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Chemistry 2 Marks Questions

Electrochemistry · Rajasthan Board (RBSE) STD 12 Science (Rajasthan - English Medium). 25 questions.

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2 Marks Questions

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Question 12 Marks
What are reduction potential and oxidation potential?
Answer
The reduction half-potential in a metabolic reaction is known as reduction potential and the potential of oxidation half reaction is called oxidation potential. For a half reaction, the values of oxidation potential and reduction potential are same but their signs are opposite.
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Question 22 Marks
$Fe ^{2+} \rightarrow Fe ^{3+}\left( E ^{\circ}=-0.77 V\right)$ and$Cu \rightarrow Cu ^{2+}\left( E ^{\circ}=-0.34 V\right) On$ the basis of these values, find out whether $Cu ^{2+}$ ion can be reduce by $Fe ^{2+}$ ion or not?
Answer
The given values are of oxidation potential, hence the value of reduction potential of $Fe ^{2+} \rightarrow Fe ^{3+}$ will be 0.77 V and 0.34 V for $Cu \rightarrow Cu ^{2+}$. Since the reduction potential for $Fe ^{3+}$ is higher than that of $Cu ^{2+}$. Therefore, $Cu ^{2+}$ ion cannot be reduced by $Fe ^{2+}$ ion.
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Question 32 Marks
Explain concentration cell.
Answer
The cell in which two identical electrodes are immersed in solutions with different concentration of their ions. Due to this, the potential of these electrodes also varies. Therefore the potential is produced in the cell. The cell potential at 298 K is $E _{\text {cell }}=\frac{0.059}{ n } \log \frac{ C _2}{ C _1}$
$C _1=$ concentration of anode, $C _2=$ concentration of cathode
Here $\left( C _1< C _2\right)$.
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Question 42 Marks
What happens when copper is reacted with HCl and $HNO _3$ ?
Answer
There is no reaction of copper with HCl because $H ^{+}$ions cannot oxidize Cu , that is Cu can't give electrons to $H ^{+}$as it is less reactive than hydrogen. But when Cu is reacted with $HNO _3$, it gets oxidized by $NO _3{ }^{-}$ ions instead of reacting with $H ^{+}$.
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Question 52 Marks
Cell $Z n _{( s )}\left| Z n _{( aq )}^{2+} \| C u _{( aq )}^{+ 2 }\right| C u _{( s )}$. The value of cell potential for $Cu ( s )$ is positive. What conclusion is drawn from this?
Answer
This is a galvanic cell and the Nernst equation written for this is a follows :
$
E_{cell}=E_{cell}^{o}-\frac{0.059}{2} \log \frac{\left[Zn^{2+}\right]}{\left[Cu^{2+}\right]}
$
This equation proves that it is possible for the cell value to be positive only when the concentration of $Cu ^{2+}$ is high. In this situation, Zn gets oxidized to give $Zn ^{2+}$ and $Cu ^{2+}$ ion is reduced to get Cu .
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Question 62 Marks
On what factors do the products obtained from electrolysis a solution of an electrolyte depend?
Answer
Factors affecting the products obtained during electrolysis are:
(i) Nature of the electrolyte (Aqueous or molten)
(ii) Concentration of the electrolyte
(iii) Charge density flowing during electrolysis
(iv) Nature of the electrode (inert or active)
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Question 72 Marks
How does electrolysis of molten NaCl occurs?
Answer
Molten NaCl contains only $Na ^{+}$and $Cl ^{-}$ions, hence $Na ^{+}$is reduced at the cathode to form Na and $Cl ^{-}$at the anode get oxidised to form $Cl _2$.
At cathode: $2 Na ^{+}+2 e ^{-} \rightarrow 2 Na _{( s )}$
At anode : $\quad 2 Cl ^{-}-2 e ^{-} \rightarrow Cl _2(g)$
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Question 82 Marks
Write the reactions taking place in the electrolysis of dilute and concentrated $H _2 SO _4$.
Answer
On electrolysis of dilute $H _2 SO _4$, the following reaction ocurs :
$
2 H_2 O(l) \rightarrow O_2(g)+4 H^{+}(aq)+4 e^{-} ; E_{cell}^{\circ}=+1.23 V
$
And on taking concentrated $H _2 SO _4$, following reaction will take place :
$
2 SO_4^{2-}(aq) \rightarrow S_2 O_8^{2-}(aq)+2 e^{-} ; E_{cell}^{\circ}=+1.96 V
$
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Question 92 Marks
How is the solubility of a sparingly soluble salt determined by measuring conductivity?
