I believe your answer is correct. A) Solar Cell
Answer:
b and c
Explanation:
the problem was solved through the experiment and tested
Answer:
I believe the answer would be chemical reaction.
Explanation:
We have to fhnd the standard cell potencial for this electrochemical cell:have to fi
Cd²⁺ (aq) + Mg (s) ----> Cd (s) + Mg²⁺ (aq)
We are given these electrode potentials at 25 °C.
Cd²⁺ (aq) + 2 e- ----> Cd (s) E∘(V) = - 0.40
Mg²⁺ (aq) + 2 e− ----> Mg (s) E∘(V) - 2.37
In our reaction, Cd is being reduced from i ou to solid Cd. Solid Mg is being oxidized from solid Mg to Mgd . In electrochemistry, the anode is where oxidation occurs and the cathode is where reduction occurs.
Cd²⁺ (aq) + 2 e- ----> Cd (s) Reduction = Cathode
Mg (s) ----> Mg²⁺ (aq) + 2 e− Oxidation = Anode
The standard cell potential of an electrochemicalectro will be hdard to reduction potential of the cathode minus that of the anode.²s) Cd (s) ²⁺
E°cell = E°cathode - E°anode
E°cell = -0.40 V - (- 2.37 V)
E°cell = 1.97 V
Answer: the standard cell potential of the electrochemical cell is 1.97 V.
Answer:
Explanation:
Hello,
In this case, since the ionization of the given HA acid is:
The equilibrium expression is:
![Ka=\frac{[H^+][A^-]}{[HA]}](https://tex.z-dn.net/?f=Ka%3D%5Cfrac%7B%5BH%5E%2B%5D%5BA%5E-%5D%7D%7B%5BHA%5D%7D)
Whereas the concentration of hydrogen ions is compute from the pH=
![[H^+]=10^{-pH}=10^{-2.00}=0.01M](https://tex.z-dn.net/?f=%5BH%5E%2B%5D%3D10%5E%7B-pH%7D%3D10%5E%7B-2.00%7D%3D0.01M)
Which also equals the concentration of 
and the in general the ionization extent, therefore, the acid ionization constant, Ka, turns out:
Regards.
