \dfrac{{1000}}{{{\text{C}}\left( {{\text{mol}}\,{{\text{L}}^{ - 1}}} \right)}}$$
$$x + y = \dfrac{{{\text{K}} \times 1000}}{{{\text{C}}\left( {{\text{mol}}\,{{\text{L}}^{ - 1}}} \right)}}$$
(Solubility) C = $$\left( {\dfrac{{K \times 1000}}{{x + y}}} \right){\text{mol}}\,.{{\text{L}}^{ - 1}}$$
No. of mol = $$\dfrac{{{\text{mass(g)}}}}{{{\text{molar}}\,\,{\text{mass}}\left( {{\text{g}}\,{\text{mo}}{{\text{l}}^{{\text{ - 1}}}}} \right)}}$$
molar mass of AgBr = 188 $${{\text{g}}\,{\text{mo}}{{\text{l}}^{{\text{ - 1}}}}}$$
So, $$\boxed{{\text{C}} = \left( {\dfrac{{{\text{K}} \times 1000}}{{x + y}}} \right) \times 188\,\,{\text{g}}{\text{.}}{{\text{L}}^{ - 1}}}$$
Hence, option (C) is correct.
For the cell Pt | $${{\text{H}}_{\text{2}}}$$ (0.4 atm) | $${{\text{H}}^ + }$$(pH = 1) || (pH = 2) | $${{\text{H}}_{\text{2}}}$$ 90.1 atm) | Pt. The measured potential at $${\text{25}}^\circ {\text{C}}$$ is:
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$$-0.1 V$$
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$$-0.5 V$$
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$$-0.041 V$$
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$$-0.030 V$$
Explanation
$$answer is C$$
$$at anode (0.4 atm)H_{2}\rightarrow 2H^{+}+2e^{-}$$
$$at cathode 2H^{+}+2e^{-}\rightarrow H_{2}(0.1 atm)$$
$$pH=1=-\log H^{+}$$
$$at anode [H^{+}]=10^{-1}=0.1 M$$
$$at cathode [H^{+}]=10^{-2}=0.01 M$$
$$E_{cell}=E^{0}_{cell}-\frac{0.0591}{2} \log \frac{[H^{+}]^{2}\times P_{H_{2}}}{[H^{+}]^{2}\times P_{H_{2}}}$$
$$E_{cell}=0 -\frac{0.0591}{2} \log \frac{10^{-2}\times 0.1}{10^{-4}\times 0.4}$$
$$E_{cell}=-0.041 V$$
$$Cd$$ amalgam is prepared by electrolysis of a solution of $$CdCl_2$$ using a mercury cathode. Current of how much ampere must be passed for $$100$$ seconds in order to prepare $$20\%$$ $$Cd-Hg$$ amalgam on a cathode of $$2$$ g mercury? (At. wt. of Cd$$=112.40$$)
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$$37.77$$ ampere
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$$8.58$$ ampere
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$$4.29$$ ampere
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$$17.16$$ ampere
Given below are the half-cell reactions:
$$Mn^{2+} + 2e^- \to Mn; E^o = -1.18V$$
$$2(Mn^{3+} + e^- \to Mn^{2+}); E^o = +1.51V$$
The $$E^o$$ for $$3Mn^{2+} \to Mn +2Mn^{3+}$$ will be:
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$$-0.33V$$; the reaction will not occur
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$$-0.33V$$; the reaction will occur
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$$-2.69V$$; the reaction will not occur
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$$-2.69V$$; the reaction will occur
Explanation
Standard electrode potential of reaction will not change due to multiply the half-cell reactions with some numbers,
To get the main eq we have to reverse $$2nd$$ equation and add them
So $$E_3 = E_2+E_1$$
$$E_3 = -1.18+(-1.51)$$
$$E_3 = -2.69V$$
The reaction is not possible as the $$\Delta G$$ will come +ve for this case and that indicates reaction is non-spontaneous.
In the electrolysis of aqueous sodium chloride solution which of the half-cell reaction will occur at the anode?
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$$Na^+_{(aq)}+e^-\rightarrow Na_{(s)}; E^o_{cell}=-2.71$$V
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$$2H_2O_{(l)}\rightarrow O_{2(g)}+4H^+_{(aq)}+4e^-; E^o_{cell}=1.23$$V
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$$H^+_{aq}+e^-\rightarrow \dfrac{1}{2}H_{2(g)}; E^o_{cell}=0.00$$V
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$$Cl^-_{(aq)}\rightarrow \dfrac{1}{2}Cl_{2(g)}+e^-; E^o_{cell}=1.36$$V
Explanation
At anode, oxidation takes place.
