MCQ 11 Mark
Calculate standard cell potential for a cell having following reaction $2 Al _{( s )}+3 Ni ^{2+} \longrightarrow 2 Al ^{3+}+3 Ni _{( s )}$
$\left( E _{ Ni }^{\circ}=-0.25 V , E _{ Al }^{\circ}=-1.66 V \right)$
View full question & answer→MCQ 21 Mark
Calculate the quantity of electricity required to produce $0.42 g Ag$ at cathode during electrolysis of $AgNO _3$ solution. (molar mass $Ag =108 g mol ^{-1}$ )
- A
$257 \cdot 1 C$
- B
$470 \cdot 0 C$
- C
$965 \cdot 0 C$
- ✓
$375 \cdot 3 C$
AnswerCorrect option: D. $375 \cdot 3 C$
View full question & answer→MCQ 31 Mark
Which element from following has highest negative standard reduction potential?
MCQ 41 Mark
For which of the following electrolyte, Kohlrausch law of independent migration of ions is used to calculate molar conductivity at zero concentration?
- A
$NaNO _3$
- B
$Na _2 SO _4$
- ✓
$NH _4 OH$
- D
$KCl$
AnswerCorrect option: C. $NH _4 OH$
View full question & answer→MCQ 51 Mark
A conductivity cell has two electrodes $18 mm$ apart and having cross sectional cises area $2.0 cm ^2$. What is the value of cell constant?
- A
$3.6 cm ^{-1}$
- ✓
$0.9 cm ^{-1}$
- C
$0.18 cm ^{-1}$
- D
$0.2 cm ^{-1}$
AnswerCorrect option: B. $0.9 cm ^{-1}$
View full question & answer→MCQ 61 Mark
In acetylene molecule, $C - H$ sigma bond is formed by
- A
- B
$p-p$ overlap
- ✓
$s p-s$ overlap
- D
$s-p$ overlap
AnswerCorrect option: C. $s p-s$ overlap
View full question & answer→MCQ 71 Mark
The resistance of a conductivity cell containing $0.001 M KCl$ solution at $300 K$ is $150 \Omega$. What is the cell constant if conductivity of $KCl$ solution is $1.5 \times 10^{-3} \Omega^{-1} cm ^{-1}$ ?
- A
$0.112 cm ^{-1}$
- B
$0.450 cm ^{-1}$
- C
$0.337 cm ^{-1}$
- ✓
$0.225 cm ^{-1}$
AnswerCorrect option: D. $0.225 cm ^{-1}$
View full question & answer→MCQ 81 Mark
What is the $SI$ unit of molar conductivity?
- A
$S cm ^2 mol ^{-1}$
- B
$S m ^2$
- C
$S dm ^3 mol ^{-1}$
- ✓
$S m ^2 mol ^{-1}$
AnswerCorrect option: D. $S m ^2 mol ^{-1}$
View full question & answer→MCQ 91 Mark
What is the change in potential of following cell $\left.2 n_{(s)}\right|_{I M} ^{C n^{-}}||_{I M}^{P 0^{-}} \mid P b_{(s)}$, if concentration of lons at anode is increased 10 times ?
View full question & answer→MCQ 101 Mark
How many coulomb of electricity is required to produce $1 g ~ N a$ metal from it's ions ? (atomic mass $Na =23$ ) prenes
View full question & answer→MCQ 111 Mark
How many moles of electrons are required for the reduction of $3$ moles of $Zn ^{2+}$ to $Zn ( s )$ ?
View full question & answer→MCQ 121 Mark
Calculate molar conductivity at infinite dilution for $NaBr$ if molar conductivity at infinite dilution for $NaCl , KBr$ and $KCl$ are 126,152 and $150 \Omega^{-1} cm ^2 mol ^{-1}$ respectively.
- A
$302 \Omega^{-1} cm ^2 mol ^{-1}$
- ✓
$128 \Omega^{-1} cm ^2 mol ^{-1}$
- C
$176 \Omega^{-1} cm ^2 mol ^{-1}$
- D
$278 \Omega^{-1} cm ^2 mol ^{-1}$
AnswerCorrect option: B. $128 \Omega^{-1} cm ^2 mol ^{-1}$
View full question & answer→MCQ 131 Mark
Identify the reaction taking place at anode during electrolysis of molten $NaCl$ ?
- A
$Na ^{+}+1 e ^{-} \longrightarrow Na _{( s )}$
- ✓
$2 Cl ^{-} \longrightarrow Cl _{2( g )}+2 e ^{-}$
- C
$2 H ^{+}+2 e ^{-} \longrightarrow H _{2( g )}$
- D
$2 OH ^{-} \longrightarrow H _2 O +\frac{1}{2} O _2+2 e ^{-}$
AnswerCorrect option: B. $2 Cl ^{-} \longrightarrow Cl _{2( g )}+2 e ^{-}$
View full question & answer→MCQ 141 Mark
How long will it take to produce $5.4 g$ of $Ag$ from molten $AgCl$ by passing $5 amp$ current? (molar mass $Ag =108 g mol ^{-1}$ )
View full question & answer→MCQ 151 Mark
Calculate $\Lambda_0$ of $CH _2 ClCOOH$ if $\Lambda_0$ for $HCl , KCl$ and $CH _2 ClCOOK$ are 4.2 .1 .5 and $1-1 \Omega^{-1} cm ^2 mol ^{-1}$ respectively?
- A
$2.7 \Omega^{-1} cm ^2 mol ^{-1}$
- ✓
$3.8 \Omega^{-1} cm ^2 mol ^{-1}$
- C
$1.9 \Omega^{-1} cm ^2 mol ^{-1}$
- D
$4 \cdot 2 \Omega^{-1} cm ^2 mol ^{-1}$
AnswerCorrect option: B. $3.8 \Omega^{-1} cm ^2 mol ^{-1}$
View full question & answer→MCQ 161 Mark
Which of the following is NOT an example of secondary voltaic cell?
MCQ 171 Mark
How many faraday of electrictty is required to produce $5 g$ of magnesium from magnesium chloride ? (molar mass Mg $=24 g mol ^{-1}$ )
View full question & answer→MCQ 181 Mark
Calculate the molar conductivity of $CH _2 ClCOOH$ at zero concentration if mola conductivities of $HCl , KCl$ and $CH _2 ClCOOK$ at zero concentration are $4 \cdot 2,1 \cdot 4$ and $1 \cdot 1 \Omega^{-1} cm ^2 mol ^{-1}$ respectively.
- A
$1.7 \Omega^{-1} cm ^2 mol ^{-1}$
- ✓
$3.9 \Omega^{-1} cm ^2 mol ^{-1}$
- C
$6.6 \Omega^{-1} cm ^2 mol ^{-1}$
- D
$4 \cdot 5 \Omega^{-1} cm ^2 mol ^{-1}$
AnswerCorrect option: B. $3.9 \Omega^{-1} cm ^2 mol ^{-1}$
View full question & answer→MCQ 191 Mark
Identify representation of standard hydrogen electrode?
