Revision - Electrochemistry (Cbse 2023)
Revision - Electrochemistry (Cbse 2023)
Revision - Electrochemistry (Cbse 2023)
1. Define Electrochemical cell.What happens when applied external opposite potential becomes greater
than Eocell of electrochemical cell.
2. Write the Nernst equation and emf of the following cells at 298K:
a) Mg(s)/Mg2+(0.001M)//Cu2+(0.0001)/Cu(s); EoCu2+/Cu = +0.34V, EoMg2+/Mg = -2.36
b) Fe(s)/Fe2+(0.001M)//H+(1M)/H2(g)/Pt(s) EoFe2+/Fe= -0.44 V
c) Cu/Cu2+(2M)//Ag+(0.05M)/Ag ; EoCu2+/Cu = +0.34V , EoAg+/Ag = + 0.80V
d) Mg(s)/Mg2+(10-3M)//Cu2+(10-4M)/Cu(s); EoCu2+/Cu = +0.34V, EoMg2+/Mg = -2.36
e) Sn/Sn2+(0.050M)//H+(0.020M)/H2(g)/Pt(s) EoSn2+/Sn= - 0.14V
3. Calculate the emf of the cell at 25oC for the following :
a) Mg(s) + 2Ag+(0.0001M)→ Mg+2(0.130M) + 2Ag(s). if Eocell =3.17V.
b) Ni(s) + 2Ag+ (0.002M)→ Ni+2(0.160M) + 2Ag(s),Given Eocell =1.05 V
c) 2Cr(s) + 3Fe2+(0.1M) → 2Cr3+(0.01M) + 3Fe(s) EoCr3+/Cr= - 0.74V , EoFe2+/fe= - 0.44V
4. A voltaic cell is constructed at 25oC with the following half cell Ag+(0.001M)/Ag and Cu+2 (0.01M)/Cu
what would be the voltage of this cell? Given EoAg+/Ag = + 0.80V, EoCu2+/Cu = +0.34V).
5. Calculate ΔrG0 & value of equilibrium constant for the following :
a) Mg(s)/Mg2+//Cu2+/Cu(s); EoCu2+/Cu = +0.34V, EoMg2+/Mg = -2.36
b) Zn(s) + Cu2+ Zn2+ + Cu EoCu2+/Cu = +0.34V, EoZn2+/Zn = -0.76V
c) Cu (s) + 2Ag+ Cu +2+ 2Ag(s) EoCu2+/Cu = +0.34V, EoAg+/Ag = + 0.80V
d) 2Fe3+ + 2I- → 2Fe2+ + I2 has Eocell =0.236 V
e) 2Cr(s) + 3Cd2+ → 2Cr3+ + 3Cd(s) EoCr3+/Cr= - 0.74V , EoCd2+/Cd= - 0.40V
6. Given the standard electrode potentials,:K+/K = –2.93V, Ag+/Ag = 0.80V, Hg2+/Hg = 0.79V Mg2+/Mg = –
2.37 V, Cr3+/Cr = – 0.74V.Arrange these metals in their increasing order of reducing power
7. Electrochemical cell gives an electrical potential of
when concentration and ions is unity. State the direction of flow of current and also specify
whether zinc and copper are deposited or dissolved at their respective electrodes when: (i) an
external opposite potential of less than 1.1 V is applied. (ii) an external potential of is applied. (iii)
an external potential of greater than is applied.
8. Define conductivity,molar conductivity &. limiting molar conductivity.
9. Express the relation among the cell constant ,the resistance of the solution in the cell and the
conductivity of the solution .How is the conductivity of a solution related to its molar conductivity.
10. The resistance of 0.01M NaCl solution at 25oC is 200ohm.The cell constant of the conductivity cell is
unity .calculate the molar conductivity of the solution.
11. The conductivity of 0.20M solution of KCl at 298 K is 0.0248SCm-1.Calculate its molar conductivity.
12. The Molar conductivity of a 1.5M solution of an electrolyte is found to be 138.9 SCm 2 mol-1 . Calculate
the conductivity of this solution.
13. The resistance of conductivity cell containing 0.001M KCl at 298K is 1500 ohm. What is cell constant if
conductivity of 0.001M KCl at 298K is 0.146X 10-3Scm-1.
14. When a certain conductance cell was filled with 0.1M KCl solution it has resistance of 85 ohm at
25oC.When the same cell was filled with an aqueous solution of 0.052M of unknown electrolyte the
resistance was 96 ohm. Calculate the molar conductivity of electrolyte. The conductivity of 0.1 M
solution of KCl is 1.29 x 10-2Scm-1.
15. The electrical resistance of a column of 0.05 molL-1 NaOH solution of diameter 1 cm and length 50 cm
is 5.55 X 103 ohm. Calculate its resistivity, conductivity & molar conductivity.
16. State Kohlrausch law and its application. Limiting molar conductivity of NaCl, HCl and NaAc are 126.4,
425.9 &91 SCm2 mol-1.Calculate Limiting molar conductivity of HAc.
17. Calculate the degree of dissociation of acetic at 298K,given that Λm(CH3COOH)=11.7 SCm2 mol-1 ,
Λ0m(CH3COO-)=40.9 SCm2 mol-1 , Λ0m(H+)=349.1 SCm2 mol-1