On the basis of the standard electrode potential values stated for acid solution. Predict whether Ti4+ species may be used to oxidise FeII to FeIII.
Reaction: E°/V TiIV + e– → Ti3+ + 0.01
Fe3+ + e– → Fe2+ + 0.77
Predict the products of electrolysis obtained at the electrodes in each case when the electrodes used are of platinum.
(i) An aqueous solution of AgNO3.
(ii) An aqueous solution of H2SO4.
a)
electrolytes |
nonelectrolyte |
An electrolyte dissociates in solution and thus produce ion.
|
A nonelectrolyte does not dissociate at all in solution and therefore does not produce any ions. |
Electrolytes are ionic substance that dissolve in water |
Nonelectrolytes are typically polar covalent substances that do dissolve in water as molecules instead of ions. |
C)
Primary cell |
Secondary |
Lower initial cost. |
Higher Initial Cost |
Higher life-cycle cost ($/kWh). |
Lower life-cycle cost ($/kWh) if charging in convenient and inexpensive |
Disposable. |
Regular maintenance required. |
Typically lighter and smaller thus traditionally more suited for portable applications. |
Traditionally less suited for portable applications, although recent advances in Lithium battery technology have lead to the development of smaller/lighter secondary batteries. |
Molar conductivity |
Specific conductivity |
Molar Conductivity of a solution at a given concentration is the conductance of the volume V of solution containing one mole of electrolyte kept between two electrodes with area of cross section A and distance of unit length. Therefore, Distance is unit so l = 1 Volume = area of base × length So V = A × 1 = A Λm =κA/l Λm = κV
|
Conductivity of a solution is equal to the conductance of a solution of 1 cm length and cross section area of 1 square cm. it may also be define as the conductance of ine centimeter cube of the conductor . It is represented by the symbol Kappa (κ). mathematically we can write κ = 1/ p here ρ is resistivity the unit of K is ohm –1 cm –1 or S cm–1 The conductivity, κ, of an electrolytic solution depends on the concentration of the electrolyte, nature of solvent and temperature.
|
The conductivity of 0.01 M solution of acetic acid at 25°C is 1.63 x 10–4 s cm–1. Given:
Λ°m (HCl) = 426 s cm2 mol–1, Δ°m (Na AC) = 91.5 cm2 mol–1
Λ°m (NaCl) = 126.5 cm2 mol–1 Calculate:
(a) the molar conductivity of acetic acid
(b) the degree of dissociation of acetic acid.
(c) the dissociation constant.
(d) the pH of 0.01 M solution of acetic acid.