Answer : The energy removed must be, 29.4 kJ
Explanation :
The process involved in this problem are :
The expression used will be:
where,
= heat released for the reaction = ?
m = mass of benzene = 94.4 g
= specific heat of solid benzene =
= specific heat of liquid benzene =
= enthalpy change for fusion =
Now put all the given values in the above expression, we get:
Negative sign indicates that the heat is removed from the system.
Therefore, the energy removed must be, 29.4 kJ
Answer:
-1.05 V
Explanation:
A detailed diagram of the setup as required in the question is shown in the image attached to this answer. The electrolytes chosen are SnCl2 for the anode half cell and MnCl2 for the cathode half cell. Tin rod and manganese rod are used as the anode and cathode materials respectively. Electrons flow from anode to cathode as indicated. The battery connected to the set up drives this non spontaneous electrolytic process.
Oxidation half equation;
Sn(s) ------> Sn^2+(aq) + 2e
Reduction half equation:
Mn^2+(aq) + 2e ----> Mn(s)
Cell voltage= E°cathode - E°anode
E°cathode= -1.19V
E°anode= -0.14 V
Cell voltage= -1.19 V - (-0.14V)
Cell voltage= -1.05 V
