Students can read the important questions given below for Electrochemistry Class 12 Chemistry. All Electrochemistry Class 12 Notes and questions with solutions have been prepared based on the latest syllabus and examination guidelines issued by CBSE, NCERT and KVS. You should read all notes provided by us and Class 12 Chemistry Important Questions provid ed for all chapters to get better marks in examinations. Chemistry Question Bank Class 12 is available on our website for free download in PDF.
Important Questions of Electrochemistry Class 12
Very Short Answer Questions
Question. Express the relation between conductivity and molar conductivity of a solution held in a cell?
Answer.
Question. Express the relation among the conductivity of solution in the cell, the cell constant and the resistance of solution in the cell.
Answer.
where k is the conductivity R is resistance and l/A is the cell constant.
Question. Following reactions occur at cathode during the electrolysis of aqueous silver chloride solution :
On the basis of their standard reduction electrode potential (E°) values, which reaction is feasible at the cathode and why?
Answer. The species that get reduced at cathode is the one having higher value of standard reduction potential. Hence, the reaction that will occur at cathode is Ag+(aq) + e– → Ag(s).
Question. How much charge is required for the reduction of 1 mol of Zn2+ to Zn?
Answer.
One mole of Zn2+ requires 2 moles of electrons for reduction i.e.
Q = 2 × F = 2 × 96500 = 193000 C
Question. Following reactions occur at cathode during the electrolysis of aqueous copper (II) chloride solution :
On the basis of their standard reduction electrode potential (E°) values, which reaction is feasible at the cathode and why?
Answer. The species that get reduced at cathode is the one which have higher value of standard reduction potential. Hence, the reaction that will occur at cathode is
Question. State the Faraday’s rst law of electrolysis.
Answer. Faraday’s first law of electrolysis : During electrolysis the amount of any substance deposited or liberated at an electrode is directly proportional to the quantity of electricity passed through the electrolyte i.e.,
w a Q or w ∝ I × t [∴ Q = I × t]
w = Z × I × t
where, Z is a constant of proportionality known as electrochemical equivalent of the substance deposited.
Question. Given that the standard electrode potential (E°) of metals are :
K+/K = –2.93 V, Ag+/Ag = 0.80 V,
Cu2+/Cu = 0.34 V,
Mg2+/Mg = –2.37 V, Cr3+/Cr = –0.74 V,
Fe2+ /Fe = –0.44 V.
Arrange these metals in an increasing order of their reducing power.
Answer. The reducing power increases with decreasing value of electrode potential. Hence, the order is
Ag < Cu < Fe < Cr < Mg < K.
Question. Formulate the galvanic cell in which the following reaction takes place :
Zn(s) + 2Ag+(aq) → Zn2+ (aq) + 2Ag(s)
State :
(i) Which one of its electrodes is negatively charged.
(ii) The reaction taking place at each of its electrode.
(iii) The carriers of current within this cell.
Answer. The cell reaction is
Zn(s) + 2Ag+(aq) Zn2+(aq) + 2Ag(s)
The cell is represented as
Zn(s) | Zn2+(aq) || Ag+(aq) | Ag(s)
(i) Anode i.e., zinc electrode will be negatively charged.
(ii) At anode :
Zn(s) Zn2+(aq) + 2e– (oxidation)
At cathode :
Ag+(aq) + e– Ag(s) (Reduction)
(iii) Ions are the carriers of current within the cell.
Question. The conductivity of 0.001 M acetic acid is 4 × 10–5 S/cm. Calculate the dissociation constant of acetic acid, if molar conductivity at infinite dilution for acetic acid is 390 S cm2/mol.
Answer. C = 0.001 M, k = 4 × 10–5 S cm–1,
Λcm = 390 S cm2/mol
Question. The molar conductivity of a 1.5 M solution of an electrolyte is found to be 138.9 S cm2 mol–1. Calculate the conductivity of this solution.
Answer.
Question. The resistance of a conductivity cell containing 0.001 M KCl solution at 298 K is 1500 Ω. 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. Here, conductivity (k) = 0.146 × 10–3 S cm–1,
resistance (R) = 1500 Ω
Question. Define molar conductivity of a substance and describe how for weak and strong electrolytes, molar conductivity changes with concentration of solute. How is such change explained?
Answer. Strong electrolyte : The molar conductivity of strong electrolyte decreases slightly with the increase in concentration. This increase is due to increase in attraction as a result of greater number of ions per unit volume. With dilution the ions are far apart, interionic attractions become weaker and conductance increases.
Weak electrolyte : When the concentration of weak electrolyte becomes very low, its degree of ionisation rises sharply. There is sharp increase in the number of ions in the solution. Hence the molar conductivity of a weak electrolyte rises steeply at low concentration.
Question. Explain with examples the terms weak and strong electrolytes.
