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
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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.)
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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
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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
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.
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