Q: Resistance and conductivity of a cell containing 0.001 M KCl solution at 298 K are 1500 \Omega and 1.46 \times 10^{-4} S.cm^{-1} respectively. What is the cell constant ?

R=1500 \ \Omega (Resistance) k= 1.46 \times 10^{-4} \ Scm^{-1} (Conductivity) T=298 \ K We will use the formula, \boxed {k=Gx} where k is conductivity G is \cfrac 1R x is cell constant 1.46 \times 10^{-4} = \cfrac 1{1500} \times x 1.46 \times 10^{-4} \times 1500 =x \boxed {0.219 \ cm^{-1}=x} The cell constant is 0.219 \ cm^{-1}

Q: How many electrons are there in one coulomb of electricity?

Change on 1e^-=1.602\times 10^{-19} C If total charge =1 C \therefore Total no. of e^-=\cfrac{1}{1.602\times 10^{-19}}=6.24\times 10^{18}e^- Hence, the correct option is (C) .

Question 1

Calculate the emf of the cell.  Zn | Zn^{2+} (0.001 M) || Cu^{2+} (0.1 M) | Cu The standard potential of  Cu/Cu^{2+}  half-cell is +0.34 and  Zn/Zn^{2+}  is -0.76 V.

Accepted Answer

Given  Zn|Zn^{2+}(0.001M)||Cu^{2+}(0.1M)|Cu Overall cell reaction: Zn\longrightarrow { Zn }^{ 2+ }+2{ e }^{ - }\ \underline { { Cu }^{ 2+ }+2{ e }^{ - }\longrightarrow Cu\quad \quad \quad  } \ \underline { Zn+{ Cu }^{ 2+ }\longrightarrow { Zn }^{ 2+ }+Cu }  E_{cell}^{o}= standard reduction potential of cathode + standard oxidation potential of anode E_{cell}^{o}= 0.34 to 0.76 V E_{cell}^{o}= 1.1 V { K }_{ C }=\cfrac { \left[ { Zn }^{ 2+ } ight]  }{ \left[ { Cu }^{ 2+ } ight]  } =\cfrac { 10^{ -3 } }{ 10^{ -1 } } =10^{ -2 } EMF of the cell at any electrode concentration is: E=E^{ o }-\cfrac { 0.059 }{ n } \log  \left( { K }_{ C } ight) =1.1-\cfrac { 0.059 }{ 2 } \log  \left( 10^{ -2 } ight) =1.1-\cfrac { 0.059 }{ 2 } 	imes (2)=1.1-0.059=1.041V

Question 2

Calculate the potential of hydrogen electrode in contact with a solution whose  pH  is  10 .

Accepted Answer

given  pH=10 we known that  pH=-\log[H^+] \Rightarrow [H^+]=10^{-10}M also for hydrogen electrode :  E_{cell}=E^0-\dfrac{0.0591}{1}\log \dfrac{|H_2|}{|H^+|} (no of electron included  =1 ) \boxed{E=0} E_{cell}=\dfrac{-0.0591}{1}\log\left(\dfrac{1}{10^{-10}}\right) \boxed{E_{cell}=-0.591V}

Question 3

The conductivity of 0.20 M solution of  KCl  at 298 K is  0.0248 S cm^{-1} . Calculate its molar conductivity.

Accepted Answer

k  (Conductivity)  =0.0248 \ Scm^{-1} C  (molarity)  =0.20 \ M T  (temperature)  =298 \ K  \Lambda _m =?  (molar conductivity) The formula is  \Lambda_m = \cfrac {1000 	imes k}{C} = \cfrac {1000 	imes 0.0248 \ Scm^{-1}}{0.20 \ mol\ cm^{-3}} =124 \ S \ cm^2 \ mol^{-1} The molar conductivity is  124 \ S \ cm^2 \ mol^{-1}

Question 4

Calculate the emf of the cell in which of the following reaction takes place: Ni(s) + 2Ag^+ (0.002 M) \rightarrow Ni^{2+} (0.160 M) + 2Ag(s) [Given that  E^{\circ}_{cell} = 1.05 V ]

Accepted Answer

Given, [Ag^+]=0.002M [Ni^{2+}]=0.160M n=2 According to Nernst equation: E_{cell}=E^0-\dfrac{0.0591}{2}\log \dfrac{[Ni^{2+}]}{[Ag^+]^2} 	herefore E_{cell}=1.05-\dfrac{0.0591}{2}\log \dfrac{0.160}{(0.002)^2} 	herefore E_{cell}=1.05-0.02955\, \log{(4	imes 10^4)} 	herefore E_{cell}=0.91V Hence, the correct option is  	ext{B}

Question 5

Depict the galvanic cell in which the reaction  Zn(s) + 2Ag^{+} (aq) \rightarrow Zn^{2+}(aq) + 2Ag(s)  takes place. Further show:(i) Which of the electrode is negatively charged?(ii) The carriers of the current in the cell.(iii) Individual reaction at each electrode.

