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**Solved Example**

**Ex.1** **The reaction : Zn ^{2+ } (aq) + 2e^{-} → Zn (s) has a electrode potential of - 0.76 V. This means-**

**Ex.2 Certain quantity of current is passed through 2V connected in series and containing XSO _{4}(aq) and Y_{2}SO_{4}(aq) respectively. If the atomic masses of X and Y are in the ratio of 2 : 1 the ratio of the masses of Y liberated to that of X is :**

2 mol e^{-} produce X = 1 mol = 1 × M g

2 mol e^{-} produce Y = 2 mol = 2 × M/2 = M g

Hence, ratio of the masses of Y:X is M:M or 1:1

**Ex.3** **The equivalent conductivities at infinite dilution of the cation and the anion of a salt A _{2}B are 140 and 80 ohm^{-1} cm^{2} eq^{-1 }respectively. The equivalent conductivity of the salt at infinite dilution is- **

= 140 + 80 = 220 ohm^{-1} cm^{2} eq^{-1}

**Ex.4** **The specific conductance of a 0.20 mol L ^{-1} solution of an electrolyte at 20ºC is **

= 1.24 ohm^{-1} cm^{2} mol^{-1}**Ex.5** **When an electric current is passed through acidulated, water, 112 mL of hydrogen gas at N.T.P. collects at the cathode in 965 seconds. The current passed, in amperes, is-****(A) 1.0 ****(B) 0.5 ****(C) 0.1 ****(D) 2.0****Sol.** **(A) **22,400 mL of hydrogen at STP(or NTP)=2g

Therefore,112 mL of hydrogen at

STP = = 10^{-2} g

Therefore,2H^{ } 2e^{-} → H_{2 }

2F 1 mol

= 2 × 96,500 C = 2 g, 2 g hydrogen is deposited by 2 × 96,500 C

Therefore,10^{-2} g hydrogen will be deposited by = = 965 C

Q = i × t = 965 = i × 965

= i = 1

**Ex.6** **The charge required to deposit 40.5 g of Al (atomic mass = 27.0 g) from the fused Al _{2} (SO_{4})_{3} is-**

3F 1 mol = 27.0 g

to deposite 27g required charge =3×96,500 C

Therefore,to deposite 40.4g required charge

= = 4.34 × 10

**Ex.7** **The same amount of electricity was passed through two separate electrolytic cells containing solutions of nickel nitrate and chromium nitrate respectively. If 0.3 g of nickel was deposited in the first cell, the amount of chromium deposited is (At. wt. Ni= 59, Cr=52)****(A) 0.1 g ****(B) 0.176 g ****(C) 0.3 g ****(D) 0.6 g****Sol.** **(B)** =

For Ni^{2+ } and Cr^{3+ }, we have : = m_{Cr} = = 0.176 g

**Ex.8** **Electrolytic conduction differs from metallic conduction. In case of metallic conduction -****(A) The resistance increases with increasing temperature****(B) The resistance decreases with increasing temperature****(C) The flow of current does not generate heat****(D) The resistance is independent of the length of electrolytic conductor****Sol. (A)** With increase in temperature vibration of Kernal (Cation) increases and therefore, conduction decreases and hence, resistance of the metallic conductor increases.

**Ex.9** **Three faraday of electricity is passed through molten solutions of AgNO _{3}, NiSO_{4} and CrCl_{3} kept in three vessels using inert electrodes. The ratio in mol in which the metals Ag, Ni and Cr will be deposited is-**

1 mol = 1F 1 mol

3F = 3 mol**(ii)**

Ni^{2+ }(aq) + 2e^{-} → Ni (s)

2 mol = 2F 1 mol

3 F =3/2 mol**(iii)**

Cr^{3+ }(aq) + 3e^{-} → Cr(s)

3 mol = 3F 1 mol

The required ratio of moles of Ag, Ni and Cr is :

3 mol Ag : 3/2 mol Ni : 1 mol Cr

or **6 mol Ag : 3 mol Ni : 2 mol Cr. **

**Ex.10** **In the reaction :****4 Fe + 3O _{2} → 4 Fe^{3+} + 6O^{2-}**

In this reaction, Fe is oxidized to Fe^{3+ } and O_{2} is reduced to O^{2_}.

