Answer:
Cathode: Ag
Anode: Br₂
Explanation:
In the cathode must occur a reduction, so it's more likely to a metal atom be in the cathode. For the metals given the reduction reactions and the potential of reduction are:
Ag⁺ + e⁻ ⇒ Ag⁰ E° = + 0.80 V
Fe⁺² + 2e⁻ ⇒ Fe⁰ E° = - 0.44 V
Al⁺³ + 3e⁻ ⇒ Al⁰ E° = -1.66 V
As the potential for Ag is the higher, the reduction will occur for it first, so in the cathode will produce Ag.
For the anode an oxidation must occurs, so the reactions for the nonmetals are:
F₂ + 2e⁻ ⇒ 2F⁻ E° = +2.87 V
Cl₂ + 2e⁻ ⇒ 2Cl⁻ E° = +1.36 V
Br₂ + 2e⁻ ⇒ 2Br⁻ E° = +1.07 V
For oxidation, the less the E°, the faster the reaction will occur, so Br₂ will be formed in the anode.
Answer:
Option B, Because of the reversible nature of crystallizing and dissolving
Explanation:
Solution containing the maximum amount of solute that can be dissolved in the given solvent at the particular temperature is called saturated solution.
Reversible reaction is the reaction which can go in reverse and forward direction both on varying reaction condition.
In the saturated NaCl solution, on lowering temperature, The the dissolved NaCl molecules may crystallize. Likewise on increasing temperature, the crystallized crystals may dissolved. As the reaction moves in both the direction, therefore its considered to be equilibrium system.
Therefore, amog given, option B is correct.
Because of the reversible nature of crystallizing and dissolving
Given the equilibrium reaction: 2 A (aq) + 3 B (aq) <— —> 2 C (aq) + D (aq) and equilibrium concentrations of [A] = 0.60M, [B] = 0.30 M, [C] = 0.10 M and [D] = 0.50 M. The Kc value will be:
a. 1.9 c. 2.4 b. 0.15 d. 0.51
Answer . A
Answer:
D. The electron moved up to an energy level and has an energy of 21.72 x 10−19 J.