According to preferential discharge theory, more is the $$\varepsilon_{cell}$$ of anode potential more is the tendency for reaction to take place.
hence, $$Cl_2$$ gas is evolved.
$$Cl^{\circleddash}\longrightarrow \cfrac{1}{2}Cl_2+e^-\quad\quad \varepsilon^o_{cell}=1.36$$ $$V$$
In electrolysis of NaCl when Pt electrode is taken then $${ H }_{ 2 }$$ is liberated at the cathode while with Hg cathode it forms sodium amalgam, because:
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Hg is more inert than Pt
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more voltage is required to reduce $${ H }^{ + }$$ at Hg than at Pt
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Na is dissolved in Hg while it does not dissolve in Pt
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concentration of $${ H }^{ + }$$ ions is larger when Pt electrode is taken
Explanation
Sodium chloride in water dissociates as :
$$NaCl \rightleftharpoons Na^+ + Cl^- $$
$$H_2O \rightleftharpoons H^+ + OH^- $$
When electric current is passed through this solution using platinum electrodes, $$Na^+$$ and $$H^+ $$ move towards cathode and
$$Cl^-$$ and $$OH^- $$ ions move towards anode.
If mercury is used as a cathode,
$$H^+ $$ ions are not discharged at mercury cathode because mercury has high hydrogen overvoltage.
$$Na^+$$ ions are discharged at the cathode in preference of
$$H^+$$ ions yielding sodium, which dissolves in mercury to form sodium amalgam.
$$Al_2O_3$$ is reduced by electrolysis at low potentials and high currents. If $$4.0\times 10^4$$ amperes of current is passed through molten $$Al_2O_3$$ for $$6$$ hours, what mass of aluminium is produced?
(Assume $$100\%$$ current efficiency and atomic mass of $$Al$$ $$=27$$ g $$mol^{-1}$$)
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$$1.3\times 10^4$$ g
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$$9.0\times 10^3$$ g
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$$8.05\times 10^4$$ g
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$$2.4\times 10^5$$ g
Explanation
$$n=3$$ $$ Al^{3+} +3e^- \rightarrow Al$$
$$Molar \ mass = 27 \ gmol^{-1}$$
$$t=6 \ hours=6 \times 60 \times 60 sec$$
$$i=4 \times 10^4 \ A$$
$$\boxed {m= \cfrac {Atomic \ Mass}{n \times F} \times i \times t}$$
$$m= \cfrac {27}{3 \times 96500} \times 4 \times 10^4 \times 6 \times 60 \times 60$$
$$=\cfrac {23328}{3 \times 965} \times 10^4$$
$$= 8.05 \times 10^4 \ g$$
Mass of $$Al$$ produced is $$8.05 \times 10^4 \ g$$
A constant electric current flows for 4 hours through two electrolytic cells connected in series. One contains $$AgNO_3$$ solution and second contains $$CuCl_2$$ solution. During this time, 4 grams of $$Ag$$ is deposited in the first cell. How many grams of Cu is deposited in the second cell?
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2.176 g
0%
0.176 g
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1.176 g
0%
None of the above
Explanation
since $$ Cl $$ & $$ Ag $$ both are in series,
no. of eq. of $$ cl $$= no. of eq. of $$Ag$$
No./ eq of $$ Ag = \dfrac{4gm}{107}$$
& no. of eq.$$ cl =$$ no. of eq. of $$ Ag = \dfrac{4}{107}$$
eq. wt. of $$ cl = \dfrac{4}{107} \times \dfrac{1}{2} \times 63.5$$
$$ \boxed{eq. wt of $$ Cl = 1.176 gm}$$
If a $$100 \; mL$$ solution of $$0.1M$$. $$HBr$$ is titrated using a very concentrated solution of $$NaOH$$, then the conductivity (specific conductance) of this solution at the equivalence point will be:
(assume volume change is negligible due to the addition of $$(NaOH)$$. Report your answer after multiplying it with 10 in $$Sm^{-1}$$.