- A
$Pt ,\left[ H ^{+}\right]\left| H _{2( g )} P _{ H _2}\right| H ^{+}(1 m )$
- ✓
$H ^{+}(1 M )\left| H _2( g ) 1 atm \right| Pt$
- C
$H_{a q}^{+}|H_2 [ 1M]| P t,\left[H^{+}\right]$
- D
$H _{2( g )}\left| H _{ aq }^{+}\right| Pt , H ^{+}$
AnswerCorrect option: B. $H ^{+}(1 M )\left| H _2( g ) 1 atm \right| Pt$
View full question & answer→MCQ 201 Mark
Molar conductivity of $0.01 M CH _3 COOH$ is $19.5 \Omega^{-1} cm ^2 mol ^{-1}$. Calculate its degree of dissociation if molar conductivity at zero concentration is $390 \Omega^{-1} cm ^2 mol ^{-1}$ ?
View full question & answer→MCQ 211 Mark
In a certain electrolysis experiment, $0.650 g$ of zinc is deposited in a cell having $ZnSO _4$ solution. Calculate mass of $Cu$ deposited in other cell having CuSO, solution arranged in series with first cell ? (molar mass $Zn =65, Cu =63.5 g mol ^{-1}$ )
View full question & answer→MCQ 221 Mark
What will be the weight of Al deposited at cathode, when 0.5 faraday of electricity is passed through aqueous solution of $AlCl _3$ ? (at. mass of $Al =27$ )
View full question & answer→MCQ 231 Mark
What is the value of conductivity of $0.01 M$ solution of an electrolyte having molar conductivity $141 \Omega ^{-1} cm ^2 mol ^{-1}$ ?
- A
$5.64 \times 10^{-3} \Omega^{-1} cm ^{-1}$
- B
$4.23 \times 10^{-3} \Omega^{-1} cm ^{-1}$
- ✓
$1.41 \times 10^{-3} \Omega^{-1} cm ^{-1}$
- D
$7.09 \times 10^{-3} \Omega^{-1} cm ^{-1}$
AnswerCorrect option: C. $1.41 \times 10^{-3} \Omega^{-1} cm ^{-1}$
View full question & answer→MCQ 241 Mark
Calculate mass of a divalent metal produced at cathode by passing $5 amp$ current through it's salt solution for 100 minute. (molar mass of metal $=x$ )
- ✓
$\frac{30}{193} x$
- B
$\frac{193}{30} x$
- C
$\frac{193}{15} x$
- D
$\frac{15}{193} x$
AnswerCorrect option: A. $\frac{30}{193} x$
View full question & answer→MCQ 251 Mark
Which among the following concentration of $KCl$ solution is not used to determine cell constant of conductivity cell?
View full question & answer→MCQ 261 Mark
Calculate standard potential of a cell having electrode reactions as
$Cd _{( aq )}^{2+}+2 e ^{-} \rightarrow Cd _{( s )} E ^{\circ}=-0.403 V Zn _{( aq )}^{2+}+2 e ^{-} \rightarrow Zn _{( s )} E ^{\circ}=-0.763 V $
View full question & answer→MCQ 271 Mark
What current strength is required to deposit $36 g$ of $Ag$ in 7 minute from $AgNO _3$ solution by electrolysis? (atomic mass Ag $=108$ )
- A
$5.72 amp$
- B
$38 \cdot 3 amp$
- C
$11.44 amp$
- ✓
$76 cdot 6 amp$
AnswerCorrect option: D. $76 cdot 6 amp$
View full question & answer→MCQ 281 Mark
Identify the molecule containing triple bond.
MCQ 291 Mark
Which of the following is NOT a function of salt bridge
- A
Provide electrical contact
- B
Maintain electrical neutrality
- C
Prevent mixing of solutions
- ✓
Convert electrical energy to chemical energy
AnswerCorrect option: D. Convert electrical energy to chemical energy
View full question & answer→MCQ 301 Mark
Which of the following expressions represent molar conductivity of $Al _2\left( SO _4\right)_3 ?$
- A
$3 \lambda_{ Al ^{3+}}^0+2 \lambda_{ SO _4{ }^{2-}}^0$
- ✓
$2 \lambda_{ Al ^{3+}}^0+3 \lambda_{ SO _4{ }^{2-}}^0$
- C
$\frac{1}{3} \lambda_{ Al ^{3+}}^0+\frac{1}{2} \lambda_{ SO _4{ }^{2-}}^0$
- D
$\lambda_{ Al ^{3+}}^0+\lambda_{ SO _4{ }^{2-}}^0$
AnswerCorrect option: B. $2 \lambda_{ Al ^{3+}}^0+3 \lambda_{ SO _4{ }^{2-}}^0$
View full question & answer→MCQ 311 Mark
The oxidation reaction that takes place in lead storage battery during discharge is
- A
$Pb _{( aq )}^{2+}+ SO _{4( aq )}^{2+} \longrightarrow PbSO _{4( s )}$
- B
$\begin{aligned} PbSO _{4( s )}+2 H _2 O _{( l )} \longrightarrow & PbO_2( s )+4 H _{( aq )}^{+}+ SO _{4( aq )}^{2-}+2 e ^{-}\end{aligned}$
- ✓
$Pb _{( s )}+ SO _{4( aq )}^{2-} \longrightarrow PbSO _{4( s )}+2 e _{ }^{-}$
- D
$PbSO _{4( s )}+2 e ^{-} \longrightarrow Pb _{( s )}+ SO _{4( aq )}^{2-}$
AnswerCorrect option: C. $Pb _{( s )}+ SO _{4( aq )}^{2-} \longrightarrow PbSO _{4( s )}+2 e _{ }^{-}$
View full question & answer→MCQ 321 Mark
Which of the following is not correct ?
- A
Gibbs energy is an extensive property
- B
Electrode potential or cell potential is an intensive property.
- C
- ✓
If half reaction is multiplied by a numerical factor, the corresponding E0 value is also multiplied by the same factor.
AnswerCorrect option: D. If half reaction is multiplied by a numerical factor, the corresponding E0 value is also multiplied by the same factor.
If half reaction is multiplied by a numerical factor, the corresponding E0 value is also multiplied by the same factor.