**Ex.11** **Calculate current strength in ampere required to deposit 10 g Zn in 2hrs. At wt. of Zn = 65.****Sol.** Q w = ; Therefore,i = i = = 4.12 ampere

**Ex.12** **How many hour are required for a current of 3.0 ampere to decompose 18 g water.****Sol.** H_{2}O → H_{2} O_{2}

Therefore,Eq. of H_{2}O = Equivalent weight of H_{2}O = 18/2 as two electrons are used for

1 mole H_{2}O to decompose in H_{2} and O_{2}.

Therefore, =

Therefore,t = 64333.3 sec = 1072.2 minute = 17.87 hr

**Ex.13** **Calculate the Avogadro's number using the charge on the electron 1.60 × 10 ^{-19} C and the fact that 96500 C deposits 107.9 g silver from its solution.**

Here eq. wt = Atomic weight

because Ag is monovalent. Thus 96500 coulomb charge means charge on N electrons

where N in Av. no.

Thus N × e = 96500

N =

= 6.03 × 10

**Ex.14** **Calculate the volume of gases liberated at anode and cathode at NTP from the electrolysis of Na _{2}SO_{4}(aq.) solution by a current of 2 ampere passed**

At anode : 2H

Therefore,At anode = = 8 Therefore,= = = 0.0995 g

At NTP : Volume of O

Similarly at cathode = = = 0.0124 g

At NTP : Volume of H

**Ex.15** **Fused Ni(NO _{3})_{2} is electrolysed between platinum electrodes using a current of 5 ampere for 20 minute. What mass of Ni is deposited at the cathode ?**

= = **= **0.0622

or w_{Ni} = 0.0622 × 58.71/2 = 1.825 g

(Ni^{2+ }+ 2e → Ni)

**Ex.16** **A current of 3.7 ampere is passed for 6 hrs. between Ni electrodes in 0.5****litre of 2M solution of Ni(NO _{3})_{2}. What will be the molarity of solution at the**

At anode : Ni → Ni

At cathode : Ni

Eq. of Ni^{2+ } formed = Eq. of Ni^{2+ } lost

Thus, there will be no change in conc. of Ni(NO_{3})_{2} solution during electrolysis i.e. ,

It will remain 2M

**Ex.17** **How long a current of 3 ampere has to be passed through a solution of AgNO _{3} to coat a metal surface of 80 cm^{2} with a thickness of 0.005 mm ?**

Also Volume covered by

Ag = 80 × 0.005 × 10

Therefore,Weight of Ag used = 0.04 × 10.5 g

w

0.04 × 10.5 =

Therefore,t = 125.09 sec.

**Ex.18** **Calculate e.m.f of half cells given below :****(a) Eº _{OP} = 0 V**

= 0- log = 0.100 V

= 0.44- log[0.2] = 0.4606 volt.**(c) **2Cl^{-} → Cl_{2} 2e Therefore,E_{OP} = Eº_{OP}- log

=-1.36- log =- 1.49 volt

**Ex.19** **If the oxidation of oxalic acid by acidified MnO _{4}^{-} solution were carried out in a reversible cell, what would be the electrode reaction ? Also calculate the equilibrium constant of the reaction. Given**

** = 1.51 V and =- 0.49 V.**

**Sol.** = 1.51 V ** ** =- 1.51 V

=- 0.49 V = 0.49 V

More is Eº_{OP}, more is the tendency to get oxidise

→ 2CO_{2} 2e; Eº_{OP} = 0.49 v 8H^{ } 5e → Mn^{2 } 4H_{2}O; Eº_{RP }= 1.51 V

16H^{ } → 10CO_{2} 2Mn^{ } 8H_{2}O; n = 10 Q 10 electron are used in redox change.