[Given: $$\lambda^{\circ}_{(Na^+)} = 8 \times 10^{-3} Sm^2 mol^{-1},
\lambda^{\circ}_{(Br^-)} = 4 \times 10^{-3} Sm^2 mol^{-1}$$]
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6
0%
12
0%
15
0%
24
The number of coulombs necessary to deposit 1 g of potassium metal (molar mass $$39 g mol^{-1}$$) from $$K^+ ions$$ is:
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$$96500^oC$$
0%
$$1.93 \times 10^5 $$$$^oC$$
0%
$$1237$$$$^o C$$
0%
$$2474$$$$ ^oC$$
Explanation
$$Mass \ Deposited = 1 \ g$$
$$n=1$$ $$(K^+ \ ions)$$; $$ K^+ +e^- \rightarrow K$$
$$Molar \ mass = 39 \ gmol^{-1}$$
$$\boxed {m= \cfrac {Atomic \ Mass}{n \times F} \times i \times t}$$
$$1= \cfrac {39}{1 \times 96500} \times i \times t$$
$$\cfrac {96500}{39}= i \times t$$
$$i \times t = 2474.35 \ C$$
The number of coulombs necessary to deposit $$1 \ g$$ of potassium metal from $$K^+$$ ions is $$2474 \ ^oC$$
At equimolar concentrations of $$Fe^{2+}$$ and $$Fe^{3+}$$, what must [$$Ag^{+}$$] be so that the voltage of the galvanic cell made from the ($$Ag^{+}|Ag)$$ and ($$Fe^{3+}|Fe^{2+})$$ electrodes equals zero ?
$$Fe^{2+}+Ag^{+}\rightleftharpoons Fe^{3+}+Ag$$
$$E^{o}_{Ag^{o}|Ag}=0.7991 ; E^{o}_{Fe^{3+}|Fe^{2-}}=0.771$$
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$$0.34$$
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$$0.44$$
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$$0.47$$
0%
$$0.61$$
Explanation
$${ Fe }^{ 2+ }+{ Ag }^{ + }\rightleftharpoons { Fe }^{ 3+ }+Ag$$
Given $$\left[ { Fe }^{ 2+ } \right] =\left[ { Fe }^{ 3+ } \right] \quad ,\quad { E }_{ cell }=0$$
$${ E }_{ cell }={ { E }^{ 0 } }_{ cell }-\dfrac { 0.059 }{ n } log\dfrac { \left[ { Fe }^{ 3+ } \right] }{ \left[ { Fe }^{ 2+ } \right] \left[ { Ag }^{ + } \right] } $$
$$\downarrow $$
Neunsl equation.
$${ { E }^{ 0 } }_{ cell }={ E }_{ { Ag }^{ + }/Ag }-{ E }_{ { Fe }^{ 3+ }/{ Fe }^{ 2+ } }$$
$$=0.7991-0.771=0.0281$$
$$0=0.0281-\dfrac { 0.059 }{ 1 } log\left[ \dfrac { 1 }{ \left( { Ag }^{ + } \right) } \right] $$
$$\dfrac { 0.0281 }{ 0.059 } =log\left[ \dfrac { 1 }{ \left( { Ag }^{ + } \right) } \right] $$
$$0.4754=log\left( \dfrac { 1 }{ { Ag }^{ + } } \right) $$
$$2.988=1/\left[ { Ag }^{ + } \right] $$
$$\left[ { Ag }^{ + } \right] =0.34M$$
How long (approximate) should water be electrolysed by passing through $$100$$ amperes current so that the oxygen released can completely burn $$27.66\ g$$ of diborane?
[Atomic weight of $$B=10.8\ u$$]
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$$6.4$$ hours
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$$0.8$$ hours
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$$3.2$$ hours
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$$1.6$$ hours
Explanation
$$B_2H_6 + 3O_2 \longrightarrow B_2O_3 +3H_2O $$
27.66 gm of $$B_2H_6$$ i.e 1 mole of $$B_2H_6$$ requires 3 moles of $$O_2$$ .
Now oxygen is produced by electrolysis of water.
$$2H_2O \overset{4F}\longrightarrow2H_2+ O_2$$
1 mole of $$O_2$$ is produced by 4F charge therefore 3 mole of $$O_2$$ will produced by 12F charge
By applying ,Q= It
$$12\times96500 =100\times t$$
$$t = \dfrac{12\times96500}{100\times3600}$$ hours
$$t = 3.2 $$ hours
Hence option C is correct option.
