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.
![E^0_{[Ag^{+}/Ag]}=+0.80V](https://tex.z-dn.net/?f=E%5E0_%7B%5BAg%5E%7B%2B%7D%2FAg%5D%7D%3D%2B0.80V)
![E^0_{[Cu^{2+}/Cu]}=+0.34V](https://tex.z-dn.net/?f=E%5E0_%7B%5BCu%5E%7B2%2B%7D%2FCu%5D%7D%3D%2B0.34V)
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.
Add water to mixture to dissolve sugar then filter. sugared water will be the filtrate while the salt & iron fillings will be the residue. use magnet to attract iron fillings and remaining will be salt.
Answer: 2.71 moles of solute for every 1 kg of solvent.
Explanation: As you know, the molality of a solution tells you the number of moles of solute present for every 1 kg of the solvent.This means that the first thing that you need to do here is to figure out how many grams of water are present in your sample. To do that, use the density of water.500.mL⋅1.00 g1mL=500. g Next, use the molar mass of the solute to determine how many moles are present in the sample.115g⋅1 mole NanO385.0g=1.353 moles NaNO3So, you know that this solution will contain 1.353moles of sodium nitrate, the solute, for 500. g of water, the solvent.In order to find the molality of the solution, you must figure out how many moles of solute would be present for 1 kg=103g of water.103g water⋅1.353 moles NaNO3500.g water=2.706 moles NaNO3You can thus say that the molality of the solution is equal to molality=2.706 mol kg−1≈2.71 mol kg−1 The answer is rounded to three sig figs.