Answer:
3I^-/I3^-//Tl3^+//Tl^+
Explanation:
Based on the given reduction potentials, the half reaction equation having greater positive value will be the cathode while the half reaction equation having less positive reduction potential will be the anode.
This implies that this equation:
Tl3+ (aq) + 2e- → Tl+ (aq) ξo = 1.24 occurs at the cathode
While:
3 I- (aq) → I3- (aq) + 2e- ξo = -0.55 occurs at the anode
Cell notation,
Anode//cathode
Hence for this system under consideration:
3I^-/I3^-//Tl3^+//Tl^+
Use the General Gas Law
PV = nRT => n = PV/ RT
P= <span>10130.0 kPa
V= </span><span>50 L
R= </span><span>R = 8.314 L∙kPa/K∙mol
T= </span><span>300°C + 273 = 573 K
n = </span>10130.0 kPa 50 L / 8.314 L∙kPa/K∙mol <span>573 K
n = </span><span>106.32 mol</span>
B- 90 grams bc 45 mL times 2 equals 90
Hydrogen can be used as a byproduct in the electrolysis of fuel