The charge required to deposit $$40.5 \,g$$ of $$Al$$ (atomic mass $$= 27.0 \,g$$) from the fused $$Al_2(SO_4)_3$$ is:
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$$4.34 \times 10^5 C$$
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$$43.4 \times 10^5 C$$
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$$1.44 \times 10^5$$
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None of these
Explanation
$$Al^{3+} + 3e^- \rightarrow Al$$
3F 1 mol = 27.0 g
to deposite 27 g required charge = $$3 \times 96,500 c$$
Therefore, to deposite 40.4 g required charge
$$4.34 \times 10^5 C$$
Based on the cell notation for spontaneous reaction, at the anode:
$$Ag(s)|AgCl(s)|{ Cl }^{ - }(aq)\parallel { Br }^{ - }(aq)|{ Br }_{ 2 }(I)|C(s)$$
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$$AgCl$$ gets reduced
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$$Ag$$ gets oxidized
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$${ Br }^{ - }$$ gets oxidized
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$${ Br }_{ 2 }$$ gets reduced
Explanation
$$Ag\left( s \right) \left| AgCl\left( s \right) \right| { Cl }^{ \left( - \right) }\left| { Br }^{ - } \right| \left| { Br }_{ 2 } \right| C\left( s \right) $$
Anode : $$Ag+{ Cl }^{ \left( - \right) }\rightarrow AgCl+{ e }^{ - }$$
Cathode : $$\dfrac { 1 }{ 2 } { Br }_{ 2 }+{ e }^{ \left( - \right) }\rightarrow { Br }^{ \left( - \right) }$$
$$\therefore $$ at anode, oxidation takes place.
Hence, $$Ag$$ gets oxidized.
Alizarin belongs to the class of
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0%
Vat dyes
0%
Mordant dye
0%
Basic dye
0%
Reactive dye
Given below are the half-cell reactions:
$${Mn}^{2+}+2{e}^{-}\rightarrow Mn;{E}^{o}=-1.18V$$
$$2({Mn}^{+3}+{e}^{-}\rightarrow {Mn}^{2+});{E}^{o}=+1.51V$$
The $${E}^{o}$$ for $$3{Mn}^{2+}\rightarrow Mn+2{Mn}^{3+}$$ will be:
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$$+0.33V$$; the reaction will occur
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$$-2.69V$$; the reaction will not occur
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$$+2.69V$$; the reaction will occur
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$$-0.33V$$; the reaction will not occur
Explanation
Standard potential of reaction $$[E^0]$$ is given by,
$$E_0^{cell}=E_R-E_P$$
$$Mn^{2+}+2e^-\longrightarrow Mn\quad E^0=-1.18V\\2Mn^{2+}\longrightarrow 2Mn^{3+}+2e^-\quad E^0=-1.51V\\\underline{\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad}$$
$$3Mn^{2+}\longrightarrow Mn+2Mn^{3+}$$ $$E^0=-1.18+(-1.51)\\=-2.69V$$
The reaction is non spontaneous, $$E^0=-2.69V$$
Arrange the following metals in the order of their decreasing reactivity?
$$Fe, Cu, Mg, Ca, Zn, Ag$$
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$$Ca> Zn> Mg> Cu> Ag. Fe$$
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$$Ca> Zn>Cu> Mg> Ag> Fe$$
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$$Ca> Mg> Zn> Fe> Cu> Ag$$
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$$Ca> Mg> Cu> Zn> Fe> Ag$$
If $${E}_{Au^{+}/Au}^{0}$$ is $$1.69\ V$$ and $${E}_{Au^{3+}/Au}^{0}$$ is $$1.40\ V$$, then $${E}_{Au^{+}/Au^{3+}}^{0}$$ will be :
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$$0.19\ V$$
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$$2.945\ V$$
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$$1.255\ V$$
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$$none\ of\ these$$
Explanation
$$Au^{+} + e^{-} \rightarrow Au(s) $$ $$E^{\circ} = 1.69\,V$$ .............(1) $$ \Delta G^{\circ}_{1} $$
$$Au^{3+} + 3e^{-} \rightarrow Au(s) $$ $$E^{\circ} = 1.40\,V $$ ..................(2) $$ \Delta G^{\circ}_{2} $$
From (2) - (1)
$$ Au^{3+} + 2e^{-} \rightarrow Au^{+} $$ ;
$$ \Delta G^{\circ}_{3} = \Delta G^{\circ}_{2} - \Delta G^{\circ}_{1} $$
$$ -2 \times F \times F^{\circ} = -3 \times F \times 1.40 + 1 \times 1.69 \times F ; E^{\circ} = 1.255\,V$$
$$ \therefore $$ $$E^{\circ}_{Au^{+} | Au^{3+}} = -1.255\,V $$
Option (C) is correct.
