AgF consists of Ag+ and F- ions, which are fully dissociated in aqueous solution. When solving electrolysis problems, it is important to remember that water itself may also be a subject to electrolysis. Therefore, determining which species is oxidized and which species is reduced depends on selecting the processes that are the most energetically favorable. The most preferred reduction reaction will be Ag+ + e- = Ag (Emf=0.7996 V) which will occur at the cathode, on the other hand, the most favorable oxidation reaction will be
2H2O = O2 +4H+ + 4e- (Emf = -1.3 V) that will occur at the anode. Thus, the product at the anode is oxygen gas and at the cathode electrode is silver metal.
Answer:
(C) Mass of KCl(s), mass of H20, initial temperature of the water, and final temperature of the solution
Explanation:
molar enthalpy of solution of KCl(s) is heat evolved or absorbed when one mole of KCl is dissolved in water to make pure solution . The heat evolved or absorbed can be calculated by the following relation.
Q = msΔt where m is mass of solution or water , s is specific heat and Δt is change in temperature of water .
So data required is mass of water or solution , initial and final temperature of solution , specific heat of water is known .
Now to know molar heat , we require mass of solute or KCl dissolved to know heat heat absorbed or evolved by dissolution of one mole of solute .
Answer: option C) II < III < I
i.e [OH−] < [H3O+] < I
Explanation:
First, obtain the pH value of I and II, then compare both with III.
For I
Recall that pH = -log (H+)
So pH3O = -log (H3O+)
= - log (1x10−5)
= 4
For II
pOH = - log(OH-)
= - log(1x10−10)
= 9
For III
pH = 6
Since, pH range from 1 to 14, with values below 7 to be acidic, 7 to be neutral, above 7 to be alkaline: then, 9 < 6 < 4
Thus, the following solutions from least acidic to most acidic is II < III < I
<span>C.) Protons, neutrons. Hope it helps :)</span>
