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
Answer:
1,31÷2 =10,11
Explanation:
c10h22+31÷2o2=10co2+11h2o
Answer:
ΔH = 2.68kJ/mol
Explanation:
The ΔH of dissolution of a reaction is defined as the heat produced per mole of reaction. We have 3.15 moles of the solid, to find the heat produced we need to use the equation:
q = m*S*ΔT
<em>Where q is heat of reaction in J,</em>
<em>m is the mass of the solution in g,</em>
<em>S is specific heat of the solution = 4.184J/g°C</em>
<em>ΔT is change in temperature = 11.21°C</em>
The mass of the solution is obtained from the volume and the density as follows:
150.0mL * (1.20g/mL) = 180.0g
Replacing:
q = 180.0g*4.184J/g°C*11.21°C
q = 8442J
q = 8.44kJ when 3.15 moles of the solid react.
The ΔH of the reaction is:
8.44kJ/3.15 mol
= 2.68kJ/mol