$${{\text{E}}^{\text{o}}}_{{\text{C}}{{\text{u}}^{{\text{2 + }}}}{\text{|Cu = }}}\,{\text{ + 0}}{\text{.337}}\,{\text{V}}\,$$, $${{\text{E}}^{\text{o}}}_{{\text{Z}}{{\text{n}}^{{\text{2 + }}}}{\text{|Zn = }}}\,{\text{-0.762}}\, {\text{V}}\,$$. The EMF of the cell, $${\text{Zn|Z}}{{\text{n}}^{{\text{2 + }}}}\left( {{\text{0}}{\text{.1M}}} \right)||\,{\text{C}}{{\text{u}}^{{\text{2 + }}}}\left( {{\text{0}}{\text{.01M}}} \right){\text{|Cu}}$$ is :
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$$+1.099 V$$
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$$-1.099 V$$
0%
$$+1.069 V$$
0%
$$-1.069 V$$
Which one of the following metals could not be obtained on electrolysis of aqueous solution of its salts ?
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Ag
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Mg
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Cu
0%
Cr
Explanation
The reduction potential of $$Mg$$ is less than that of water $$(E^o=-0.83\ V)$$. Hence their ions in the aqueous solution cannot be reduced instead water will be reduced
$$2H_2O +2e^- \to H_2+2OH^-$$.
$$\left. {Ag} \right|\left. {AgCl} \right|\left. {C{l^ - }({C_2})} \right\|\left. {C{l^ - }({C_1})} \right|\left. {AgCl} \right|Ag$$ for this cell $$\Delta G$$ is negative if:
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$${C_{1 = }}{C_2}$$
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$${C_1} > {C_2}$$
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$${C_2} > {C_1}$$
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Both (1 ) and (3)
Explanation
Given cell is electrolyte concentration cell
$$E_{cell} = \dfrac{RT}{F} ln \dfrac{[Ag^{+}]_{RHS}}{[Ag^{+}]_{LHS}} $$
where $$ [Ag^{+}] = \dfrac{K_{sp}}{[Cl^{-}]} $$. For spontaneous chemical change conc. of $$Ag^{+}$$ of cathodic compartment must be greater than anodic compartment.
Hence option (C) is correct.
In the electrolysis of aqueous $$NaCl$$, what volume of $$Cl_2(g)$$ is produced in the time that it takes to liberate $$5.0$$ liter of $$H_2(g) ?$$ Assume that both gases are measured at STP.
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$$5.0$$
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$$2.50$$
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$$7.50$$
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$$10.0$$
$$\frac{1}{2}F_{2}+e^{-}\rightarrow F^{-};\ E^{\circ}=+3.02V$$
The electrode potential for given reaction: $$F_{2}+2e^{-}\rightarrow 2F^{-}$$
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3.02 V
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6.04 V
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1.5 V
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-3.02 V
$$Co|{ Co }^{ 2+ }({ C }_{ 2 })\parallel { Co }^{ 2+ }({ C }_{ 1 })|Co$$ for this cell, $$\triangle$$G is negative if :
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$$C_2$$ > $$C_1$$
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$$C_1$$ > $$C_2$$
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$$C_1$$ = $$C_2$$
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unpredictable
Find the potential of a half-cell having reaction, $$Ag_{2}S+2e\rightarrow2Ag+S^{2-}$$ in a solution buffered
$$pH=3$$ and which is also saturated with $$0.1\ M\ H_{2}S$$. For $$H_{2}S:\ K_{1}=10^{-8}$$ and $$K_{2}=2\times 10^{-13},\ K_{sp}(Ag_{2}S)=2\times 10^{-48},\ E_{Ag,ag}^{o}=0.8$$
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$$0.432\ V$$
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$$1.658\ V$$
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$$-0.245\ V$$
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$$-0.144\ V$$
Given the $$E^0_{cell}$$ for the reaction is $$-0.34V$$.
$$Cu(s)+2H^+(aq)⇌Cu^{2+}(aq)+H_2(g)$$
Find the equilibrium constant at 25 degrees celsius for the above reaction.
