Data
Electrodes: V
Electrolyte: V(NO3)2
Amps: 6.0
Time (mins): 800
1. Calculate the charge in coulombs that passes through the cell during electrolysis.
2. What is the anode and cathode reaction of this cell?
3. Calculate the mass of metal deposited on the cathode for your cell.
4. How many moles of electrons are transferred at the anode and cathode?
I am sorry for making you do this but please show a detailed explanation and work so I am able to understand what I should be able to do.
Given V(NO₃)₂(aq) + 6-amps for 800-min(=48,000-sec)
a. Amps = Charge/time => amps = Coulombs/seconds => Coulombs = amps x seconds
=> Coulombs = (6-amps)(800-min)(60-sec∙minˉ¹) = 288,000-Coulombs
………………………………………………………………………………………
b. The Vanadium electrode will undergo oxidation by the nitrate ion (undergoing reduction) to produce Vanadium(2) and Nitrogen dioxide.
…………Oxidation Rxn (Anode)* => Vᵒ(s-electrode) => V⁺²(aq) + 2eˉ
*Eᵒ(V) = -1.13-v < Eᵒ(H₂O) = +1.23-v (The reaction with the more negative electrode potential dominates at the anode in aqueous electrolysis => i.e., Vanadium Metal Oxidation is dominate over water oxidation.)
…………Reduction Rxn (Cathode)* => NO₃ˉ + 2H⁺ + 1eˉ => NO₂ + H₂O
*E(NO₃ˉ) = +0.94-v > E(H₂O) = -0.83-v (The reaction with the more positive electrode potential dominates at the cathode in aqueous electrolysis => Nitrate Reduction is dominate over water reduction.)
……………………………………………………………………………..
c. Using the Faraday Law equation for mass transfer
mass (m) = (Coulombs)(formula wt)/(Faraday Constant)(Oxidation State)
mass of Vanadium (mᵥ = (228,000-C)(51 g/mol)/(96500-C/mol)(2) = 76.1-g Vanadium
……………………………………………………………………………..
d. Balancing the net oxidation-reduction equation reactions by the half-reaction method gives 2-moles electrons transferred during the redox process.
Oxidation: ………. Vᵒ(s-electrode) => V⁺²(aq) + 2eˉ
Reduction: ……….. 2(NO₃ˉ + 2H⁺ + 1eˉ => NO₂ + H₂O)
Net Redox: ……….Vᵒ + 2NO₃ˉ + 4H⁺ => V⁺² + 2NO₂ + 2H₂O
2 moles eˉ lost =2 moles eˉ gained => Charge Balance & Mass Balance
No problem! I'm here to help. Let's go through each question step by step.
1. To calculate the charge in coulombs that passes through the cell during electrolysis, you need to use the formula Q = I * t, where Q is the charge in coulombs, I is the current in amps, and t is the time in seconds. However, since the time given is in minutes, we need to convert it to seconds.
Given:
Current (I) = 6.0 amps
Time (t) = 800 mins
First, we need to convert the time to seconds:
800 mins * 60 seconds/min = 48,000 seconds
Now, we can calculate the charge:
Q = (6.0 amps) * (48,000 seconds) = 288,000 coulombs
Therefore, the charge that passes through the cell during electrolysis is 288,000 coulombs.
2. To determine the anode and cathode reactions, we need to know the electrolyte being used. In this case, the electrolyte is V(NO3)2. Looking at the chemical formula, we see that it contains vanadium and nitrate ions.
The general half-reactions for electrolysis in an electrochemical cell are:
Anode (oxidation): M → M^y+ + y e^-
Cathode (reduction): N^z+ + z e^- → N
In the anode half-reaction, M represents the metal being oxidized and M^y+ represents the metal ion formed when the metal loses electrons. In the cathode half-reaction, N^z+ represents the cation being reduced, and N represents the substance formed when the cation gains electrons.
Since we know that V(NO3)2 is the electrolyte, the anode and cathode reactions depend on how the vanadium cations and nitrate anions behave during electrolysis. Without further information, we cannot determine the specific reactions.
3. To calculate the mass of metal deposited on the cathode, we need to use Faraday's law of electrolysis, which states that the mass of a substance formed at an electrode is directly proportional to the charge (Q) and the molar mass (M) of the substance, and inversely proportional to the Faraday constant (F).
The formula is given by: Mass = (Q * M) / (F * z)
Given:
Q = 288,000 coulombs (computed from question 1)
M = molar mass of the metal (need to be given)
F = Faraday constant = 96,485 coulombs/mol (constant value)
z = number of moles of electrons involved in the reaction (need to be determined from question 4)
4. To determine the number of moles of electrons transferred at the anode and cathode, we need to use Faraday's law again. The formula for moles of electrons (n) is given by: n = Q / (F * y)
Given:
Q = 288,000 coulombs (computed from question 1)
F = Faraday constant = 96,485 coulombs/mol (constant value)
y = number of moles of electrons involved in the reaction (need to be determined)
Once we have determined the number of moles of electrons at the anode and cathode (y), we can use that information to calculate the mass of metal deposited on the cathode (question 3).
Please provide the molar mass and any other relevant information related to the electrolysis reaction, so that we can proceed with calculating the mass of metal deposited and determining the anode and cathode reactions.