View full question & answer→MCQ 331 Mark
For the reaction: $ Ni _{( s )}+ Cu ^{2+}(1 M ) \rightarrow Ni ^{2+}(1 M )+ Cu _{( s )}, E_{\text {cell }}^0=0.57 V. $$\text { Hence } \Delta G ^0 \text { of the reaction is }$
View full question & answer→MCQ 341 Mark
Consider the half reactions with standard potentialsi. $Ag _{( aq )}^{+}+ e ^{-} \longrightarrow Ag _{( s )} \quad E^0=0.8 V$
ii. $I _{2( s )}+2 e ^{-} \longrightarrow 2 l _{( aq )}^{-} \quad E^0=0.53 V$
iii. $Pb _{( aq )}^{2+}+2 e ^{-} \longrightarrow Pb _{( s )} E^0=-0.13 V$
iv. $Fe _{( aq )}^{2+}+2 e ^{-} \longrightarrow Fe _{( s )} E^0=-0.44 V$
The strongest oxidising and reducing agents respectively are
- A
Ag and $Fe ^{2+}$
- ✓
$Ag ^{+}$and Fe
- C
$Pb ^{2+}$ and $I ^{-}$
- D
$I _2$ and $Fe ^{2+}$
AnswerCorrect option: B. $Ag ^{+}$and Fe
$Ag ^{+}$and Fe
View full question & answer→MCQ 351 Mark
For the cell, $Pb _{( s )}\left| Pb ^{2+}(1 M ) \| Ag ^{+}(1 M )\right| Ag _{( s )}$, if concentration of an ion in the anode compartment is increased by a factor of 10 , the emf of the cell will
View full question & answer→MCQ 361 Mark
The standard potential of the cell in which the following reaction occurs $H _{2+}( g , 1 \ atm )+ Cu ^{2+}(1 \ M ) \rightarrow 2 H (1\ M )+ Cu _{( s )}\left(E_{ Cu }^0=0.34 V \right)$ is
View full question & answer→MCQ 371 Mark
$15 m^2 mol^{-1}$ is equal to
- A
$10^{-4} S m ^2 mol^{-1}$
- ✓
$10^4 \Omega^{-1} cm^2 mol^{-1}$
- C
$10^{-2} S cm ^2 mol^{-1}$
- D
$10^2 \Omega^{-1} cm^2 mol^{-1}$
AnswerCorrect option: B. $10^4 \Omega^{-1} cm^2 mol^{-1}$
View full question & answer→MCQ 381 Mark
On diluting the solution of an electrolyte,
- A
both $\lambda$ and $k$ increase
- B
both $\lambda$ and $k$ decrease
- ✓
$\lambda$ increases and $k$ decreases
- D
$\lambda$ decreases and $k$ increases
AnswerCorrect option: C. $\lambda$ increases and $k$ decreases
View full question & answer→MCQ 391 Mark
Two solutions have the ratio of their concentrations 0.4 and ratio of their conductivities 0.216. The ratio of their molar conductivities will be
MCQ 401 Mark
What is the conductivity of $0.05 M BaCl _2$ solution if its molar conductivity is $220 \Omega^{-1} cm ^2 mol ^{-1}$ ?
- A
$0.011 \Omega^{-1} cm ^{-1}$
- B
$0.022 \Omega^{-1} cm ^{-1}$
- C
$0.033 \Omega^{-1} cm ^{-1}$
- D
Answer
$\begin{aligned} & \text {(a): } \Lambda_m=220 \Omega^{-1} \mathrm{~cm}^2 \mathrm{~mol}^{-1} \\ & C=0.05 \mathrm{M}, \kappa=? \\ & \quad \Lambda_m=\frac{1000 \times \kappa}{C} \Rightarrow 220=\frac{1000 \times \kappa}{0.05} \\ & \Rightarrow \quad \kappa=\frac{220 \times 0.05}{1000}=0.011 \Omega^{-1} \mathrm{~cm}^{-1}\end{aligned}$
View full question & answer→MCQ 411 Mark
A reaction $\mathrm{Ni}_{(s)}+\mathrm{Cu}^{++}(1 \mathrm{M}) \rightarrow \mathrm{Ni}^{++}(1 \mathrm{M})+\mathrm{Cu}_{(s)}$ occurs in a cell. Calculate $E_{\text {cell }}^{\circ}$ if $E_{\mathrm{Cu}}^{\circ}=0.337 \mathrm{~V}$ and $E_{\mathrm{Ni}}^{\circ}=-0.257 \mathrm{~V}$
- A
$0.594 V$
- B
$-0.594 V$
- C
$-0.08 V$
- D
$0.08 V$
Answer
$\begin{aligned} & \text {(a): } \mathrm{Ni}_{(s)}+\mathrm{Cu}^{2+}(1 \mathrm{M}) \rightarrow \mathrm{Ni}^{2+}(1 \mathrm{M})+\mathrm{Cu}_{(s)} \\ & E_{\mathrm{Cu}^{2+} / \mathrm{Cu}}^{\circ}=0.337 \mathrm{~V}, E_{\mathrm{Ni}^{2+} / \mathrm{Ni}}^{\circ}=-0.257 \mathrm{~V} \\ & E_{\text {cell }}^{\circ}=\left(E_{\text {red }}^{\circ}\right)_{\text {cathode }}-\left(E_{\text {red }}^{\circ}\right)_{\text {anode }} \\ & \quad=0.337 \mathrm{~V}-(-0.257 \mathrm{~V})=0.594 \mathrm{~V}\end{aligned}$
View full question & answer→MCQ 421 Mark
Calculate $\Delta G^{\circ}$ for the cell $\mathrm{Sn}_{(s)}\left|\mathrm{Sn}^{++}(1 \mathrm{M}) \| \mathrm{Ag}^{+}(1 \mathrm{M})\right| \mathrm{Ag}_{(s)}$ at $25^{\circ} \mathrm{C}\left(E_{\text {cell }}^{\circ}=0.90 \mathrm{~V}\right)$.
- A
$-173.7 kj$
- B
$-225.3 kJ$
- C
$-100.2 kl$
- D
$-290.8 kJ$
Answer
$\begin{aligned} & \text {(a) : } \mathrm{Sn}\left|\mathrm{Sn}_{(1 \mathrm{M})}^{2+} \| \mathrm{Ag}_{(1 \mathrm{M})}^{+}\right| \mathrm{Ag} \\ & \Delta G=-n F E_{\text {cell }}^{\mathrm{o}} \quad(n=2) \\ & =-2 \times 96500 \times 0.90=-173700 \mathrm{~J}=-173.7 \mathrm{~kJ}\end{aligned}$
View full question & answer→MCQ 431 Mark
Identify the gas produced due to reduction of $\mathrm{NH}_4^{+}$ ions at cathode during working of dry cell.
Answer(a) : At cathode:
$
\mathrm{MnO}_2+\mathrm{NH}_4^{+}+e^{-} \longrightarrow \mathrm{MnO}(\mathrm{OH})+\mathrm{NH}_3 \uparrow
$
View full question & answer→MCQ 441 Mark
Calculate current in ampere required to deposit $4.8 g$ $Cu$ from its salt solution in 30 minutes. (Molar mass of $Ca =63.5 g mol ^{-1}$ )
Answer(a): Mass of $Cu =4.8 g$
Molar mass of $Cu =63.5 g mol ^{-1}$
$t=30$ minutes $=1800 s$
Stoichiometry for formation of $Cu$ is $Cu ^{2+}+2 e^{-} \rightarrow Cu$
According to Faraday's first law of electrolysis, $W=Z I t$
where, $Z=\frac{E}{96500}$
$
E_{ Cu }=\frac{\text { Mol. weight }}{2}=\frac{63.5}{2}
$
Heace, $Z=\frac{63.5}{2 \times 96500}$
Sobstituting the values,
$
4.8 g =\frac{I \times 1800}{96500} \times \frac{63 .5}{2} \therefore I=\frac{4.8 \times 96500 \times 2}{63.5 \times 1800}=8.1 A
$
View full question & answer→MCQ 451 Mark
A conductivity cell containing $0.001 \mathrm{M} \mathrm{AgNO}_3$ solution develops resistance $6530 \mathrm{ohm}$ at $25^{\circ} \mathrm{C}$. Calculate the electrical conductivity of solution at same temperature if the cell constant is $0.653 \mathrm{~cm}^{-1}$.