Therefore,Eº_{Cell} = Eº_{OP} Eº_{RP} = 0.49 1.51 = 2.0 V Also Eº = log K_{C}

Therefore,2 = log K_{C }Therefore,K_{C} = 10^{338.98}

**Ex.20** **The e.m.f. of cell, ****Ag|AgI _{(s)}, 0.05 MKI || 0.05 M AgNO_{3} | Ag, is 0.788 V.**

**Sol.** K_{sp} of AgI = [Ag^{ }] [I^{-}] = [Ag^{ }] [0.05]

For given cell E_{Cell} = E_{OPAg} E_{RPAg} ….(i)

= -log [Ag^{ }]_{L.H.S.}log[Ag^{ }]_{R.H.S. }E_{Cell} = log

0.788 = log Therefore,[Ag^{ }]_{L.H.S.} = 2.203 × 10^{-15 }

By equation (i) K_{sp} = [2.203 × 10^{-15}] [0.05] = 1.10 × 10^{-16}

**Ex.21** **Calculate the reduction potential of a half cell consisting of a platinum electrode immersed in 2.0 MFe ^{2 } and 0.02 M Fe^{3 } solution. Given = 0.771 V.**

Thus = log

Therefore,= 0.771 + log = 0.771 + log 10

**Ex.22** **An electrochemical cell has two half cell reactions as, A ^{2 } 2e^{-} → A; E° = 0.34 V, **

**Ex.23** **Eº of some oxidants are given as :****I _{2} + 2e → 2I^{-} , Eº = 0.54 V**

Therefore, since maximum Eº

Sn

Therefore,strongest reductant : Sn

Therefore, since maximum Eº

MnO

Therefore,strongest oxidant : MnO

=-0.77 + 0.1

Fe^{2+ } oxidizes and Sn^{4+ } reduces in change, Eº_{Cell} =-0.67 V, Eº_{Cell} is negative

Therefore,**(i) Is non-spontaneous change**

For (ii) Eº_{Cell} = =-0.77 0.54 =- 0.23 V

Therefore,**(ii) Is non-spontaneous change**

For (iii) Eº_{Cell} = =-0.54 0.1 =-0.44 V

Therefore,**(iii) Is non-spontaneous change**

For (iv) Eº_{Cell} = =-0.1 0.54 = 0.44 V**(iv) Is spontaneous change **

**Ex.24** **Calculate the standard cell potentials of galvanic cell in which the following reactions take place :(Given Eº _{OP} Cr, Cd, Fe^{2+ }, Ag are 0.74V, 0.40**

[2Cr → 2Cr

= 0.74 - 0.40) = 0.34 V

Six electrons (n = 6) are used in redox change

-Δ

or Δ

Also-Δ

Therefore,196860 = 2.303 × 8.314 × 298 log K

K = 3.17 × 10

[Fe

= -0.77 + 0.80 = 0.03 V

Also -Δ

or Δ

Also -Δ

2895 = 2.303 × 8.314 × 298 log K

K= 3.22

**Ex.25** **A cell is constituted as follows ****Pt, H _{2 (1 atm)} | HA_{1} | HA_{2} | H_{2 (1 atm)}, Pt**

Now, E

**Ex.26** **The standard EMF of the cell reaction**

** Cu(s) + Cl _{2}(g) → Cu^{2+ } + Cl^{- }**

=- 98430 J

or =- 98.43 kJ

**Ex.27** **A current of 1.70 ampere is passed through 300 mL of 0.160 M solution of ZnSO _{4} for 230 sec with a current efficiency of 90%. Find the molarity of Zn^{2 } after the deposition of Zn. Assume the volume of the solution remains constant during electrolysis.**

Therefore,Eq. of Zn

Therefore,Meq. of Zn

Initial Meq. of Zn

[M × 2 = N for Zn

Therefore,Meq. of Zn

Therefore,[ZnSO

**Ex.28** **If 0.01 M solution of an electrolyte has a resistance of 40 ohms in a cell having a cell constant of 0.4 cm ^{-1} then its molar conductance in ohm^{-1} cm^{2} mol^{-1} will be :**

L_{m} = = 1000 = 10^{3} ohm^{-1} cm^{2} mol^{-1}

**Ex.29** **Calculate the quantity of electricity that would be required to reduce 12.3 g of nitrobenzene to aniline, if current efficiency is 50%. If the potential drops across the cell is 3.0 volt, how much energy will be consumed ****Sol.** C_{6}H_{5}NO_{2} + 6H^{+} +^{ }6e → C_{6}H_{5}NH_{2} + 2H_{2}O

N^{3+ } + 6e → N^{3-}

Therefore,Eq. wt. of nitrobenzene = = , Since current efficiency is 50% Therefore,i =

