in the electrolysis of Cuso4 solution using copper electrodes what happens to the negative cathode electrode ,, does its substance get reduced ?​

Copper (ii) ions are reduced to metallic copper which are then deposited at the negative electrode irrespective of whether the negative electrode is  carbon or copper.

Explanation:

Electrolysis is the chemical decomposition of a compound when an electric current is passed through either a solution or a molten form of the compound. In electrolysis, the compound that is decomposed when electric current passes through it is known as the electrolyte. The conductors which carries electric current in and out of the electrolyte is known as electrodes. The positive electrode is the anode while the negative electrode is the cathode.

In the electrolysis of copper(ii) tetraoxosulfate (vi) solution, the positive ions present in the solution, copper (ii) Cu²⁺ and hydrogen, H⁺ ions are attracted to the negative electrode. However, on getting to the negative electrode (be it a carbon or copper rod/plate), Cu²⁺ ions are preferentially discharge due to their being lower than hydrogen in the electrochemical series. The Cu²⁺ ions accept two electrons each and are reduced to metallic copper which are deposited on the negative electrode. The reduction half-equation is shown below:

Cu²⁺(aq) + 2e⁻ > Cu(s)

idk what the question is about

the density of gold = 19.3 g/cm3

the relation between density and mass and volume is

density = mass / volume

given :

mass of gold = 13 g

density = 13g / volume = 19.3

volume = 13 /19.3 = 0.67 cm^3

thus volume of gold nugget = 0.67 cm^3