AnswerCorrect option: D. $1.0 \times 10^{-4} \Omega^{-1} cm ^{-1}$
(d) : $kappa =\frac{\text { Cell constant }}{\text { Resistance }}=\frac{0.653}{6530}=1.0 \times 10^{-4} \Omega^{-1} cm ^{-1}$
View full question & answer→MCQ 461 Mark
Calculate $E_{\text {cell }}^o$ for $\mathrm{Cd}_{(s)}\left|\mathrm{Cd}^{++}(1 \mathrm{M}) \| \mathrm{Ag}^{+}(1 \mathrm{M})\right| \mathrm{Ag}_{(s)}$
$
\left[E_{\mathrm{Cd}}^{\circ}=-0.403 \mathrm{~V} ; E_{\mathrm{Ag}}^{\circ}=0.799 \mathrm{~V}\right]
$
- ✓
$1.202 V$
- B
$-1.202 V$
- C
- D
$-0.396 V$
AnswerCorrect option: A. $1.202 V$
(a) $\begin{aligned}: E_{\text {cell }}^{\circ} & =E_{\text {cathode }}^{\circ}-E_{\text {anode }}^{\circ} \\ & =0.799-(-0.403)=1.202 \mathrm{~V}\end{aligned}$
View full question & answer→MCQ 471 Mark
Which from following is not true about voltaic cell?
- A
The anode acts as negative electrode.
- B
- ✓
It converts electrical energy into chemical energy.
- D
Dry cell is an example of volitaic cell.
AnswerCorrect option: C. It converts electrical energy into chemical energy.
(c) : Voltaic or galvanic cell converts chemical energy into electrical energy.
View full question & answer→MCQ 481 Mark
Calculate molar conductivity of $NH _4 OH$ at infinite dilution if molar conductivities of $Ba ( OH )_2$, $BaCl _2$ and $NH _4 Cl$ at infinite dilution are 520,280 , $129 \Omega^{-1} cm ^2 mol ^{-1}$ respectively.
- A
$249.0 \Omega^{-1} cm ^2 mol ^{-1}$
- B
$498.0 \Omega^{-1} cm ^2 mol ^{-1}$
- C
$125.0 \Omega^{-1} cm ^2 mol ^{-1}$
- D
$369.0 \Omega^{-1} cm ^2 mol ^{-1}$
Answer
$\begin{aligned} & \text {(a) : } \lambda_{\mathrm{Ba}^{2+}}^{\circ}+2 \lambda_{\mathrm{OH}^{-}}^{\circ}=520 \Omega^{-1} \mathrm{~cm}^2 \mathrm{~mol}^{-1}. . . . . .(i) \\ & \lambda_{\mathrm{Ba}^{2+}}^{\circ}+2 \lambda_{\mathrm{Cl}^{-}}^{\circ}=280 \Omega^{-1} \mathrm{~cm}^2 \mathrm{~mol}^{-1}......(ii) \\ & \lambda_{\mathrm{NH}_4^{+}}^{\circ}+\lambda_{\mathrm{Cl}^{-}}^{\circ}=129 \Omega^{-1} \mathrm{~cm}^2 \mathrm{~mol}^{-1} ......(iii)\\ & \mathrm{By} \frac{\text { (i) }}{2}-\frac{\text { (ii) }}{2}+\text { (iii), we get } \\ & \lambda_{\mathrm{NH}_4^{+}}^{\circ}+\lambda_{\mathrm{OH}^{-}}^{\circ}=\frac{520}{2}-\frac{280}{2}+129 \\ & \quad=260-140+129=249 \Omega^{-1} \mathrm{~cm}^2 \mathrm{~mol}^{-1}\end{aligned}$
View full question & answer→MCQ 491 Mark
Which from following expressions is used to find the cell potential of $\mathrm{Cd}_{(s)}\left|\mathrm{Cd}_{(a q)}^{++}\right|\left|\mathrm{Cu}_{(a q)}^{++}\right| \mathrm{Cu}_{(s)}$ cell at $25^{\circ} \mathrm{C}$ ?
Answer
$\begin{aligned} & \text {(a) : Anode reaction: } \mathrm{Cd}-2 e^{-} \longrightarrow \mathrm{Cd}^{++} \\ & \text {Cathode reaction: } \mathrm{Cu}^{++}+2 e^{-} \longrightarrow \mathrm{Cu} \\ & \text { Net reaction : } \mathrm{Cd}+\mathrm{Cu}^{++} \longrightarrow \mathrm{Cd}^{++}+\mathrm{Cu} \\ & \therefore \quad E_{\text {cell }}=E_{\text {cell }}^{\circ}-\frac{0.059}{2} \log \frac{\left[\mathrm{Cd}^{++}\right]}{\left[\mathrm{Cu}^{++}\right]} \\ & =E_{\text {cell }}^0-0.0295 \log \frac{\left[\mathrm{Cd}^{++}\right]}{\left[\mathrm{Cu}^{++}\right]} \\ & \end{aligned}$
View full question & answer→MCQ 501 Mark
Which of the following cannot be considered as application of electrochemical series?
- A
Relative strength of reducing agent
- ✓
Use of SHE in automobile battery
- C
Relative strength of oxidising agent
- D
Spontaneity of redox reaction
AnswerCorrect option: B. Use of SHE in automobile battery
(b) : Use of SHE in automobile battery cannot be considered as application of electrochemical series.
View full question & answer→MCQ 511 Mark
Calculate mass of a divalent metal produced at cathode by passing $5 \mathrm{amp}$ current through its salt solution for 100 minute. $($ Molar mass of metal $=x)$
Answer(c) : $t=100 min =100 \times 60 sec =6000 sec$
Charge $=$ current $\times$ time $=5 \times 6000=30000 C$
According to the reaction, $M _{\text {(aq) }}^{2+}+2 e ^{-} \longrightarrow M _{(s)}$
( $\because$ Metal is divalent.)
$2 F$ or $2 \times 96500 C$ is required to deposit 1 mole or $x$ $g$ of $M$.
(Given, atomic mass of metal $=x$ )
$\therefore$ For $30000 C$, the mass of metal deposited
$
=\frac{x \times 30000}{2 \times 96500}=\frac{x \times 300}{2 \times 965}=\frac{30 x}{193}
$
View full question & answer→MCQ 521 Mark
The unit of conductivity is
Answer(a) : Conductivity of an electrolytic solution is a measure of its ability to conduct electricity. The SI unit of conductivity is Siemens per metre ( $S / m$ ).
View full question & answer→MCQ 531 Mark
How many hours are required for a current of $3 \mathrm{~A}$ to decompose electrolytically one mole of water?
- A
$9$ hours
- B
$12$ hours
- ✓
$18$ hours
- D
$20$ hours
AnswerCorrect option: C. $18$ hours
$\ce{H_2O \Rightarrow H_2 + \frac{1}{2} O_2}$
$\therefore 18 g (1$ mole) $H _2 O$ liberates $2 g$ of $H _2$.
$E=\frac{\text { At mass }}{\text { Valency }}=\frac{1}{1}=1$
$\therefore Z=\frac{1}{96500} ; t=?, I=3 A , W=2 g$
$W=\frac{E}{96500} \times I \times t$
$t=\frac{W \times 96500}{E \times I}$
$=\frac{2 \times 96500}{3}$
$=64333 s$
$=\frac{64333}{3600} \text { hours }$
$=17.9 \text { hours }$
$\approx 18 \text { hours }$
View full question & answer→MCQ 541 Mark
Is dry cell, what acts as negative electrode?
Answer(a) : In dry cell, zinc acts as a negative electrode.
