- Standard reduction potential of Ag/Ag⁺ is 0.80 v and that of Cu⁺²(aq)/Cu⁰ is +0.34 V.
- The couple with a greater value of standard reduction potential will oxidize the reduced form of the other couple.
Ag⁺ will be reduced to Ag(s) and Cu⁰ will be oxidized to Cu²⁺
Anode reaction: Cu⁰(s) → Cu²⁺ + 2 e⁻ E⁰ = +0.34 V
Cathode reaction: Ag⁺(aq) + e → Ag(s) E⁰ = +0.80 V
Cell reaction: Cu⁰(s) + 2 Ag⁺(aq) → Cu⁺²(aq) + 2 Ag⁰(s)
E⁰ cell = E⁰ cathode + E⁰ anode
= 0.80 + (-0.34) = + 0.46 V
Answer:
0.6743 M
Explanation:
HC₂H₃O₂ + NaOH → NaC₂H₃O₂ + H₂O
First we <u>calculate how many NaOH moles reacted</u>, using the <em>definition of molarity</em>:
- Molarity = moles / volume
- moles = Molarity * volume
- 0.4293 M * 39.27 mL = 16.86 mmol NaOH
<em>One NaOH moles reacts with one acetic acid mole</em>, so <u>the vinegar sample contains 16.86 mmoles of acetic acid as well</u>.
Finally we <u>calculate the concentration (molarity) of acetic acid</u>:
- 16.86 mmol HC₂H₃O₂ / 25.00 mL = 0.6743 M
Answer: S8(s) + 8O2(g) → 8SO2(g) ΔH = -2368 kJ a.