Now w = 12.3 = Therefore,i × t = 115800 coulomb

Now energy used = Q × V = 115800 × 3 = 347.4 kJ**Ex.30** **A cell is containing two H electrodes. The negative electrode is in contact with a solution of 10 ^{-6} M H^{ } ion. The e.m.f. of the cell is 0.118 volt at 25ºC. Calculate [H^{ }] at positive electrode.**

(negative polarity) [H

**Cathode :** 2H^{+} + 2e → H_{2}

(positive polarity). [H^{ }] → aM

Therefore,E_{cell} = = - log_{10}[H^{ }]^{ 2}_{Anode}log_{10}[H^{ }]^{ 2}_{Cathode}

= log_{10} , 0.118=log_{10}=log_{10}, Therefore,[H^{ }]_{Cathode} = 10^{-4} M

**Ex.31** **A current of 2.0 A passed for 5 hours through a molten metal salt deposits 22.2 g of metal (At wt. = 177). The oxidation state of the metal in the metal salt is :****(A) 1****(B) 2****(C) 3****(D) 4****Sol. (C)** E = × 96500 = × 96500 = 59.5

Oxi. state = = 3

**Ex.32** **Cost of electricity for the production of x L H _{2} at NTP at cathode is Rs x; then cost of production of x L O_{2} at NTP at anode will be (assume 1 mole of electron as one unit of electricity)**

Thus, L O_{2} requires Rs x for its production. i.e., x L O_{2} will require Rs x for its production.

**Ex.33** **In which direction can the reaction, 2Hg(l) + 2Ag ^{+}^{ }(aq.) 2Ag(s) + Hg_{2}^{2+ }(aq.) proceed spontaneously at the following concentrations of the ions participating in the reactions (i) and (ii) ?**

**Sol.** (i) Q = = = 10^{7}, Eº = - = 0.80- 0.79 = 0.01 V

E = Eº- log Q = 0.01- log 10^{7} =- 0.1965 V

Negative value shows that the reaction will proceed from right to left, i.e. in backward direction.

(ii) Q = = = 10^{-2}, Eº = 0.01 volt, E = Eº- log_{10} Q = 0.01- log_{10} 10^{-2}

= 0.01 0.059 V = 0.069 V

Since, the value of cell potential is positive, the reaction will proceed spontaneously in forward direction.

**Ex.34** **Two students use same stock solution of ZnSO _{4} and a solutions of CuSO_{4}. The e.m.f. of one cell is 0.03 V higher than the other. The conc. of CuSO_{4} in the cell with higher e.m.f. value is 0.5 M. Find out the conc. of CuSO_{4} in the other cell [(2.303 RT/F) = 0.06].**

Cell I : ZnCu, E

Cell II : ZnCu E¢

If E

By Eq. (i) and (ii)

E

**Ex.35** **How much will the reduction potential of a hydrogen electrode change when its solution initially pH = 0 is neutralized to pH = 7.****(A) Increase by 0.059 V ****(B) Decrease by 0.058 V ****(C) Increase by 0.41 V ****(D) Decrease by 0.41 V****Sol. (D) **Eº = E log 10^{-7}** **= Eº = Eº- 0.41 V

**Ex.36 The time required to coat a metal surface of 80 cm ^{2} with 5 × 10^{-3} cm thick layer of silver (density 1.05 g cm^{-3}) with the passage of 3A current through a silver nitrate solution is.**

= 80 × 5 × 10

Q w = Therefore,0.42 = Therefore,t = 125 sec

**Ex.37** **Standard electrode potentials are****Fe ^{2+}/Fe (E° =- 0.44 V), Fe^{3+ }/Fe^{2+ } (E° = 0.77 V)**

So spontaneous reaction is Fe + Fe

**Ex.38** **Conductivity (Unit : siemen's S) is directly proportionally to the area of the vessel and the concentration of the solution in it and is inversely proportional to the length of vessel, then the unit of constant of proportionality is****(A) S m mol ^{-1} **

**Ex.39** **Passage of three faraday of charge through aqueous solution of AgNO _{3}, CuSO_{4}, Al(NO_{3})_{3} and NaCl respectively will deposit the metals in the ratio (molar).**

Note electrolysis of NaCl_{aq.} and AlCl_{3aq.} does not give Na and Al metal at cathode.

