Answer:
The oxidation state of the reducing agent has changed from 0 to +2.
Explanation:
reducing agent is anything that loses electron or gains oxygen
in this case, zinc
Answer: When an acidic solution is diluted with water the concentration of H + ions decreases and the pH of the solution increases towards 7.
Explanation:
Answer:
The 1st and 4th options are correct
I.the oxidized form has a higher affinity for electrons
IV. the greater the tendency for the oxidized form to accept electrons
Explanation:
Half reaction can be described as the oxidation or reduction reaction in a redox reaction.it is In the redox rection there is a change in the oxidation states of Chemical species involved. the oxidized form in the redox has a higher affinity for electrons and the greater the tendency for the oxidized form to accept electrons.
Standard reduction potential which is also referred to as standard cell potential can be described as the potential difference that exist between cathode and anode of the cell. In the standard reduction potential most times the species will be reduced which is usually analysed in a reduction half reaction.
(Standard Hydrogen Electrode) is utilized when determining the Standard reduction or potentials of a chemical specie. this is because of Hydrogen having zero reduction and oxidation potentials, as a result of this a measured potential of any species is compared with that of Hydrogen, the difference helps to know the potential reduction of that particular specie.
Answer:
Change in entropy for the reaction is
ΔS° = -268.13 J/K.mol
Explanation:
To calculate the change in entropy for the balanced reaction, we require the natural entropy of all the reactants and products in the reaction.
3 NO₂(g) + H₂O(l) → 2 HNO₃(l) + NO(g)
From Literature.
S°(NO₂) = 240.06 J/K.mol
S°(H₂O) = 69.91 J/K.mol
S°(HNO₃) = 155.60 J/K.mol
S°(NO) = 210.76 J/K.mol
These are the entropies of the reactants and products under standard conditions of 298.15 K and 1 atm.
Note that
ΔS° = Σ nᵢS°(for products) - Σ nᵢS°(for reactants)
Σ nᵢS°(for products) = [2 × S°(HNO₃)] + [1 × S°(NO)]
= (2 × 155.60) + (1 × 210.76) = 521.96 J/K.mol
Σ nᵢS°(for reactants) = [3 × S°(NO₂)] + [1 × S°(H₂O)]
= (3 × 240.06) + (1 × 69.91) =790.09 J/K.mol
ΔS° = Σ nᵢS°(for products) - Σ nᵢS°(for reactants)
ΔS° = 521.96 - 790.09 = -268.13 J/K.mol
Hope this Helps!!