Answer. Weak electrolytes : The electrolytes which are not completely dissociated into ions in solution are called weak electrolytes e.g., CH3COOH, NH4OH, HCN, etc.
Strong electrolytes : The electrolytes which are completely dissociated into ions in solution are called strong electrolytes. e.g., HCl, KCl, NaOH, NaCl, etc.
Question. A solution of Ni(NO3)2 is electrolysed between platinum electrodes using a current of 5.0 ampere for 20 minutes. What mass of nickel will be deposited at the cathode?
(Given : At. mass of Ni = 58.7 g mol–1, 1F = 96500 C mol–1)
Answer. Given : Current I = 5 A; t = 20 × 60 s, w = ?
Q = I × t = 5 × 20 × 60 = 6000 C
Reaction for deposition of Ni,
Question. Predict the products of electrolysis in each of the following :
(i) An aqueous solution of AgNO3 with platinum electrodes.
(ii) An aqueous solution of H2SO4 with platinum electrodes.
Answer. (i) At cathode : The following reduction reactions compete to take place at the cathode.
Ag+(aq) + e– → Ag(s) ; E° = 0.80 V
The reaction with a higher value of E° takes place at the cathode. Therefore, the deposition of silver will take place at the cathode. Since, Pt electrodes are inert, the anode is not attacked by NO–3 ions. Therefore, OH– or NO–3 ions can be oxidized at the anode. But OH– ions having a lower discharge potential get preference and decompose to liberate O2.
Question. How much electricity in terms of Faradays is required to produce 20 g of calcium from molten CaCl2?
Answer. Reaction for production of Ca from molten
CaCl2 :
CaCl2 → Ca2+ + 2Cl–
Ca2+ + 2e– → Ca
2F 40 g
Electricity required to produce 40 g = 2 F
∴ Electricity required to produce 20 g = 0.5 × 2 F = 1 F
Question. Silver is uniformly electrodeposited on a metallic vessel of surface area of 900 cm2 by passing a current of 0.5 ampere for 2 hours.
Calculate the thickness of silver deposited.
Given : the density of silver is 10.5 g cm–3 and atomic mass of Ag = 108 amu.
Answer. Calculation of mass of Ag deposited
The electrode reaction is Ag+ + e– → Ag The quantity of electricity passed = Current × Time
= 0.5 (amp) × 2 × 60 × 60 (sec) = 3600 C From the electrode reaction, it is clear that 96500 C of electricity deposit Ag = 108 g 3600 C of electricity will deposit Ag
Let the thickness of deposit be x cm
Mass = volume × density = Area × thickness × density
[Q volume = area × thickness]
∴ 4.03 g = 900 (cm2) × x (cm) × 10.5 (g cm–3)
Question. Mention the reactions occurring at (i) anode, (ii) cathode, during working of a mercury cell.
Why does the voltage of a mercury cell remain constant during its operation?
Answer. Mercury Cell : It is a miniature cell which finds a frequent use these days to supply energy for watches, video cameras, hearing aids and other compact devices. In mercury cell the anode is zinc-mercury amalgam, and the cathode is a paste of mercury (II) oxide and carbon, electrolyte is a moist paste of KOH – ZnO.
The cell reactions are as follows :
Anode :
Zn(Hg) + 2OH– → ZnO(s) + H2O(l) + 2e– (Amalgam)
Cathode :
HgO(s) + H2O(l) + 2e– → Hg(l) + 2OH–
Net reaction :
Zn(Hg) + HgO(s) → ZnO(s) + Hg(l)
The cell potential remains constant during its life as the overall reaction does not involve any ion in solution whose concentration can change during its use.
Question. Write two advantages of H2 O2 fuel cell over ordinary cell.
Answer. (i) It is pollution free.
(ii) It has high efficiency of 70 – 75% and its rate can be controlled.
Question. The chemistry of corrosion of iron is essentially an electrochemical phenomenon. Explain the reactions occurring during the corrosion of iron in the atmosphere.
Answer. According to electrochemical theory of rusting the impure iron surface behaves like small electrochemical cell. In this any point of iron acts as anode and other iron surface acts as cathode. Moisture having dissolved CO2 or O2 acts as an electrolyte. The reactions are given below. At anode :
Short Answer Questions
Question. What is a nickel-cadmium cell? State its one merit and one demerit over lead storage cell.
Write the overall reaction that occurs during discharging of this cell.
Answer. Nickel cadmium cell is a secondary battery which consists of a cadmium anode, nickel hydroxide as cathode and sodium or potassium hydroxide acts as electrolyte.
Merit : It has longer life than lead storage battery.
Demerit : It is more expensive than lead storage battery.