Anode reaction: $Zn _{(s)} \rightarrow Zn _{(aq)}^{2+}+2 e^{-}$
View full question & answer→MCQ 551 Mark
Two electrolytic cells are connected in series containing $\mathrm{CuSO}_4$ solution and molten $\mathrm{AlCl}_3$. If in electrolysis 0.4 moles of 'Cu' are deposited on cathode of first cell, the number of moles of 'Al' deposited on cathode of the second cell is
Answer(b) : $\mathrm{Cu}^{2+}+2 e^{-} \longrightarrow \mathrm{Cu}$
Thus for 1 mole of $\mathrm{Cu}$, charge required $=2 \mathrm{~F}$
For 0.4 mole of $\mathrm{Cu}$, charge required $=(2 / 1) \times 0.4=0.8 \mathrm{~F}$
For $\mathrm{Al}, \mathrm{Al}^{3+}+3 e^{-} \longrightarrow \mathrm{Al}$
For 1 mole of $\mathrm{Al}$, charge required $=3 \mathrm{~F}$
Thus from $0.8 \mathrm{~F}, \mathrm{Al}$ will deposit $=(1 / 3) \times 0.8=0.27$ mole
View full question & answer→MCQ 561 Mark
The conductivity of an electrolytic solution decreases in dilution due to
- ✓
decrease in number of ions per unit volume
- B
increase in ionic mobility of ions
- C
increase in percentage ionisation
- D
increase in number of ions per unit volume.
AnswerCorrect option: A. decrease in number of ions per unit volume
View full question & answer→MCQ 571 Mark
Standard hydrogen elctrode (SHE) is a
- ✓
primary reference electrode
- B
secondary reference electrode
- C
metal - sparingly soluble salt electrode
- D
metal metal ion electrode.
AnswerCorrect option: A. primary reference electrode
(a) : The electrode potential of a single electrode can only be measured by using some reference electrode. The reference electrode used is the standard or normal hydrogen electrode (S.H.E or N.H.E.).
View full question & answer→MCQ 581 Mark
The correct representation of Nernst equation for halfcell reaction $\mathrm{Cu}_{(a q)}^{2+}+e^{-} \rightarrow \mathrm{Cu}_{(a q)}^{+}$is
- A
$E_{ Cu ^{+}, Cu ^{ T }}=E_{ Cu ^2, Cu ^{2+}}=\frac{0.0592}{2} \log \frac{\left[ Cu ^{+}\right]}{\left\{ Cu ^{2+}\right]}$
- B
- C
$E_{\mathrm{Cu}^{+}, \mathrm{Cu}^{2+}}^{\circ}=E_{\mathrm{Cu}^{+}, \mathrm{Cu}^{2+}}^{\circ}+\frac{0.0592}{2} \log \frac{\left[\mathrm{Cu}^{+}\right]}{\left[\mathrm{Cu}^{2+}\right]}$
- D
$E_{\mathrm{Cu}^{+}, \mathrm{Cu}^{2+}}=E_{\mathrm{Cu}^{+}, \mathrm{Cu}^{2+}}^{\circ}+\frac{0.0592}{2} \log \frac{\left[\mathrm{Cu}^{+}\right]}{\left[\mathrm{Cu}^{2+}\right]}$
Answer
$\begin{aligned} & \text {(None): } \mathrm{Cu}_{(a q)}^{2+}+e^{-} \rightarrow \mathrm{Cu}_{(a q)}^{+} \\ & E_{\mathrm{Cu}^{2+} / \mathrm{Cu}^{+}}=E_{\mathrm{Cu}^{2+} / \mathrm{Cu}^{+}}^{\circ}-\frac{2.303 R T}{n F} \log \frac{\left[\mathrm{Cu}^{+}\right]}{\left[\mathrm{Cu}^{2+}\right]} \\ & E_{\mathrm{Cu}^{2+} / \mathrm{Cu}^{+}}=E_{\mathrm{Cu}^{2+} / \mathrm{Cu}^{+}}^{\circ}-\frac{0.0591}{1} \log \frac{\left[\mathrm{Cu}^{+}\right]}{\left[\mathrm{Cu}^{2+}\right]}\end{aligned}$
Note : In question both options (c) and (d) are same.
View full question & answer→MCQ 591 Mark
which of the following acts as oxidising agent in hydrogen-oxygen fuel cell?
Answer(b) : Hydrogen-oxygen fuel cell :
At anode: $\quad 2 \mathrm{H}_{2(g)}+4 \mathrm{OH}_{(a q)}^{-} \rightarrow 4 \mathrm{H}_2 \mathrm{O}_{(b)}+4 e^{-}$
At cathode : $\quad \mathrm{O}_{2(g)}+2 \mathrm{H}_2 \mathrm{O}_{(l)}+4 e^{-} \rightarrow 4 \mathrm{OH}_{(a q)}^{-}$
Overall reaction: $2 \mathrm{H}_{2(g)}+\mathrm{O}_{2(g)} \rightarrow 2 \mathrm{H}_2 \mathrm{O}_{(l)}$
View full question & answer→MCQ 601 Mark
During galvanization of iron, which metal is used for coating iron surface?
Answer(b): As zinc is more reactive than iron, zinc gets oxidised first when comes in contact with moisture and hence, iron surface is protected from corrosion.
View full question & answer→MCQ 611 Mark
What is the density of solution of sulphuric acid used as an electrolyte in lead accumulator?
MCQ 621 Mark
The number of moles of electrons passed when current of $2 A$ is passed through a solution of electrolyte for 20 minutes is
Answer(c) : $Q=I \times t=2 A \times 20 \times 60 sec =2400 C$
1 mole of electrons $=96500 C$
$1 C =\frac{1}{96500}$ mole of electrons
$2400 C =\frac{1 \times 2400}{96500}$ moles of electrons
$=0.02487=2.487 \times 10^{-2}$ moles of electrons
View full question & answer→MCQ 631 Mark
What is the Si unit of conductivity?
- A
$8 m$
- ✓
$5 m ^{-4}$
- C
$5 m ^2$
- D
$5 an ^{-7}$
AnswerCorrect option: B. $5 m ^{-4}$
(b) : St unit of conductivity is $5 m ^{-1}$.
View full question & answer→MCQ 641 Mark
Which among the following equations represents the reduction reaction taking place in lead accumulator at positive electrode, while it is being used as a source of electrical energy?
Answer(d): Reactions occurring in lead storage battery during discharging:
(-ve electrode)
(t) eiectrode)
$
\stackrel{+2}{ PhSO _{4 H }}+2 H _2 O
$
View full question & answer→MCQ 651 Mark
In the cell represented by $Pb _{(s)}\left| Pb ^{2+}(1 M ) \| Ag ^{+}(1 M )\right| Ag _{(s)}$, the reducing agent is
Answer(a) : In the cell represented by
$\begin{array}{cc} Pb _{(s)} \mid Pb ^{2+}(1 M ) & \| Ag ^{+}(1 M ) \mid Ag _{(s)} \\ \text { Anode } & \text { Cathode } \\ \text { (oxidation occurs) } & \text { (reduction occurs) }\end{array}$
The reducing agent is $Pb$ that undergoes oxidation.
View full question & answer→MCQ 661 Mark
How many faradays of electricity are required to deposit $10 g$ of calcium from molten calcium chloride using inert electrodes?
