At the anode, half-cell oxidation occurs in a voltaic cell.
<h3>Voltaic Cell Principle</h3>
A voltaic cell generates electricity due to the Gibbs free energy of spontaneous redox processes occurring inside the cell, which is the basis for the voltaic cell's operating principle.
Two half-cells plus a salt bridge make up the voltaic cell. An electrolyte-immersed metallic electrode is present on each side of the cell. These two half-cells are wired together to form a connection to a voltmeter.
<h3>Voltaic Cell Parts</h3>
- Copper makes comprises the cathode of a photovoltaic cell. This electrode serves as the cell's positive terminal, where reduction takes place.
- Anode: Zink metal makes up this electrode. It creates the cell's negative electrode, where oxidation takes place.
- Oxidation and reduction are divided into two discrete parts in two half-cells.
- Salt Bridge: It contains the electrolytes needed to finish the circuit in the voltaic cell.
- The flow of electrons between the electrodes occurs via the external circuit.
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Answer:
(a) 0.047 g (b) 0.0016 oz (c) 0.0001 lb
Explanation:
The given mass of the sodium in the slice = 47 mg
(a) Mass has to be calculated in grams
The conversion of mg to g is shown below as:
1 mg = 10⁻³ g
So,
<u>Mass of sodium = 47 × 10⁻³ g = 0.047 g</u>
(b) Mass has to be calculated in ounces
The conversion of ounces to g is shown below as:
453.6 g = 16 oz
Or,
1 g = 16 / 453.6 oz
So,
<u>Mass of sodium = (0.047 × 16) / 453.6 oz = 0.0016 oz</u>
(c) Mass has to be calculated in pounds
The conversion of pounds to g is shown below as:
1 lb = 453.6 g
Or,
1 g = 1/ 453.6 lb
So,
<u>Mass of sodium = (0.047 × 1) / 453.6 oz = 0.0001 lb</u>
Answer:
Equation of reaction:
a) 2HCl + Ba(OH)2 ==> CaCl2 + 2H2O
b) Molarity of base = 0.042 M.
Explanation:
Using titration equation
CAVA/CBVB = NA/NB
Where NA is the number of mole of acid = 2
NB is the number of mole of base = 1
CA is the molarity of acid =0.15M
CB is the molarity of base = to be calculated
VA is the volume of acid = 25 ml
VB is the volume of base = 44.45mL
Substituting
0.15×25/CB×44.45 = 2/1
Therefore CB =0.15×25×1/44.45×2
CB = 0.042 M.
Answer:
(A) N4H6 (B) H2O (C) LiH (D) C12H26
Explanation:
The given compounds have been arranged from left to right in order of increasing percentage by mass of hydrogen.
The percent by mass of hydrogen can be calculated by mass of hydrogen in that compound divided by total mass of that compound and finally multiplying the result with 100 to obtain the required percentage.
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