Answer:
Part A. The half-cell B is the cathode and the half-cell A is the anode
Part B. 0.017V
Explanation:
Part A
The electrons must go from the anode to the cathode. At the anode oxidation takes place, and at the cathode a reduction, so the flow of electrons must go from the less concentrated solution to the most one (at oxidation the concentration intends to increase, and at the reduction, the concentration intends to decrease).
So, the half-cell B is the cathode and the half-cell A is the anode.
Part B
By the Nersnt equation:
E°cell = E° - (0.0592/n)*log[anode]/[cathode]
Where n is the number of electrons being changed in the reaction, in this case, n = 2 (Sn goes from S⁺²). Because the half-reactions are the same, the reduction potential of the anode is equal to the cathode, and E° = 0 V.
E°cell = 0 - (0.0592/2)*log(0.23/0.87)
E°cell = 0.017V
The answer to fill in the blank is (igneous)
I hope I helped.
Answer:
232.5 g C2H6O2
Explanation:
The equation you need to use here is ΔTf = i Kf m
Since pure water freezes at 0 C, your ΔTf is just 4.46 C
i = 1 (ethylene glycol is a weak electrolyte)
Kf = molal freezing constant, which for water is 1.86 C/m
m = molality = x mols C2H6O2 / 1.15 kg H2O (don't know the moles of ethylene glycol we're dissolving yet)
Than,
4.46 C = 1.86 C/m (x mol C2H6O2 / 1.15 kg H2O)
Solve for x, you should get x = 2.75 mol C2H6O2
3.75 mol C2H6O2 (62 g C2H6O2 / 1 mol C2H6O2) = 232.5 g C2H6O2
