Complete this example: A 0.67 molal solution was
1 answer:
You might be interested in
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
Answer:
1) The order of the reaction is of FIRST ORDER
2) Rate constant k = 5.667 × 10 ⁻⁴
Explanation:
From the given information:
The composition of a liquid-phase reaction 2A - B was monitored spectrophotometrically.
liquid-phase reaction 2A - B signifies that the reaction is of FIRST ORDER where the rate of this reaction is directly proportional to the concentration of A.
The following data was obtained:
t/min 0 10 20 30 40 ∞
conc B/(mol/L) 0 0.089 0.153 0.200 0.230 0.312
For a first order reaction:
where :
K = proportionality constant or the rate constant for the specific reaction rate
t = time of reaction
= initial concentration at time t
= final concentration at time t
= concentration at time t
To start with the value of t when t = 10 mins
When t = 20
When t = 30
When t = 40
We can see that at the different time rates, the rate constant of all have similar constant values
As such :
Rate constant k = 0.034 min⁻¹
Converting it to seconds ; we have :
60 seconds = 1 min
∴
0.034 min⁻¹ =(0.034/60) seconds
= 5.667 × 10 ⁻⁴ seconds
Rate constant k = 5.667 × 10 ⁻⁴