Short Answer TypeOn 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.
Long Answer TypeIf we connect the zinc and copper by means of a metallic conductor, the excess electrons that remain when Zn2+ ions emerge from the zinc in the left cell would be able to flow through the external circuit and into the right electrode, where they could be delivered to the Cu2+ ions which become "discharged", that is, converted into Cu atoms at the surface of the copper electrode. The net reaction is the oxidation of zinc by copper(II) ions:
but this time, the oxidation and reduction steps (half reactions) take place in separate locations:
left electrode : Zn----> Zn2+ +2e-
| galvanic cell | Electrolytic cell |
|---|---|
| A Galvanic cell converts chemical energy into electrical energy. | An electrolytic cell converts electrical energy into chemical energy. |
| Here, the redox reaction is spontaneous and is responsible for the production of electrical energy. | The redox reaction is not spontaneous and electrical energy has to be supplied to initiate the reaction. |
| The two half-cells are set up in different containers, being connected through the salt bridge or porous partition. | Both the electrodes are placed in a same container in the solution of molten electrolyte. |
| Here the anode is negative and cathode is the positive electrode. The reaction at the anode is oxidation and that at the cathode is reduction. | Here, the anode is positive and cathode is the negative electrode. The reaction at the anode is oxidation and that at the cathode is reduction. |
| The electrons are supplied by the species getting oxidized. They move from anode to the cathode in the external circuit. | The external battery supplies the electrons. They enter through the cathode and come out through the anode. |
Short Answer TypeLong Answer TypeThe 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.
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