Answer
The solubility of sparingly soluble salts like BaSO4 in water is very low, hence the concentration ions in its solution will be very low. Therefore, its solution can be considered infinitely dilute and the concentration of its saturated solution will be its solubility. By knowing the molar conductivity of the solution, the solubility can be calculated by following formula:<br>$S =\frac{\kappa \times 1000}{\Lambda_{ m }^{\circ}}$
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Question 102 Marks
Explain the difference between electrolytic conduction and metallic conduction.
Answer
(i) Electrolytic conduction occurs through ions whereas metallic conduction occurs through electrons.
(ii) In electrolytic conduction, when electricity is passed, ions takes part in reaction, that is, electrolysis takes place, whereas there is no effect on metallic conduction.
(iii) Conductivity of electrolytic conductors increases on increasing the temperature whereas metallic conductors conductivity decreases with increase in temperature.
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Question 112 Marks
What is the function of zinc chloride $\left( ZnCl _2\right)$ in dry cell?
Answer
$Zinc \left( Zn ^{2+}\right)$ is obtained from $ZnCl _2$ reacts with $NH _3$ produced in the reaction to form $\left[ Zn \left( NH _3\right)_4\right]^{2+}$ complex ion due to which pressure is not produced and there is no possibility of seal breaking.
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Question 122 Marks
What is the effect on the $H _2 SO _4$ present in a lead storage cell if it is used continously?
Answer
When a lead storage cell is used continously, the density of $H _2 SO _4$ present in cell decreases because during the reaction, sulphate $\left( SO _4{ }^{2-}\right)$ ions react with $Pb ^{+2}$ to form $PbSO _4$.
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Question 132 Marks
How are $Na , Mg$ and Al are obtained from electrolysis?
Answer
Sodium and Magnesium are obtained by electrolysis of their molten chlorides and aluminium is obtained by electrolysis of aluminium oxide in presence of cryolite $\left( Na _3 AlF _6\right)$.
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Question 142 Marks
What is called electrochemical equivalent (Z)?
Answer
The amount of substance deposited on the electrode when a current of one ampere is passed for one second (one coulomb charge) in solution of an electrolyte is called electro-chemical equivalent.
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Question 152 Marks
What is called overpotential?
Answer
Some electrochemical processes are possible but their speed is so low that they do not happen easily at low potential. In this situation, additional potential has to be applied to make the process happen, it is called overpotential.
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Question 162 Marks
How many Coulomb charge is required to obtain one mole of Al from $Al ^{3+}$ ?
Answer
According to the reaction,
$
Al^{3+}+3 e^{-} \longrightarrow Al,
$
$3 F=3 \times 96500=289500$ coulomb charge is required to obtain 1 mole of $Al ^2$ from $Al ^{3+}$.
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Question 172 Marks
Calculate the value of $\text{EMF}$ for the cell reaction :
$Cu_{(s)}+2 Ag_{(aq)}^{+} \rightarrow Cu_{(aq)}^{2+}+2 Ag_{(s)}$
$E_{Cu^{+2} / Cu}^{o}=0.34 V \text { and } E_{Ag^{+} / Ag}^{o}=0.80 V$
Answer
$ E _{\text {cell }}^{\circ} = E _{ R }- E _{ L }$
$ =0.80-(0.34)$
$E _{\text {cell }}^{\circ} =0.46 V$
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Question 182 Marks
Explain the direction of current and flow of electrons in Daniell cell.
Answer
Flow of current in Daniell cell is from Cu to Zn electrode and the flow of electrons is from Zn towards Cu electrode, that is, the effect of current and electrons is in opposite direction to each other.
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Question 192 Marks
Can aqueous solution of copper sulphate $\left( CuSO _4\right)$ be kept in iron vessel or not?
Answer
 Aqueous solution of $CuSO _4$ can not be kept in an iron vessel because its reactivity is more than Cu . $\left( E _{ Fe ^{+2} / Fe }^{ o }< E _{ Cu ^{+2} / Cu }^{ o }\right)$, hence, Fe reduces $Cu ^{2+}$ to Cu in which the following reaction occurs :
$
Cu_{(aq)}^{2+}+Fe_{(s)} \rightarrow Fe_{(aq)}^{+2}+Cu_{(s)}
$
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Question 202 Marks
When a zinc metal rod is placed in aqueous solution of $CuSO _4$ and is immersed, the blue colour of the solution disappears. Why?
