Answer:
1.18 V
Explanation:
The given cell is:
Half reactions for the given cell follows:
Oxidation half reaction: 
Reduction half reaction: 
Multiply Oxidation half reaction by 2 and Reduction half reaction by 3
Net reaction: 
Oxidation reaction occurs at anode and reduction reaction occurs at cathode.
To calculate the 
of the reaction, we use the equation:
Putting values in above equation, we get:
To calculate the EMF of the cell, we use the Nernst equation, which is:
![E_{cell}=E^o_{cell}-\frac{0.059}{n}\log \frac{[Al^{3+}]^2}{[Fe^{2+}]^3}](https://tex.z-dn.net/?f=E_%7Bcell%7D%3DE%5Eo_%7Bcell%7D-%5Cfrac%7B0.059%7D%7Bn%7D%5Clog%20%5Cfrac%7B%5BAl%5E%7B3%2B%7D%5D%5E2%7D%7B%5BFe%5E%7B2%2B%7D%5D%5E3%7D)
where,
= electrode potential of the cell = ?V
= standard electrode potential of the cell = +1.21 V
n = number of electrons exchanged = 6
Putting values in above equation, we get:
i think it's A. cause CH is 1:1 and if you reduce C2H2, the ratio would also be 1:1
H will definitely be positive because a bond is always more stable than no bond surely if it is a sigma bond.
For G you can't really know because you don't know how much energy is provided by the bond and if it outways the loss in disorder.
The reaction will become more spontaneous with a lower temperature because H tells you the reaction is exotherm
Using the ideal gas law: PV=nRT
P is pressure; V is volume; n is the amount in moles; R=0.082; T is temperature in K.
(4.68)*(4.95)=(16.45)*(0.0821)*T
Solve for T.
T=17.15
Answer:
The answer is "![\bold{\log \frac{[0] mole}{[R]mole}}](https://tex.z-dn.net/?f=%5Cbold%7B%5Clog%20%5Cfrac%7B%5B0%5D%20mole%7D%7B%5BR%5Dmole%7D%7D)
"
Explanation:
![E_{cell} =E_{cell}^{\circ} - \frac{0.0591}{n}= \log\frac{[0]}{[R]}\\](https://tex.z-dn.net/?f=E_%7Bcell%7D%20%3DE_%7Bcell%7D%5E%7B%5Ccirc%7D%20-%20%5Cfrac%7B0.0591%7D%7Bn%7D%3D%20%5Clog%5Cfrac%7B%5B0%5D%7D%7B%5BR%5D%7D%5C%5C)
In the above-given equation, we can see from 
, of both oxidant 
as well as the reactant were connected. however, weight decreases oxidant and reduction component concentration only with volume and the both of the half cells by the very same factor and each other suspend
![\to {\log \frac{[0] mole}{[R]mole}}](https://tex.z-dn.net/?f=%5Cto%20%7B%5Clog%20%5Cfrac%7B%5B0%5D%20mole%7D%7B%5BR%5Dmole%7D%7D)