$\left(\right.$ molar mass of calcium $\left.=40 g mol ^{-1}\right)$
AnswerCorrect option: A. $0.5 F$
(a) : $Ca ^{2+}+2 e^{+} \longrightarrow Ca$
So, to deposit 1mol of $C$, it requires $2 F$ of efectricity. To deposit $\frac{10}{40}$ mol of $Ca$, it requires $2 \times \frac{10}{40} F =0.5 F$ of electricity.
View full question & answer→MCQ 671 Mark
Which among the following solutions is not used in determination of the cell constant?
- A
- B
$10^{-1} M KCl$
- C
$1 M KCl$
- D
Saturated $KCl$
View full question & answer→MCQ 681 Mark
The overall reaction taking place at anode during electrolysis of fused sodium chloride using suitable electrode is
- ✓
- B
- C
- D
oxidation of sodium atoms.
Answer(a) : Electrolysis of fused sodium chloride :$\begin{array}{c} NaCl _{(s)} \longrightarrow Na _{(l)}^{+}+ Cl _{(l)}^{-} \\\text {At cathode: } Na _{(l)}^{+}+e^{-} \longrightarrow Na _{(I)} \text { (Reduction) } \\\text { At anode: } Cl _{(l)}^{-} \longrightarrow Cl _{(g)}+e^{-} \quad \text { (Oxidation) } \\Cl _{(g)}+ Cl _{(g)} \longrightarrow Cl _{2(g)}\end{array}$
View full question & answer→MCQ 691 Mark
Which among the following metals is employed to provide cathodic protection to iron?
Answer(a): Zinc is traditionally used for cathodic protection of iron because it is cheap, adheres well to steel, with a relatively low galvanic potential and a small tendency to corrode. However, the corrosion rate is low in moist environments.
View full question & answer→MCQ 701 Mark
The molar conductivity of a $1.5 \mathrm{M}$ solution of an electrolyte is found to be $138.9 \mathrm{~S} \mathrm{~cm}^2 \mathrm{~mol}^{-1}$. The conductivity of this solution is
- A
- ✓
$0.208 \mathrm{~S} \mathrm{~cm}^{-1}$
- C
$0.350 \mathrm{~S} \mathrm{~cm}^{-1}$
- D
$2.085 \mathrm{~S} \mathrm{~cm}^{-1}$
AnswerCorrect option: B. $0.208 \mathrm{~S} \mathrm{~cm}^{-1}$
(b): Molar conductivity, $\Lambda_m= \kappa \times \frac{1000}{\text { Molarity }}$
$
\begin{aligned}
\text { Conductivity } (\kappa) & =\frac{138.9\left( S cm ^2 mol ^{-1}\right) \times 1.5( mol/L )}{1000\left( cm ^3 / L \right)} \\
& =0.208 cm ^{-1}
\end{aligned}
$
View full question & answer→MCQ 711 Mark
How is electrical conductance of a conductor related with length and area of cross section of the conductor?
- A
- B
$G=\kappa. l. a^{-1}$
- ✓
$G=\kappa. a. l^1$
- D
$G=\kappa. l. a^{-2}$
AnswerCorrect option: C. $G=\kappa. a. l^1$
(c): Conductance, $G=\frac{1}{R}$ and, $\kappa=\frac{l}{a} \times G \quad$ or, $G=\kappa a l^{-1}$
View full question & answer→MCQ 721 Mark
If standard reduction potentials for $Pb , K , Zn$ and $Cu$ are $-0.126 V ,-2.925 V ,-0.763 V$ and 0.337 $V$, the decreasing order of reducing power is
- A
$Zn > Pb > K > Cu$
- B
$Cu > Pb > Zn > K$
- ✓
$K > Zn > Pb > Cu$
- D
$K > Pb > Cu > Zn$
AnswerCorrect option: C. $K > Zn > Pb > Cu$
View full question & answer→MCQ 731 Mark
The standard reduction potentials of $Sn , Hg$ and $Cr$ are $-1.36 V , 0.854 V$ and $-0.746 V$ respectively. The increasing order of oxidising power of the given elements is
- A
$Sn < Hg < Cr$
- B
$Hg < Cr < Sn$
- ✓
$Sn < Cr < Hg$
- D
$Cr < Hg < Sn$
AnswerCorrect option: C. $Sn < Cr < Hg$
View full question & answer→MCQ 741 Mark
The cell potential of the following cell is $\begin{aligned} & \left( E ^0 Al ^{3+} / Al =-1.66 V \right) \\ & Al ^2 Al _2\left( SO _4\right)_{3( aq )} \| HCl _{\text {(aq }}\left| H _{2( g )}\right| Pt \\ & 0.5 M \end{aligned}$
- ✓
$1.66 V$
- B
$-1.66 V$
- C
$0.5533 V$
- D
$2.14 V$
AnswerCorrect option: A. $1.66 V$
View full question & answer→MCQ 751 Mark
The reaction, $2 Br _{( aq )}^{-}+ Sn _{( aq )}^{2+} \longrightarrow Br _{2( l )}+ Sn _{( s )}$
with the standard potentials, $E _{ Sn }^0=-0.114 V , E _{ Br _2}^0=+1.09 V$, is
- ✓
spontaneous in reverse direction
- B
spontaneous in forward direction
- C
- D
non-spontaneous in reverse direction
AnswerCorrect option: A. spontaneous in reverse direction
spontaneous in reverse direction
View full question & answer→MCQ 761 Mark
The strongest oxidizing agent among the species $\ln ^{3+}$ $\left( E ^0=-1.34 V \right), Au ^{3+}\left( E ^0=1.4 V \right)$ $Hg ^{2+}\left( E ^0=0.86 V \right), Cr ^{3+}\left( E ^0=-0.74 V \right)$ is
- A
$Cr ^{3+}$
- ✓
$Au ^{3+}$
- C
$Hg ^{2+}$
- D
$\ln ^{3+}$
AnswerCorrect option: B. $Au ^{3+}$
View full question & answer→MCQ 771 Mark
The efficiency of the hydrogen$-$oxygen fuel cell is about
- A
$20 \%$
- B
$40 \%$
- ✓
$70 \%$
- D
$90 \%$
AnswerCorrect option: C. $70 \%$
View full question & answer→MCQ 781 Mark
In lead accumulator, anode and cathode are
- A
$\left( Pb + PbO _2\right), Pb$
- B
$Pb , PbO _2$
- C
$PbO _2, Pb$
- ✓
$Pb ,\left( Pb + PbO _2\right)$
AnswerCorrect option: D. $Pb ,\left( Pb + PbO _2\right)$
View full question & answer→MCQ 791 Mark
During the discharging of a lead storage battery,
AnswerCorrect option: A. $H _2 SO _4$ is consumed
View full question & answer→MCQ 801 Mark
In the Lead storage battery during discharging
- ✓
pH of the electrolyte increases
- B
- C
- D
pH increases or decreases depends on the extent of discharging
AnswerCorrect option: A. pH of the electrolyte increases
pH of the electrolyte increases
View full question & answer→MCQ 811 Mark
The metal which cannot displace hydrogen from dil. $H _2 SO _4$ solution is
View full question & answer→MCQ 821 Mark
The standard cell potential of the following cell is $0.463 VCu C Cu ^{++}(1 M )|| Ag ^{+}(1 M ) \mid Ag$. If $E _{ Ag }^0=0.8 V$, what is the standard potential of $Cu$ electrode ?