Thus molar ratio is : for Ag : Cu

or 3 : or 6 : 3 : 0 : 0 for Ag : Cu : Al : Na

**Ex.40** **and are -0.441 V and 0.771 V respectively, Eº for the reaction.****Fe + 2Fe ^{3+ } → 3Fe^{2+ }, will be.**

**Sol. (A) **=

= 0.441 + 0.771 = 1.212 V

**Ex.41** **Efficiency of a cell with cell reaction under standard conditions,****A _{(s)} + B^{+}^{ } → A^{+} + B_{(s)}; DHº =-300 kJ is 70%. The standard electrode potential of cell is.**

**Ex.42** **Eº for F _{2} 2e 2F^{-} is 2.7 V. Thus Eº for F^{-}_{}F_{2} e is.**

**Ex.43** **Salts of A (atomic weight 7). B (atomic weight 27) and C (atomic weight 48) were electrolysed under identical condition using the same quantity of electricity. It was found that when 2.1 g of A was deposited, the weights of B and C deposited were 2.7 and 7.2 g. The valencies of A, B and C respectively.****(A) 3, 1 and 2 ****(B) 2, 6 and 3 ****(C) 3, 1 and 3 ****(D) 2, 3 and 2****Sol. (B) **gm EqA = gm Eq B = gm EqC

, 0.3 x = 0.1 y = 0.15 z

3x = y = 1.5z

x = 2, y = 6, z = 3

**Ex.44** **Standard electrode potential data are useful for understanding the suitability of an oxidant in a redox titration. Some half cell reactions and their standard potentials are given below:**** + 8H ^{ +}_{(aq.)} + 5e →**

Mn^{7+ } + 5e → Mn^{2+ } , 2Cl^{-} → Cl_{2} + 2e,

Thus Eº_{cell} = =-1.40 1.51 = 0.11 V

or reaction is feasible.

MnO_{4}^{-} will oxidize Fe^{2+ } to Fe^{3+}

Mn^{7+ } + 5e → Mn^{2+ } , Fe^{3+ } → Fe^{2+ } + e, Eº_{cell} = = -0.77 + 1.51 = 0.74 V

or reaction is feasible

Thus MnO_{4}^{-} will not oxidize only Fe^{2+ }to Fe^{3+ } in aqueous HCl but it will also oxidise Cl^{-} to Cl_{2}. Suitable oxidant should not oxidise Cl^{-} to Cl_{2} and should oxidise only Fe^{2+ } to Fe^{3+ } in redox titration.**Ex.45** **The emf of the cell Zn | Zn ^{2+ } (0.01M)|| Fe^{2+ } (0.001M)| Fe at 298 K is 0.2905, then the value of equilibrium constant for the cell reaction is.**

E_{cell} = Eº_{cell} + log, 0.2905 = Eº_{cell }+ log

Therefore,Eº_{cell} = 0.2905 + 0.0295 = 0.32 V, Now Eº_{cell} =log_{10} K_{c} , 0.32 = log_{10} K_{c}

Therefore,K_{c} = 10^{0.32/0.0295}

**Ex.46** **The rusting of iron takes place as follows :****2H ^{+} + 2e + ½O_{2} → H_{2}O_{(l)}; Eº = 1.23 V**

Therefore,**Δ**Gº =-nEº F =-2 × 1.67 × 96500 J =-322.31 kJ mol^{-1}

**Passage : (Q.47 to 52) ****Electrolysis involves electronation and de-electronation at the respective electrodes. Anode of electrolytic cell is the electrode at which de-electronation takes place whereas at cathode electronation is noticed. If two or more ions of same charge are to be electronated or de-electronated, the ion having lesser discharge potential is discharged. Discharge potential of an ion refers for Eº _{OP} or Eº_{RP} as the case may be. The products formed at either electrode is given in terms of Faraday's laws of electrolysis i.e., w =.**

**Ex.47 During electrolysis of CH _{3}COONa_{(aq.)}, the mole ratio of gases formed at cathode and anode is.**

2CH

Cathode : 2H

**Ex.48** **During electrolysis of HCOONa _{(aq.)}, the gas liberated at anode and cathode are respectively.**

Cathode : 2H

**Ex.49** **During electrolysis of CuSO _{4(aq.)}, the pH of solution becomes.**

Cathode : Cu

[OH

**Ex.50** **5 litre solution of 0.4 M CuSO _{4(aq.)} is electrolysed using Cu electrode. A current of 482.5 ampere is passed for 4 minute. The concentration of CuSO_{4} left in solution is.**

