Answer:
LiOH (aq) + VCl3(aq) ---> LiCl(aq) + V(OH)3(s) - unbalanced
3LiOH (aq) + VCl3(aq) ---> 3LiCl(aq) + V(OH)3(s) - balanced
3Li+OH- (aq) + V3+(Cl-)3(aq) ---> 3Li+Cl-(aq) + V3+(OH-)3(s) - showing ions
3Li+(aq) + 3OH- (aq) + V3+(aq) + 3Cl-(aq) ---> 3Li+(aq) + 3Cl-(aq) + V3+(OH-)3(s) (some courses don't show the charges in insoluble ionic compounds - so V(OH)3(s)) - Showing soluble ionic compounds as separate ions.
3OH- (aq) + V3+(aq) ---> V3+(OH-)3(s) (or V(OH)3(s)) - without spectator ions
Explanation:
i don't know if this is right ore not but i hope this helps even if it is just a little bit sorry if this does not help i truly apologize
This is a strong acid but the actual value of the pH will depend on its strength.
Acids have a pH of less than 7.
Ecell = E°cell - RT/vF * lnQ
R is the gas constant: 8.3145 J/Kmol
T is the temperature in kelvin: 273.15K = 0°C, 25°C = 298.15K
v is the amount of electrons, which in your example seems to be six (I'm not totally sure)
F is the Faradays constant: 96485 J/Vmol (not sure about the mol)
Q is the concentration of products divided by the concentration of reactants, in which we ignore pure solids and liquids: [Mg2+]³ / [Fe3+]²
Standard conditions is 1 mol, at 298.15K and 1 atm
To find E°cell, you have to look up the reduction potensials of Fe3+ and Mg2+, and solve like this:
E°cell = cathode - anode
Cathode is where the reduction happens, so that would be the element that recieves electrons. Anode is where the oxidation happens, so that would be the element that donates electrons. In your example Fe3+ recieves electrons, and should be considered as cathode in the equation above.
When you have found E°cell, you can just solve with the numbers I gave you.
Idek I just need free points so I can use this app for free
