Answer:
The empirical formula is the simplest form;
Given:
Oxygen O at 94.1% and
H at 5.9%
Assume 100grams.
94% = 0.941 x 100gm. = 94.1 gm x 1mole/16gm. = 5.88 moles of O
5.9% = 0.059 x 100gm. = 5.9gm. X 1moleH/1.002gm. = 5.88 moles of H
There is one mole of O for each mole of H so the empirical formula is
and written as OH.
Answer : The 'Ag' is produced at the cathode electrode and 'Cu' is produced at anode electrode under standard conditions.
Explanation :
Galvanic cell : It is defined as a device which is used for the conversion of the chemical energy produces in a redox reaction into the electrical energy. It is also known as the voltaic cell or electrochemical cell.
In the galvanic cell, the oxidation occurs at an anode which is a negative electrode and the reduction occurs at the cathode which is a positive electrode.
We are taking the value of standard reduction potential form the standard table.
In this cell, the component that has lower standard reduction potential gets oxidized and that is added to the anode electrode. The second forms the cathode electrode.
The balanced two-half reactions will be,
Oxidation half reaction (Anode) :
Reduction half reaction (Cathode) :
Thus the overall reaction will be,
From this we conclude that, 'Ag' is produced at the cathode electrode and 'Cu' is produced at anode electrode under standard conditions.
Hence, the 'Ag' is produced at the cathode electrode and 'Cu' is produced at anode electrode under standard conditions.