The following reaction takes place during discharging :
Cd(s) + 2Ni(OH)3(s) → CdO(s) + 2Ni(OH)2(s) + H2O(l)
Question. The conductivity of 0.001 mol L–1 solution of CH3COOH is 3.905 × 10–5 S cm–1. Calculate its molar conductivity and degree of dissociation (a).
Given : λ° (H+) = 349.6 S cm2 mol–1 and λ°(CH3COO–) = 40.9 S cm2 mol–1
Answer.
Question. The conductivity of 0.20 mol L–1 solution of KCl is 2.48 × 10–2 S cm–1. Calculate its molar conductivity and degree of dissociation (a).
Given λ° (K+) = 73.5 S cm2 mol–1 and λ° (Cl–) = 76.5 S cm2 mol–1.
Answer. Given : Conductivity, k = 0.0248 S cm–1
Molarity, C = 0.20 M = 0.20 mol L–1
Question. Resistance of a conductivity cell filled with 0.1 mol L–1 KCl solution is 100 Ω. If the resistance of the same cell when filled with 0.02 mol L–1 KCl solution is 520 Ω, calculate the conductivity and molar conductivity of 0.02 mol L–1 KCl solution.
The conductivity of 0.1 mol L–1 KCl solution is 1.29 × 10–2 Ω–1 cm–1.
Answer. Resistance of 0.1 M KCl solution R = 100 Ω
Conductivity k = 1.29 S m–1
Cell constant G* = k × R = 1.29 × 100 = 129 m–1
Resistance of 0.02 M KCl solution, R = 520 Ω
Question. The value of L°m of Al2(SO4)3 is 858 S cm2 mol–1 ,while l° SO2–4 is 160 S cm2 mol–1 calculate the limiting ionic conductivity of Al3+.
Answer. Λ°m Al2(SO4)3 = 2l°mAl3+ + 3λ°mSO2–4
⇒ 858 = 2λ°mAλ3+ + 3 × 160
Question. Calculate E°cell for the following reaction at 298 K.
2Al(s) + 3Cu2+(0.01M) → 2Al3+(0.01M) + 3Cu(s)
Given : Ecell = 1.98 V
Answer. Given cell,
2Al(s) + 3Cu2+ (0.01 M) → 2Al3+(0.01 M) + 3Cu(s)
Ecell = 1.98 V, E° cell = ?
Using Nernst equation at 298 K
Question. Calculate emf of the following cell at 25°C:
Fe | Fe2+(0.001 M)||H+(0.01 M)|H2(g)(1 bar)|Pt(s)
E°(Fe2+|Fe) = –0.44 V, E°(H+|H2) = 0.00 V
Answer. The cell reaction is
Question. Calculate the emf of the following cell at 25°C
Zn|Zn2+ (0.001 M) || H+(0.01M)| H2(g) (1 bar) | Pt(s)
Answer. The cell reaction is
Question. For the cell reaction
Ni(s) |Ni2+(aq)||Ag+(aq)|Ag(s)
Calculate the equilibrium constant at 25°C.
How much maximum work would be obtained by operation of this cell?
Answer. At anode : Ni → Ni2+ + 2e–
At cathode : [Ag+ + e– → Ag] × 2
Cell reaction : Ni + 2Ag+ → Ni2+ + 2Ag
E°cell = E°cathode – E°anode
= E°Ag+/Ag – E°Ni2+/Ni = 0.80 V – (– 0.25)
E°cell = 1.05 V
Question. Calculate the emf of the following cell at 298 K :
Fe(s)|Fe2+(0.001 M)||H+(1 M)|H2(g) (1 bar), Pt(s)
(Given E°cell = + 0.44 V)
Answer. Fe(s) | Fe2+(0.001 M) || H+(1 M) |H2(g)(1 bar) | Pt(s)
Reactions :
Question. Calculate the emf of the following cell at 25°C :
Ag(s)|Ag+(10–3 M)||Cu2+ (10–1 M)|Cu(s)
Given : E°cell = + 0.46 V and log 10n = n.
Answer. The cell may be represented as
Ag(s) |Ag+ (10–3 M)| |Cu2+ (10–1 M)| Cu(s)
Question. In the button cell, widely used in watches, the following reaction takes place.
Zn(s) + Ag2O(s) + H2O(l) → Zn2+(aq) + 2Ag(s)+ 2OH–(aq)
Determine E° and ΔG° for the reaction.
