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Ilya [14]
2 years ago
8

In a galvanic cell, electrons are transferred from one half cell to the other as the redox reaction progresses. what happens in

each half cell? the half cell in which the electrode gains electrons is where occurs, and the half cell in which the electrode loses electrons is where occurs.
Chemistry
1 answer:
Sidana [21]2 years ago
7 0

The half cell in which the electrode gains electrons is where reduction occurs, and the half cell in which the electrode loses electrons is where oxidation occurs.

<h3><u>What is a Galvanic cell ?</u></h3>

Voltaic or galvanic cells are electrochemical devices that use spontaneous oxidation-reduction events to generate electricity. In order to balance the overall equation and highlight the actual chemical changes, it is frequently advantageous to divide the oxidation-reduction reactions into half-reactions while constructing the equations.

Two half-cells make up most electrochemical cells. The half-cells allow electricity to pass via an external wire by separating the oxidation half-reaction from the reduction half-reaction.

<h3><u>Oxidation:</u></h3>

The anode is located in one half-cell, which is often shown on the left side of a figure. On the anode, oxidation takes place. In the opposite half-cell, the anode and cathode are linked.

<h3><u>Reduction:</u></h3>

The second half-cell, cathode, which is frequently displayed on a figure's right side. The cathode is where reduction happens. The circuit is completed and current can flow by adding a salt bridge.

To know more about processes in Galvanic cell, refer to:

brainly.com/question/13031093

#SPJ4

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A 13.1-g sample of CaCl2 is dissolved in 104 g of water, with both substances at 24.7°C. Calculate the final temperature of the
likoan [24]

Answer:

The final temperature of the solution is 44.8 °C

Explanation:

assuming no heat loss to the surroundings, all the heat of solution (due to the dissolving process) is absorbed by the same solution and therefore:

Q dis + Q sol = 0

Using tables , can be found that the heat of solution of CaCl2 at 25°C (≈24.7 °C)  is q dis= -83.3 KJ/mol . And the molecular weight is

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Qdis= -9.84 KJ

Also Qsol = ms * Cs * (T - Ti)

therefore

ms * Cs * (T - Ti) + Qdis = 0

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7 0
4 years ago
If 1.50 L of 0.780 mol/L sodium sulfide is mixed with 1.00 L of a 3.31 mol/L lead(II) nitrate solution, what mass of precipitate
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Answer:

336.1 g of PbS precipitate

Explanation:

The equation of the reaction is given as;

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Number of moles of lead II nitrate= concentration of lead II nitrate × volume of lead II nitrate

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1.17 moles of sodium sulphide also yields 1.17 moles of lead II sulphide

Hence sodium sulphide is the limiting reactant.

Thus mass of precipitate formed= amount of lead II sulphide × molar mass of sodium sulphide

Molar mass of lead II sulphide= 287.26 g/mol

Mass of lead II sulphide = 1.17 moles × 287.26 g/mol

Mass of lead II sulphide= 336.1 g of PbS precipitate

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3 years ago
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