How much chlorine will be liberated on passing one ampere current for 30 min. through NaCl solution ?
0.66 mole
0.33 mole
0.66 g
0.33 g
Three ampere currernt was passed through an aqueous solution of an unknown salt of Pd for 1 h 2.977 g of Pdn+ was deposited at cathode.The value of n is:
(given at wf of Pd= 106.4)
6
5
4
3
The metal which can displace all other metals from their salt solution among the following is:
Al
Zn
Cu
Fe
The standard reduction potentials of half reactions are given below:
F2 (g) + 2e- 2F- (aq) ; E0 = + 2.85 V
Cl2 (g) + 2e- 2Cl- (aq) ; E0 = + 1.36 V
Br2 (g) + 2e- 2Br- (aq) ; E0 = + 1.06 V
I2 (g) + 2e- 2I- (aq) ; E0 = + 0.53 V
The strongest oxidising and reducing agents respectively are:
chlorine and iodine
fluorine and iodine ion
chloride and bromine
bromine and iodine
Equivalent conductance of O .0128 N acetic acid solution, at 25°C is 14 ohm-1 eq-1 and at infinite dilution is 391 ohm-1eq-1 .The degree of dissociation of acetic acid in the solution and its dissociation constant, respectively are:
0.036 and 1.66x10-5
0.121 and 1.78x10-4
0 .213 and 1.84x10-5
O .054 and 1.94x10-4
Two metals 'A' and 'B' have E, value equal to 0.36 V and 1.23 V. What will be the correct cell equation?
A + B2+ A2+ + B
A2+ + B A + B2+
Both (a) and (b)
None of the above
Ag | AgCl | Cl- (C2) || Cl- (C1) | AgCl | Ag for this cell G is negative if:
C1 = C2
C1 > C2
C2 > C1
both (a) and (b)
Rusting of iron is a chemical reaction. The reaction is :
double decomposition
combination
displacement
analysis
Which one of the following conditions will increase the voltage of the cell represents by the equation?
Cu (s) + 2Ag+ (aq) Cu2+ (aq) + Ag (s)
Increase in the concentration of Ag+ ions
Increase in the concentration of Cu2+ ions
Increase in the dimensions of silver electrode
Increase in the dimensions of copper electrode
During the electrolysis of the aqueous solution of copper sulphate using Pt electrodes , the reaction taking place at the anode is:
Cu2+ + 2e- Cu
Cu Cu2+ + 2e-
2H2O 4H+ + O2 + 4e-
H2O + e- OH- + H2
C.
2H2O 4H+ + O2 + 4e-
CuSO4 + aq Cu2+ (aq) + S (aq)
H2O H+ + OH-
At anode OH- OH + e-
4OH 2H2O + O2
Net reaction is 2H2O 4H + O2 + 4e-