You are given a galvanic cell consists of a Ni²⁺/ Ni half-cell and a standard hydrogen electrode. Also, you are given that the half cell Ni²⁺/ Ni will act as an anode, and the standard cell potential is 0.26V. You are asked to find the standard reduction potential for the half cell Ni²⁺/ Ni.
You will have a half - reaction for both nickel and hydrogen
The conversion of the symbol Ni²⁺/ Ni half-cell is
Ni²⁺ + 2e⁻ → Ni (s) E = 0.26V
and the conversion of the standard hydrogen electrode (SHE) is
2H⁺ + 2e⁻ → H₂ (g) E = 0V
Since H⁺ ions is a it difficult to set up during the process, nickel will be deposited at the cathode side instead of the anode. Therefore, The standard electron potential of the nickel will have -0.26V.
Answer:
The equilibrium will shift in forward direction.
Explanation:
The equilibrium between the carbonic acid and bicarbonate ion is shown below as:
H₂CO₃ (aq) ⇔ H⁺(aq) + HCO₃²⁻(aq)
According to Le Chatelier's Principle, the change in any state of the equilibrium say temperature, volume, pressure, or the concentration, the equilibrium will oppose these changes and will shift in such a way that the effect cause must be nullified.
<u>If a strong base is added to the equilibrium, the base will accept hydrogen ions which are formed in the right side of the equilibrium. Thus, there will be less hydrogen ions present and to compensate this effect, the equilibrium will shift in forward direction.</u>
Answer:
335 Joules kJ of heat will be released
Explanation:
Given the balanced equation:
8 Al (s) + 3 Fe3O4 (s) ----------> 4 Al2O3 (s) + 9 Fe(s),
ΔH = -3350*KJ/mol rxn
This is the heat released when 8 moles of Al react with 3 mol Fe3O4.
We then need to calculate the moles of reactants, verify if there is a limiting reagent and proceed to answer the question based on the soichiometry of the reaction.
Atomic weight Al = 26.98 g/mol Molecular Weight Fe3O4 = 231.53 g/mol
mol Al = 47.6 g/26.98 g/mol = 1.76 mol
mol Fe3O4 = 69.12 g/ 231.53 g/mol = 0.30 mol
Limiting reagent calculation:
8 mol Al / 3 mol Fe3O4 x 0.30 mol Fe3O4 = 0.80 mol Al are required and we have 1.76 mol, therefore Fe3O4 is the limiting reagent
Amount of Heat
-3350 kJ/ 3 mol Fe3O4 x 0.30 mol Fe3O4 = -335.00 kJ
1) Higher Temperatures
2) Lower Temperatures