(Given : E°Ag+/Ag = + 0.80 V, E°Zn2+/Zn = – 0.76 V)
Answer. The cell reaction in button cell :
Zn(s) + Ag2O(s) + H2O(l) → Zn2+(aq) + 2Ag(s) + 2OH–(aq)
(i) Calculation of E°cell
Reactions :
Anode : Zn(s) → Zn2+(aq) + 2e–
Cathode :
Ag2O(s) + H2O(l) + 2e– → 2Ag(s) + 2OH–(aq) + 2e– n = 2
E°cell = E°cathode – E°anode = E°Ag2O/Ag – E°Zn2+/Zn
= + 0.80 – (– 0.76) V = 1.56 V
(ii) Calculation of ΔrG°
ΔrG° = –nFE°cell
= – 2 × 96500 C mol–1 × 1.56 V
= – 301080 C V mol–1
= – 301080 J mol–1= – 301 kJ mol–1
Question. A voltaic cell is set up at 25°C with the following half cells :
Al/Al3+ (0.001 M) and Ni/Ni2+ (0.50 M) Write an equation for the reaction that occurs when the cell generates an electric current and determine the cell potential.
E°Ni2+/Ni = – 0.25 V and E°Al3+/Al = – 1.66 V.
(log 8 × 10–6 = – 5.09)
Answer. At anode : Al(s) → Al3+(aq) + 3e–] × 2
At cathode : Ni2+ + 2e– → Ni(s)] × 3
Cell reaction : 2Al(s) + 3Ni2+(aq) → 2Al3+(aq) + 3Ni(s)
Applying Nernst equation to the above cell reaction,
Question. For the cell
Zn(s) | Zn2+ (2 M) || Cu2+ (0.5 M) | Cu(s)
(a) Write equation for each half-reaction.
(b) Calculate the cell potential at 25°C
Given :
Answer. a) Oxidation half reaction :
Zn(s) → Zn2+(aq) + 2e–
Reduction half reaction :
Cu2+(aq) + 2e– → Cu(s)
(b) E°cell =0.34 – (–0.76) = 1.10 V
Question. Calculate the equilibrium constant, K for the reaction at 298 K,
Answer.
ΔG° = –RT ln Kc = – 2.303 RT log Kc
⇒ – 212300 = – 2.303 × 8.314 × 298 × log Kc
Question. A copper-silver cell is set up. The copper ion concentration is 0.10 M. The concentration of silver ion is not known. The cell potential when measured was 0.422 V. Determine the concentration of silver ions in the cell. Given :
Answer.
The given cell may be represented as
Cu(s) |Cu2+ (0.10 M)|| Ag+ (C)| Ag(s)
E°cell = E°cathode – E°anode = 0.80 V – 0.34 V = 0.46 V
Question. Calculate the equilibrium constant for the reaction
Answer. E°cell = E°cathode – E°anode =–0.40V–(– 0.44V)= 0.04V
Question. One half-cell in a voltaic cell is constructed from a silver wire dipped in silver nitrate solution of unknown concentration. Its other half-cell consists of a zinc electrode dipping in 1.0 M solution of Zn(NO3)2. A voltage of 1.48 V is measured for this cell. Use this information to calculate the concentration of silver nitrate solution used.
Answer. The cell may be represented as
Zn(s) |Zn2+(aq)(1 M) ||Ag+(aq)|Ag(s)
E°cell = E°cathode – E°anode = 0.80 V – (– 0.76 V) = 1.56 V
Using formula,
Question. Calculate the standard cell potential of a galvanic cell in which the following reaction takes place :
2Cr(s) + 3Cd2+(aq) → 2Cr3+(aq) + 3Cd(s)
Calculate ΔrG ° and equilibrium constant, K of the above reaction at 25°C.
Given :
Answer.
T = 273 + 25°C = 298 K and n = 6
E°cell = E°cathode – E°anode = –0.40 – (– 0.74) = 0.34 V
ΔrG° = – nFE°cell = – 6 × 96500 × 0.34 = –196860 J mol–1
Again ΔrG° = – 2.303 RT log K
⇒ – 196860 = – 2.303 × 8.314 × 298 × log K
⇒ log K = 34.5014
⇒ K = antilog 34.5014 = 3.172 × 1034
Question. A cell is prepared by dipping copper rod in 1 M copper sulphate solution and zinc rod in 1 M zinc sulphate solution. e standard reduction potential of copper and zinc are 0.34 V and –0.76 V respectively.
(i) What will be the cell reaction?
(ii) What will be the standard electromotive force of the cell?
(iii) Which electrode will be positive?
Answer. (i) The cell reactions are :
Zn(s) → Zn2+(aq) + 2e– (Anode)
Cu2+(aq) + 2e– → Cu(s) (Cathode)
Net reaction :
Zn(s) + Cu2+(aq) → Zn2+(aq) + Cu(s)
(ii) E°cell = E°right – E°left = 0.34 V – (– 0.76 V) = 1.10 V
(iii) Copper electrode will be positive on which reduction takes place.
Question. Depict the galvanic cell in which the reaction
Zn(s) + 2Ag+(aq) → Zn2+(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.
Answer. The reaction is
(i) The zinc electrode is negatively charged (anode) as it pushes the electrons into the external circuit.
(ii) Ions are the current carriers within the cell.
(iii) The reactions occurring at two electrodes are :
