Answer:
E cell = +1.95 V
Explanation:
At Anode : Oxidation reaction takes place
At Cathode : Reduction reaction takes place
The reaction with lower value of reduction potential will undergo Oxidation
E = -1.18 V
This equation undergo oxidation reaction and become:
Anode(Oxidation-Half) :
E = +1.18 V
Cathode(Reduction-Half) :
E =+0.77 V
To balance the reaction multiply reduction-Half with 2.We get :

Note that E is intensive property , do not multiply E of oxidation-half with 2
Ecell = 0.77 -(-1.18)
E = +1.95 V
I think the answer will be Graphite.
Answer:
Second order
Explanation:
Given that:
When the reaction A → B + C is studied, a plot 1/[A]t vs. time gives a straight line with a positive slope.
From the integration method for the second order of reaction.
Suppose that:
rate = k₂[A]²
∴

Therefore, a plot of the linear function
versus t will be linear with a positive slope k₂ and the intercept on the concentration axis will be 
The linear plot for a second order reaction can be seen in theimage attached below.