Answer
Since, the standard reduction potential of $Zn ^{2+}$ is less than the standard reduction potential of $Cu ^{2+}, Zn$ reduces $Cu ^{2+}$ by donating electrons to it. Hence the blue colour of solution disappears because the blue colour of the solution is due to $Cu ^{2+}$ ions.
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Question 212 Marks
What will be the amount of electricity in Coulomb required to obtain $6$ moles of $H _2$ gas by electrolysis of $H ^{+}$ion solution?
Answer
$2 H ^{+}+2 e ^{-} \rightarrow H _2$
According to the reaction, the amount of electricity required to obtain 1 mole $H _2=2$ Faraday Hence, the amount of electricity required to obtain 6 mole of
$
\begin{aligned}
H_2 & =2 \times 6=12 F \\
12 F & =12 \times 96500 C \\
& =1158000 C \\
& =11.58 \times 10^5 C
\end{aligned}
$
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Question 222 Marks
The ratio of amount of $Al , Cu$ and Na obtained at the cathode by passing 3 Faraday electric current in different electrolytic cell filled with aqueous solution of molten $AlCl _3, CuSO _4$ and molten NaCl . What be the ratio of amounts of $Al , Cu$ and Na ?
Answer
(i) $Al _{( aq )}^{3+}+3 e ^{-} \rightarrow Al _{( s )}$ (cathodic reaction)
$3 \times 96500$ coulomb electricity will produce 1 mol Al .
(ii) $Cu _{\text {(aq) }}^{2+}+2 e ^{-} \rightarrow Cu _{\text {(s) }}$
2 Faraday electric produces one mole of Cu .
3 Faraday electricity will produce $3 / 2 mol$ of Cu .
(iii) $Na ^{+}+ e ^{-} \rightarrow Na _{\text {(s) }}$ 1 Faraday electricity produces 1 mol Na .
3 Faraday electricity produces 3 mol Na .
Ratio $=1: 3 / 2: 3=1: 1.5: 3=2: 3: 6$.
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Question 232 Marks
If $6 \times 10^{20}$ electron are used in reduction of an ion to form an atom, then find the equivalent of the ion.
Answer
$X^{n+}+ne^{-} \rightarrow X$
$n=6 \times 10^{20}=0.001 \text{mol}$
$6 \times 10^{23}=1 \text{mol}$
Since, one equivalent ion accepts $1$ mole electron.
Therefore, equivalent of $0.001$ mole electron accepting ion $= 0.001 .$
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Question 242 Marks
The resistance of a conductivity cell containing 0.001 M KCl solution at 298 K is 1500$\Omega$. What is the cell constant if conductivity of 0.001 M KCl solution at 298 K is 0.146 × 10-3 S cm-1?
Answer
Cell constant $\left( G ^*\right)=\frac{\text { Conductivity }}{\text { Conductance }}=\frac{\kappa}{ G }$
Resistivity $R=\frac{1}{G}$
Hence, Cell constant = Conductivity × Resistivity
Conductivity $( k )=0.146 \times 10^{-3} S cm ^{-1}$
Resistivity $R =1500 \Omega$
Hence, Cell constant $=0.146 \times 10^{-3} \times 1500$
Cell constant $=0.219 cm^{-1}$.
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Question 252 Marks
Given the standard electrode potentials,
$\begin{array}{l} K ^{+} / K =-2.93 V, Ag ^{+} / Ag =0.80 V \\ Hg ^{2+} / Hg =0.79 V \\ Mg ^{2+} / Mg =-2.37 V, Cr ^{3+} / Cr =-0.74 V\end{array}$
Arrange these metals in their increasing order of reducing power.
Answer
When the oxidation potential of the metal is high i.e. the reduction potential of the metal ion is low then the tendency of that metal to donate electron is high and it is a strong reducing agent. Therefore, on the basis of given standard electrode potential (reduction potential) values, the increaseing order of reducing capacity of these metal will be as follows : Ag < Hg < Cr < Mg < K
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2 Marks Questions - Chemistry STD 12 Science Questions - Vidyadip