Answer:
20.3-17.5=2.8ml 1ml=1cm3 vol= 2.8cm3
Explanation:
The question is incomplete, the complete question is
Which is NOT correct for when the silver and vanadium half-cells are connected via a salt bridge and a potentiometer? Ag^+ + 1 e^- rightarrow Ag Edegree = 0.7993 V V^2+ + 2e^- right arrow V E degree =-1.125 V Ag+ is reduced V is oxidized 1.924 V V2^+ is reduced Ag is oxidized I and II III, IV, and V I, II, and III III only IV and V
Answer:
only IV and V
Explanation:
If we look at the values of reduction potential for the two species, we will discover that vanadium has a negative reduction potential indicating its tendency towards oxidation.
On the other hand, solve has a positive reduction potential indicating a tendency towards reduction.
This implies that vanadium must be oxidized and silver reduced and not the not her way ground? Hence the answer above.
Answer:
The correct answer is 169.56 g/mol.
Explanation:
Based on the given information, the mass of Ag deposited is 1.24 g, and the mass of unknown metal X deposited in another cell is 0.650 g. The number of moles of electrons can be determined as,
= 1.24 g Ag * 1mol Ag/107.87 g/mol Ag * 1 mol electron/1 mol Ag ( the molecular mass of Ag is 107.87 g/mol)
= 0.0115 mole of electron
The half cell reaction for the metal X is,
X^3+ (aq) + 3e- = X (s)
From the reaction, it came out that 3 faraday will reduce one mole of X^3+.
The molar mass of X will be,
= 0.650 g/0.0115 *3 mol electron/1 mol
= 56.52 * 3
= 169.56 g/mol