- A
$1.137 V$
- ✓
$0.337 V$
- C
$0.463 V$
- D
$-0.463 V$
AnswerCorrect option: B. $0.337 V$
View full question & answer→MCQ 831 Mark
The Electromotive Force of the following Cell $Cu \left| Cu ^{++}(1 M ) \| A ^{+} g (1 M )\right| Ag$ is _________if$E _{ Cu ^{++}}=0.33 V$ and $E ^0 Ag ^{++} / Ag =0.79 V$
- ✓
$0.46 V$
- B
$-0.46 V$
- C
$1.12 V$
- D
$-112 V$
AnswerCorrect option: A. $0.46 V$
$0.46 V$
View full question & answer→MCQ 841 Mark
The emf of the cell,
$\underset{(1 \text { atm })}{ H _2} \underset{(1 m )}{ H ^{+}} \| \underset{( M m )}{ Cu ^{2+}} Cu$ is
$\left(E_{\text {red }}^0=0.34 V \right)$
- A
$-1.34$
- ✓
$0.34 V$
- C
$-0.34 V$
- D
$1.34$
AnswerCorrect option: B. $0.34 V$
$0.34 V$
View full question & answer→MCQ 851 Mark
The standard reduction potentials of metals $A$ and $B$ are $x$ and $y$ respectively. If $x>y$, the standard emf of the cell containing these electrodes would be
- A
$2 x-y$
- B
$y-x$
- ✓
$x-y$
- D
$x+y$
View full question & answer→MCQ 861 Mark
The concept of electrode potential is explained on the basis of
MCQ 871 Mark
In Nernst equation the constant $0.0592$ at $298 K$ represents the value of
- A
$\frac{R T}{n F}$
- B
$\frac{R T}{F}$
- C
$\frac{2.303 R T}{n F}$
- ✓
$\frac{2.303 R T}{F }$
AnswerCorrect option: D. $\frac{2.303 R T}{F }$
View full question & answer→MCQ 881 Mark
Which of the following species gains electrons more easily ?
- A
$Na ^{+}$
- ✓
$H ^{+}$
- C
$Mg ^{+}$
- D
$Hg ^{+}$
AnswerCorrect option: B. $H ^{+}$
View full question & answer→MCQ 891 Mark
The electrode potential of a silver electrode dipped in $0.1 M AgNO _3$ solution at $298 K$ is ( $E _{\text {red }}$ of $Ag =0.80$ volt)
- A
$0.0741 V$
- B
$0.0591 V$
- ✓
$0.741 V$
- D
$0.859 V$
AnswerCorrect option: C. $0.741 V$
View full question & answer→MCQ 901 Mark
The emf of cell is 1.3 volt. The positive electrode has potential of 0.5 volt. The potential of negative electrode is
- A
$0.8 V$
- ✓
$-0.8 V$
- C
$1.8 V$
- D
$-1.8 V$
AnswerCorrect option: B. $-0.8 V$
$-0.8 V$
View full question & answer→MCQ 911 Mark
In hydrogen$-$oxygen fuel cell, the carbon rods are immersed in hot aqueous solution of
- A
$KCl$
- ✓
$KOH$
- C
$H _2 SO _4$
- D
$NH _4 Cl$
View full question & answer→MCQ 921 Mark
The essential condition to set a standard hydrogen electrode is
MCQ 931 Mark
The standard hydrogen electrode is represented as
- A
$H _{( aq )}^{+}\left| H _2( g , 1 atm )\right| Pt$
- ✓
$H _{( aq )}^{+} 1 M \left| H _2( g , 1 atm )\right| Pt$
- C
$H _{( aq )}^{+} 1 M \left| H _{2( g )}\right| Pt$
- D
$H _{( aq )}^{+} 0.1 M \left| H _2( g , 1 atm )\right| Pt$
AnswerCorrect option: B. $H _{( aq )}^{+} 1 M \left| H _2( g , 1 atm )\right| Pt$
View full question & answer→MCQ 941 Mark
In the Daniell cell, reduction occurs at the
MCQ 951 Mark
In the representation of galvanic cell, the ions in the same phase are separated by a
MCQ 961 Mark
- A
- B
- C
primary irreversible cell
- ✓
View full question & answer→MCQ 971 Mark
Number of faradays of electricity required to liberate 12 g of hydrogen is
MCQ 981 Mark
On passing 1.5 F charge, the number of moles of aluminium deposited at cathode are [Molar mass of Al = 27 gram $mol ^{-3}$]
- A
$1.0$
- B
$13.5$
- ✓
$0.50$
- D
$0.75$
AnswerCorrect option: C. $0.50$
$0.50$
View full question & answer→MCQ 991 Mark
On calculating the strength of current in amperes if a charge of $840 C$ (coulomb) passes through an electrolyte in 7 minutes, it will be
View full question & answer→MCQ 1001 Mark
Passage of $5400 C$ of electricity through an electrolyte deposited $5.954 \times 10^{-3} kg$ of the metal with atomic mass $106.4.$ The charge on the metal ion is
View full question & answer→MCQ 1011 Mark
The quantity of electricity required to deposit $54 g$ of silver from silver nitrate solution is
View full question & answer→MCQ 1021 Mark
When a charge of 0.5 Faraday is passed through $AlCl _3$ solution, the amount of aluminium deposited at the cathode is (Atomic weight of $Al =27$ )
View full question & answer→MCQ 1031 Mark
When 0.2 Faraday of electricity is passed through an electrolytic solution, the number of electrons involved are
- A
- B
$1.603 \times 10^{-19}$
- ✓
$1.2046 \times 10^{23}$
- D
$12 \times 10^6$
AnswerCorrect option: C. $1.2046 \times 10^{23}$
$1.2046 \times 10^{23}$
View full question & answer→MCQ 1041 Mark
The number of electrons that have a total charge of $965$ coulombs is
- A
$6.022 \times 10^{23}$
- B
$6.022 \times 10^{22}$
- ✓
$6.022 \times 10^{21}$
- D
$3.011 \times 10^{23}$
AnswerCorrect option: C. $6.022 \times 10^{21}$
View full question & answer→MCQ 1051 Mark
The amount of electricity equal to $0.05 F$ is
- A
$48250 C$
- B
$3776 C$
- ✓
$4825 C$
- D
$4285 C$
AnswerCorrect option: C. $4825 C$
View full question & answer→MCQ 1061 Mark
The charge of how many coulomb is required to deposit $1.0 g$ of sodium metal (molar mass $23.0 g mol ^{-1}$ ) from sodium ions is
View full question & answer→MCQ 1071 Mark
The S.I. unit of cell constant for conductivity cell is
AnswerCorrect option: A. $m^{-1}$
$m^{-1}$
View full question & answer→MCQ 1081 Mark
What weight of copper will be deposited by passing 2 Faradays of electricity through a cupric salt? (atomic mass $=63.5$ )
- ✓
$63.5 g$
- B
$31.75 g$
- C
$127 g$
- D
$12.7 g$
AnswerCorrect option: A. $63.5 g$
$63.5 g$
View full question & answer→MCQ 1091 Mark
What is the ratio of volumes of $H _2$ and $O _2$ liberated during electrolysis of acidified water?
