What is the reaction between magnesium and copper sulphate?
The single replacement reaction of Magnesium and Copper Sulfate produce:Mg + CuSO4 → MgSO4 + CuTheoretically, if Magnesium is placed in Copper Sulfate solution, a single replacement reaction occurs. Copper solid is displaced and the magnesium loses electrons to copper ions. This is the case because Magnesium is more reactive than the metal Copper in CuSO4. For reference see the image below, I circled both metals on the reactivity series chart.**Update: Magnesium will replace Copper in Copper Sulfate because Magnesium has higher tendency to oxidize compared to Copper.
Why is hydrochloric acid stronger than sulphuric acid even though sulphuric acid should dissociate to produce more hydrogen ions?
As already written by Andrew Wolff, the strength of an acid depends on how easily a proton can break away from it. The easier it is , the stronger is the acid. HCl is a stronger acid than H2SO4 because a proton more easily separates from chloride ion than from hydrogen sulphate ion and this is reflected in their dissociation constants.The basic question is,however, why it is easier for a proton to separate from chloride ion than from hydrogensulphate ion. Let us try to answer this basic question.Let us consider an acid HX, where X may be an atom or a group of atoms. In HCl , X is Cl and in H2SO4 X is SO2(OH)O. As an acid HX dissociates as HX=> H+ +X-. We can assume that the more exothermic or the less endothermic is the dissociation, the stronger is the acid. To get an idea of the enthalpy change in dissociation we break the dissociation into steps for which either enthalpy change is known or can be easily estimated.This dissociation may be considered to occur in three stepsHX => H + X, that is homolytic breakage of the covalent bondH => H+ +e- , that is ionisation of hydrogenX + e- => X-Step 2 is independent of the nature of X , so, while comparing the strengths of different acids , we need not consider the enthalpy change of this step because it is the same for all acids. The difference in the strengths of acids therefore depends on the difference in enthalpies of steps (1) and (3) for the different acids. Since step (1) involves homolytic fission of H - X bond, the stronger the bond the weaker the acid. In HCl the bond dissociation enthalpy is 427 kJ/mol whereas in H2SO4 ,where O - H bond breaks , it is 467 kJ/mol. So the lower bond dissociation enthalpy of the H - Cl bond is in favour of HCl being a stronger acid than H2SO4.In step 3 the electron gain enthalpy of Cl is more negative than the electron gain enthalpy of the O atom which is part of HSO4. ( I couldn't find data on a rapid search, but it is well known that O has less negative electron gain enthalpy than S which in turn has lower electron affinity than Cl) So this factor is also in favour of HCl being a stronger acid than H2SO4. Hydration energy data might favour H2SO4 but they are not available ( to me) so I have not brought them into discussion. For elementary discussion like the above enthalpy data are sufficient, but for thorough discussion free energy data are required.
Minimum voltage needed to produce a reaction in the electrolysis of copper(II) sulfate using inert electrodes?
The Cathode reaction in this process is: Cu^2+ + 2e => Cu The Anode reaction in this process is: 2H2O => O2 + 4H^1+ +4e The standard reduction potentials for these are: -.34 (Sign change because it is backward from the table) -1.23 (Assuming this is at pH of 0, and since water is oxidized, and it is negative because it is backward from the table) You need to multiply the Cu equation by 2 to cancel out the electrons, which means the potential needs to be multiplied by 2 as well. So the standard reduction for the cathode ends up being (-.34*2) From here, it's just Ecell=Cathode-Anode Hope you can get the answer from that!
Which would be more conductive, 0.1 M of NaCl or 1.0 M of NaCl? Why?
If I may rephrase your question:“Which solution has higher conductivity: 0.10 M NaCl(aq) or 1.0 M NaCl(aq)”?In general, the conductivity of a solution depends on the number of ions present in this solution.For strong electrolytes, such as NaCl(aq), the number of ions depends on the concentration of the solution: strong electrolytes dissociate in 100% (however, this is true for low concentrations).To measure the conductivity [measured in S/m (siemens/metre)], you need a cell (a beaker with two electrodes, separated by a fixed distance ), a source with fixed electric potential difference - such as a battery) and an ammeter.From the reading of the ammeter and the applied voltage, you can calculate the conductance of the solution in siemens (not to be confused with conductivity measured in S/m).When you know the distance between the electrodes, in metres, you can calculate what you are looking for: CONDUCTIVITY.The more ions, the higher conductivity - the reading of an ammeter will be higher for 1.0 M solution, than for 0.10 M solution.However, you must be cautious with the extrapolation of the conductivity in terms of the concentration of the solution. For low concentrations this relation is linear, but for higher concentrations, solubility of NaCl(s) in water displays departure from the “strong electrolyte behaviour”; of course at one point you will reach the saturated solution - I think it is about 6.0 M NaCl(aq) at 25 oC.