Answer:
Anode: H₂(g) + 2 OH⁻(aq) → 2 H₂O(l) + 2 e⁻
Cathode: 2 Fe⁺³(aq) + 2 e⁻ → 2 Fe⁺²(aq)
E° = 1.60 V
Explanation:
Let's consider the reaction taking place in a galvanic cell.
2 Fe⁺³(aq) + H₂(g) + 2 OH⁻(aq) → 2 Fe⁺²(aq) + 2 H₂O(l)
The corresponding half-reactions are:
Anode (oxidation): H₂(g) + 2 OH⁻(aq) → 2 H₂O(l) + 2 e⁻ E°red = - 0.83 V
Cathode (reduction): 2 Fe⁺³(aq) + 2 e⁻ → 2 Fe⁺²(aq) E°red = 0.77 V
The standard cell potential (E°) is the difference between the standard reduction potential of the cathode and the standard reduction potential of the anode.
E° = E°red, cat - E°red, an = 0.77 V - (-0.83 V) = 1.60 V
Answer:
This 2.5061243 moles is in 4 liters.
Here's link to the answer:
tinyurl.com/wpazsebu
Milliequivalent is a unit that made from millimol unit multiplied by the valence of the atoms. Calcium has 2 valence electron so the mEq would be twice as much as the mol. The solution has 155meq/liter and the volume is 1850ml. Then, the amount of calcium in the solution should be:
1850ml * (155mEq/1000ml) / (2mEq/mmol)= 143.375mmol= 0.144 mol
