PH = -log([H+])
[H+] = 10^(-pH)
[H+] = 10^(-9)
[H+][OH-] = Kw
Kw = 1.0*10^-14 at 25 degrees celsius.
[OH-] = Kw/[H+] = (1.0*10^-14)/(1*10^-9) = 1.0*10^-5
The concentration of OH- ions is 1.0*10^-5 M.
The anode is the electrode where the oxidation occurs.
Cathode is the electrode where the reducction occurs.
Equations:
Mn(2+) + 2e- ---> Mn(s) Eo = - 1.18 V
2Fe(3+) + 2e- ----> 2 Fe(2+) 2Eo = + 1.54 V
The electrons flow from the electrode with the lower Eo to the electrode with the higher Eo yielding to a positive voltage.
Eo = 1.54 V - (- 1.18) = 1.54 + 1.18 = 2.72
Answer: 2.72 V
The answer would be B the transfer of electrons from one atom to another
Answer:
Aluminum is the 13th element on the periodic table. It is located in period 3 and family 13.
Explanation:
Answer:
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Explanation:
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