Accepted Answer

The galvanic cell in which the given reaction takes place is depicted as:  \displaystyle Zn(s)| Zn^{2+} (aq) || Ag^+ (aq) |Ag(s) (i) The negatively charged electrode is the zinc electrode. It acts as an anode.(ii) In the external circuit, the current will flow from silver to zinc. (iii) Oxidation at anode:  \displaystyle Zn(s) ightarrow Zn^{2+}(aq) +2e^- Reduction at cathode:  \displaystyle Ag^+(aq) + e^- 	o Ag(s)

Question 6

Resistance and conductivity of a cell containing 0.001 M KCl solution at 298 K are  1500 \Omega  and  1.46 	imes 10^{-4} S.cm^{-1}  respectively. What is the cell constant ?

Accepted Answer

R=1500 \ \Omega  (Resistance) k= 1.46 	imes 10^{-4} \ Scm^{-1}  (Conductivity)  T=298 \ K We will use the formula,  \boxed {k=Gx} where  k  is conductivity G  is  \cfrac 1R x  is cell constant 1.46 	imes 10^{-4} = \cfrac 1{1500} 	imes x 1.46 	imes 10^{-4} 	imes 1500 =x \boxed {0.219 \ cm^{-1}=x} The cell constant is  0.219 \ cm^{-1}

Question 7

In electrolysis of water: (a) Name the gas collected at cathode and anode.(b) Why is volume of one gas collected at one electrode is double of anode?(c) Why are few drops of dilute  H_2SO_4  added to water?

Accepted Answer

a) At cathode,  H_2  is evolved and at anode  O_2  is evolved.b) At cathode,  H^+  ion takes two electron to convert itself into  H_2  gas.  2  moles of  H^+  gives  1  mole of  H_2 . 2H^+(aq)+ 2e^- \rightarrow H_2(g) At anode,  OH^-  ion releases two electrons to convert into water and  O_2 .  2  mole  OH^-  gives  0.5  mole  O_2 , i.e why volume of one gas collected at one electron is double of another.c) Acid is added to make the water conduct electricity as the distilled water is a non-conductor of electricity.

Question 8

How many electrons are there in one coulomb of electricity?

Accepted Answer

Change on  1e^-=1.602	imes 10^{-19}   C If total charge  =1   C 	herefore  Total no. of  e^-=\cfrac{1}{1.602	imes 10^{-19}}=6.24	imes 10^{18}e^- Hence, the correct option is  (C) .

Question 9

Predict the products of electrolysis in each of the following:(i) An aqueous solution of  AgNO_{3}  with silver electrodes.(ii) An aqueous solution of  AgNO_{3}  with platinum electrodes.(iii) A dilute solution of  H_{2}SO_{4}  with platinum electrodes.(iv) An aqueous solution of  CuCl_{2}   with platinum electrodes

Accepted Answer

(i) An aqueous solution of  AgNO_{3}  with silver electrodes.At cathode: Silver ions have lower discharge potential than hydrogen ions. Hence, silver ions will be deposited in preference to hydrogen ions.At anode: Silver anode will dissolve to form silver ions in the solution. \displaystyle Ag \rightarrow Ag^+ + e^- (ii) An aqueous solution of  AgNO_{3}  with platinum electrodes.At cathode: Silver ions have lower discharge potential than hydrogen ions. Hence, silver ions will be deposited in preference to hydrogen ions.At anode: Hydroxide ions having lower discharge potential will be discharged in preference to nitrate ions. Hydroxide ions will decompose to give oxygen. \displaystyle 4OH^- (aq) \rightarrow 2H_2O(l) +O_2(g) + 4e^- (iii) A dilute solution of  H_{2}SO_{4}  with platinum electrodes.At cathode:  \displaystyle  2H^+ + 2e^- \rightarrow H_2(g) At anode: Hydroxide ions having lower discharge potential will be discharged in preference to sulphate ions. Hydroxide ions will decompose to give oxygen. \displaystyle 4OH^- (aq) \rightarrow 2H_2O(l) +O_2(g) + 4e^- (iv) An aqueous solution of  CuCl_{2}   with platinum electrodesAt cathode: Cupric ions will be reduced in preference to protons \displaystyle Cu^{2+} + 2e^- \rightarrow Cu At anode: Chloride ions will be oxidized in preference to hydroxide ions \displaystyle 2Cl^- \rightarrow Cl_2 + 2e^-

Question 10

A metal carbonate X on reacting with an acid gives a gas which when passed through a solution gives the carbonate black. On the other hand, A gas G that is obtained at anode during electrolysis of brine is passed on dry Y, it gives compound Z, used for disinfecting water. Identify X, Y, G and Z.

Accepted Answer

According to the problem:Here, X  is calcium carbonate  \left( {CaC{O_3}} \right) . Y  is slaked lime. Z  is bleaching powder  \left( {CaOC{l_2}} \right) G  is chlorine gas. Hence the reactions are involved,