The pH of the standard hydrogen electrode that has electrode potential of -0.128 V is 4.3.
The equation of the hydrogen electrode is;
2H^+(aq) + 2e ⇄ H2(g)
The standard electrode potential of hydrogen is 0.00 V
Using the Nernst equation;
Ecell = E°cell - 0.0592/n log Q
Now;
E°cell = 0.00 V
n = 2
Q = 1/[H^+]
-0.128 = 0.00 - 0.0592/2 log 1/[H^+]
-0.128 = 0.00 - 0.0296 log 1/[H^+]
-0.128 = - 0.0296 log 1/[H^+]
-0.128/ - 0.0296 = log 1/[H^+]
1/[H^+] = Antilog (4.32)
[H^+] = 4.79 × 10^-5
Now;
pH = -log[H^+]
pH = -log (4.79 × 10^-5)
pH = 4.3
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Answer:
they have the same number of electrons in their outer energy level
First solve the moles of oxgen present in the compound
mol O = 6.93 g O ( 1 mol O / 16 g O )
mol O = 0.43 mol H
then solve the moles of hydrogen present
mol H = ( 7.36 - 6.93) g H ( 1 mol H / 1 g H)
mol H = 0.43 mol H
so the O and H are in the same mole content so the molecular formula would be OH, but the molar mass will not satisfy. so the answer would be
H2O2