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:
The valence electrons are as follows for these groups of elements:
Halogen- SEVEN (halogens are group 7 elements that need one electron for the octet rule to be achieved)
Alkali Metals - ONE (these are group one elements that lose a single electron to form an octet and cation)
Alkaline Earth Metals - TWO (group two elements that lose two electrons to form 2+ cations)
Answer:
Conduct electricity when they are molten, while covalent compounds usually do not conduct electricity when they are molten.
Answer:
2K+(aq) + CO3²¯(aq) + Ca^2+(aq) + 2F¯(aq) —› Cu2CO3(s) + 2K+(aq) + 2F¯(aq)
Explanation:
K2CO3(aq) + 2CuF(aq) → Cu2CO3(s) + 2KF(aq)
The complete ionic equation for the above equation can be written as follow:
In solution, K2CO3 and CuF will dissociate as follow:
K2CO3(aq) —› 2K+(aq) + CO3²¯(aq)
CuF(aq) —› Ca^2+(aq) + 2F¯(aq)
Thus, we can write the complete ionic equation for the reaction as shown below:
K2CO3(aq) + 2CuF(aq) —›
2K+(aq) + CO3²¯(aq) + Ca^2+(aq) + 2F¯(aq) —› Cu2CO3(s) + 2K+(aq) + 2F¯(aq)