Anode : Cu → Cu

Cathode : Cu

Thus no change in conc. of Cu

**Ex.51** **5 litre solution of 0.4 M Ni(NO _{3})_{2} is electrolysed using Pt electrodes with 2.4125 ampere current for 10 hour.**

Eq. of Ni

Therefore,Eq. of Ni(NO

**Ex.52** **The volume of octane required to be used for its combustion by the oxygen liberated during electrolysis of an NaNO _{3(aq.)} by passing 9.65 ampere current for 1 hr. is.**

Cathode : 2H

Eq. of O

Therefore,Mole of O

Therefore,Mole of C

**Ex.53** **The specific conductivity of 0.02 M KCl solution at 25°C is 2.768 × 10 ^{-3 }ohm^{-1} cm^{-1}. The resistance of this solution at 25°C when measured with a particular cell was 250.2 ohm. The resistance of 0.01 M CuSO_{4} solution at 25°C measured with the same cell was 8331 ohm. Calculated the molar conductivity of the copper sulphate solution.**

For 0.01 M CuSO

Sp. conductivity = Cell constant × Conductance = 2.768 × 10

Molar conductance = Sp. cond. × =×=8.312 ohm

**Ex.54** **A decinormal solution of NaCl has specific conductivity equal to 0.0092. If ionic conductances of Na ^{ }and Cl^{-} ions at the same temperature are 43.0 and 65.0 ohm^{-1} respectively. Calculate the degree of dissociation of NaCl solution**

= Sp. conductivity × dilution = 0.0092 × 10,000 = 92 ohm

Degree of dissociation, a = = 0.85

**Ex.55** **The specific conductance of saturated solution of AgCl is found to be 1.86 × 10 ^{-6} ohm^{-1} cm^{-1} and that of water is 6 × 10^{-8 }ohm^{-1} cm^{-1}. The solubility of AgCl is ….**

, S==1.31×10^{-5} M

**Ex.56** **Resistance of a solution (A) is 50 ohm and that of solution (B) is 100 ohm, both solution being taken in the same conductivity cell. If equal volumes of solution (A) and (B) are mixed, what will be the resistance of the mixture, using the same cell ? Assume that there is no increase in the degree of dissociation of (A) and (B) mixing****Sol. **Let us suppose k_{1} and k_{2} are the specific conductance of solutions `A' and `B' respectively and cell constant is `y'. We know that,

Specific conductance=Conductance × Cell constant

For (A), For (B),

When equal volume of (A) and (B) are mixed, the volume becomes double. Then, Specific conductance of mixture =

Therefore,

R = 200/3 = 66.66 ohm

**Ex.57** **A big irregular shaped vessel contained water, specific conductance of which was 2.56 × 10 ^{-5} mho cm^{-1}, 500 g of NaCl was then added to the water and the specific conductance after the addition of NaCl was found to be 3.1 × 10^{-5} mho cm^{-1}. Find the capacity of the vessel if it was fully filled with water. ( NaCl = 149.9 ohm^{-1} cm^{2} eq^{-1})**

Volume containing 1 equivalent = =

Specific conductance of NaCl = Specific conductance of NaCl solution- Specific conductance of water

= 3.1 × 10

= k × volume containing 1 equivalent of electrolyte ….(i)

For very dilute solution, when the big vessel is fully filled

= 149.9 ohm

V = 237258.38 L

**Ex.58** **Maximum Conductivity would be of-****(A) K _{3}Fe(CN)_{6} [0.1 M solution] **

Double salt on ionization gives more ions. One molecule of the salt gives Fe

**Discussion Questions**

(i) For the reaction given below, apply Le-Chatelier principle to justify the results recorded by you and also bring out mathematical rationalisation of your results.

Zn(s) Cu^{2 } (aq) Zn^{2 } (aq) Cu(s),

(ii) Determine the slope of the graph. Match experimental value with the theoretical value. On what factors does the value of slope depend?

(iii) Devise another experiment to study the variation in cell potential with concentration of one of the ions involved in a cell reaction.

(iv) What factor is kept in mind while selecting an electrolytic solution for the construction of a salt bridge?

(v) Is it possible to measure the single electrode potential?

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