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$$3.19 \times 10^{-12}$$
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$$6.24 \times 10^{-10}$$
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$$2.46 \times 10^{-14}$$
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$$1.54 \times 10^{-10}$$
For the fuel cell reaction : $$2{H_2}\left( g \right) + {O_2}\left( g \right) \to 2{H_2}O\left( l \right);\quad {\Delta _f}H_{298}^ \circ \left( {{H_2}O,l} \right) = - 285.5\ kJ/mol$$ what is $$\Delta S_{298}^0$$ for the given fuel cell reaction?
Given: $${O_2}\left( g \right) + 4{H^ + }\left( {aq} \right) + 4{e^ - } \to 2{H_2}O\left( l \right)\,; \quad {E^ \circ } = 1.23V$$
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$$0.322 kJ/K$$
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$$-0.635kJ/K$$
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$$3.51kJ/K$$
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$$-0.322kJ/K$$
In $$Zn|Zn^{+2}|| Ag^+|Ag$$, how will cell potential be affected if KI is added to $$Ag^{+}$$ half cell?
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$$E_{cell}$$ will increase
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$$E_{cell}$$ will decrease
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$$E_{cell}$$ will remain unaffected
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$$E_{cell}$$ is first increase then decrease
A : Electrochemical cell is based on redox reaction.
R : Electrochemical cell converts electrical energy into chemical energy.
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Both Assertion & Reason are true and the reason is the correct explanation of the assertion.
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Both Assertion & Reason are true but the reason is not the correct explanation of the assertion.
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Assertion is true statement but Reason is false.
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Both Assertion and Reason are false statements.
Which of the following substance has the lowest electric resistivity at room temperature?
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Aluminium
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Iron
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Nichrome
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Diamond
The no. of electrons involved in the electro deposition of 63.5 g. of Cu from aq. $$ CuSO_4 $$ is :
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$$ 6.0 \times 10^{21} $$
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$$ 3.011 \times 10^{23} $$
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$$ 12.04 \times 10^{23} $$
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$$ 6.02 \times 10^{23} $$
The charge in coulombs of 1 mole of $$ N^{3-} $$ is (The charge on an electron is: $$ 1.602 \times 10^{-19} C ) $$
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$$ 2.894 \times 10^5 C $$
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$$ 3.894 \times 10^5 C $$
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$$ 2.894 \times 10^6 C $$
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None of these
How many faradays of charge is transferred to produce 11.2 L of $$H_2$$ at STP in the reaction, in the reaction, $$ NaH +H_2O \rightarrow NaOH +H_ 2 \uparrow $$ ?
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$$1$$
0%
$$0.5$$
0%
$$2$$
0%
$$2.5$$
2 ml ethanoic acid was taken in test tubes - A, B and C. 2 ml, 4 ml and 8 ml of water was added to the test tubes A, B and C respectively. Which test tube will show clear solution?
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only A
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only B
0%
only A and B
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all
Deduce from the following $${E^ \circ }$$ value of half cells, What combination of half cells would result in cell with the largest potential?
i) $${A^{3 - }} \to {A^2} + {e^ - };{\rm{ }}{{\rm{E}}^ \circ } = 1.5V$$
ii) $${B^{2 + }} + {e^ - } \to {B^ + };{\rm{ }}{{\rm{E}}^ \circ } = - 2.1V$$
iii) $${C^{2 + }} + {e^ - } \to + {C^ + };{\rm{ }}{{\rm{E}}^ \circ } = + 0.5V$$
iv) $$D \to {D^{2 + }} + 2{e^ - };{\rm{ }}{{\rm{E}}^ \circ } = - 1.5V$$
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(i) and (iii)
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(i) and (iv)
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(ii) and (iv)
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(iii) and (iv)
The emf of Deniell cell is 1.1 volt.If the value of Faraday is 96500 coulombs per mole, the change in free energy in kJ is :
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$$212.30$$
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$$-212.30$$
0%
$$106.15$$
0%
$$-106.15$$
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Practice Class 12 Engineering Chemistry Quiz Questions and Answers
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