- A
$1: 2$
- ✓
$2 : 1$
- C
$1 : 8$
- D
$8 : 1$
AnswerCorrect option: B. $2 : 1$
View full question & answer→MCQ 1101 Mark
If $\wedge_{ m }$ and $\wedge_0$ are the molar conductivities of a weak electrolyte at concentration $C$ and at zero concentration, then the dissociation constant $Ka$ is given by
- A
$K_{ a }=\frac{\wedge_{ m }^2 \times C }{\wedge_0-\wedge_{ m }}$
- B
$K_{ a }=\frac{\wedge_{ m }^2 \times C }{\wedge_0\left(\wedge_0-\wedge_{ m }\right)}$
- C
$K_{ a }=\frac{\wedge_0^2 \times C }{\left(\wedge_0-\Lambda_{ m }\right)}$
- D
$K_{ a }=\frac{\wedge_{ m } \times C }{\wedge_0\left(\wedge_0-\wedge_{ m }\right)}$
View full question & answer→MCQ 1111 Mark
The molar conductivity of cation and anion of salt BA are 180 and 220 mhos respectively. The molar conductivity of salt BA at infinite dilution is
- A
$90 mhos \cdot cm ^2 \cdot mol ^{-1}$
- B
$110 mhos \cdot cm ^2 \cdot mol ^{-1}$
- C
$200 mhos \cdot cm ^2 \cdot mol ^{-1}$
- ✓
$400 mhos \cdot cm ^2 \cdot mol ^{-1}$
AnswerCorrect option: D. $400 mhos \cdot cm ^2 \cdot mol ^{-1}$
$400 mhos \cdot cm ^2 \cdot mol ^{-1}$
View full question & answer→MCQ 1121 Mark
$\wedge_0$ for $CH _3 COOH$ is $390.7 \Omega^{-1} cm ^2 mol ^{-1}$. If $\wedge_0$ for $CH _3 COOK$, and $HBr$ in $\Omega^{-1} cm ^2 mol ^{-1}$ are 115 and 430.4 respectively, then $\wedge 0$ for $KBr$ is
- A
$74.6 \Omega^{-1} cm ^2 mol ^{-1}$
- B
$180.6 \Omega^{-1} cm ^2 mol ^{-1}$
- ✓
$154.7 \Omega^{-1} cm ^2 mol ^{-1}$
- D
$706.1 \Omega^{-1} cm ^2 mol ^{-1}$
AnswerCorrect option: C. $154.7 \Omega^{-1} cm ^2 mol ^{-1}$
$154.7 \Omega^{-1} cm ^2 mol ^{-1}$
View full question & answer→MCQ 1131 Mark
The degree of dissociation of a weak electrolyte is given by
- A
$\alpha=\frac{\Lambda_0}{\Lambda_{ m }}$
- B
$\alpha=\Lambda_{ m } \times \Lambda_0$
- ✓
$\alpha=\frac{\Lambda_m}{\Lambda_0}$
- D
$\alpha=\Lambda_0-\Lambda_m$
AnswerCorrect option: C. $\alpha=\frac{\Lambda_m}{\Lambda_0}$
View full question & answer→MCQ 1141 Mark
Kohlrausch’s law is represented as
- ✓
$\Lambda_0=\lambda_{+}^0+\lambda_{-}^0$
- B
$\Lambda_0=\lambda_{+}^0-\lambda_{-}^0$
- C
$\Lambda_{ m }=\lambda_{+}^0+\lambda_{-}^0$
- D
$\Lambda_0=\frac{\Lambda_{ m }}{\lambda_{+}^0+\lambda_{-}^0}$
AnswerCorrect option: A. $\Lambda_0=\lambda_{+}^0+\lambda_{-}^0$
View full question & answer→MCQ 1151 Mark
If conductivity is expressed in $\Omega^{-1} m ^{-1}$ and concentration of the electrolytic solution in $mol m ^{-3}$ then, the molar conductance is given by
- A
$\Lambda_{ m }=\frac{\kappa \times 1000}{C}$
- ✓
$\Lambda_{ m }=\frac{\kappa}{C}$
- C
$\wedge_{ m }=\frac{C}{\kappa}$
- D
$\Lambda_{ m }=\frac{C \times 1000}{\kappa}$
AnswerCorrect option: B. $\Lambda_{ m }=\frac{\kappa}{C}$
View full question & answer→MCQ 1161 Mark
The units of molar conductivity are
- A
$\Omega cm ^{-2} mol ^{-1}$
- ✓
$\Omega^{-1} cm ^2 mol ^{-1}$
- C
$\Omega^{-1} cm ^{-1} mol ^{-1}$
- D
$\Omega^{-} cm ^{-1} mol ^{-2}$
AnswerCorrect option: B. $\Omega^{-1} cm ^2 mol ^{-1}$
View full question & answer→MCQ 1171 Mark
Molar conductivity of an electrolyte is given by,
- A
$\Lambda_{ m }=\frac{C \times 1000}{\kappa}$
- B
$\Lambda_{ m }=\frac{\kappa}{C \times 1000}$
- ✓
$\Lambda_{ m }=\frac{\kappa \times 1000}{C}$
- D
$\Lambda_{ m }=\frac{1000}{\kappa \times C}$
AnswerCorrect option: C. $\Lambda_{ m }=\frac{\kappa \times 1000}{C}$
View full question & answer→MCQ 1181 Mark
The specific conductance of $0.02 M HCl$ is $8.2 \times 10^{-3} \Omega^{-1} cm ^{-1}$. Hence its molar conductivity is
- A
$164 \Omega^{-1} cm ^2 mol ^{-1}$
- B
$6.1 \times 10^3 \Omega^{-1} cm ^2 mol ^{-1}$
- C
$239.6 S \ cm ^2 mol ^{-1}$
- ✓
$410 S \ cm ^2 mol ^{-1}$
AnswerCorrect option: D. $410 S \ cm ^2 mol ^{-1}$
View full question & answer→MCQ 1191 Mark
The conductivity of $0.02 M KI$ solution is $4.37 \times 10^{-4} \Omega^{-1} cm ^{-1}$. Hence its molar conductivity is
- A
$8.74 \times 10^{-6} \Omega^{-1} cm ^2 mol ^{-1}$
- ✓
$21.85 \Omega^{-1} cm ^2 mol ^{-1}$
- C
$4.58 \times 10^{-4} \Omega^{-1} cm ^2 mol ^{-1}$
- D
$136.5 \Omega^{-1} cm ^2 mol ^{-1}$
AnswerCorrect option: B. $21.85 \Omega^{-1} cm ^2 mol ^{-1}$
View full question & answer→MCQ 1201 Mark
A conductivity cell has two platinum electrodes of area $1.2 \ cm^2$ and $0.92 \ cm$ apart. Hence the cell constant is
- A
$1.104 \ cm ^{-1}$
- B
$1.304 \ cm ^{-1}$
- C
$0.906 \ cm ^{-1}$
- ✓
$0.767 \ cm ^{-1}$
AnswerCorrect option: D. $0.767 \ cm ^{-1}$
View full question & answer→MCQ 1211 Mark
The cell constant of a conductivity cell is given by
- A
$l \times a$
- B
$\frac{a}{l}$
- C
$\frac{1}{l \times a}$
- ✓
$\frac{l}{a}$
AnswerCorrect option: D. $\frac{l}{a}$
View full question & answer→