2Au+(aq) + Cu(s) → 2Au(s) + Cu2+(aq)
That formula is correct. It represents the net ionic equation of a complex oxidation-reduction reaction. Note that mass (2Au, 1Cu) and charge (2+) are preserved on both sides. The copper is oxidized from 0 to 2+, whereas the gold is reduced from 1+ to 0. This makes copper the reducing agent and gold the oxidizing agent.
In layman's terms, when you put solid copper into a solution with gold in it, the gold will precipitate out. The reason the reaction happens at all is because gold is so resistant to being oxidized. Essentially any metal, including copper, will force gold out of solution.
That equation, as stated above, is a net ionic equation and is the simplified form of a larger equation, such as:
Molecular: 2AuNO3(aq) + Cu(s) → Cu(NO3)2(aq) + 2Au(s)
Ionic: 2Au+(aq) + 2NO3-(aq) + Cu(s) → Cu2+(aq) + 2NO3-(aq) + 2Au(s)
Note that the nitrate ions are the same on both sides and do not participate in the overall reaction, hence their removal to form the net ionic equation.
The molecular equation for the reaction between Cu(NO3)2 and Na2S is: Cu(NO3)2 + 2Na2S -> CuS + 4NaNO3
Half equations ?? The chemical formula of magnesium oxide is MgO.
Cu(NO3) (aq) + Na2SO4 (aq) ==> CuSO4 (aq) + 2NaNO3 (aq)Since all species are soluble (aq), there will be NO REACTION.
The voltage of the cell depends on the materials used and the chemical reactions happening inside. To determine the voltage, you would need more information on the cell's composition, such as the type of electrolyte and electrodes used.
copper (II) sulfate is CuSO4 ; Zinc sulfate is ZnSO4 Zn + CuSO4 --> ZnSO4 + Cu
Cu(s) | Cu2+(aq) Au+(aq) | Au(s)
The equation for the reaction between zinc nitrate and cupric nitrate is: Zn(NO3)2 + Cu(NO3)2 -> Cu(NO3)2 + Zn(NO3)2
In the cell, the half-reaction for silver will be Ag+ (aq) + e- -> Ag (s) with a standard reduction potential of +0.80 V. The half-reaction for copper will be Cu2+ (aq) + 2e- -> Cu (s) with a standard reduction potential of +0.34 V. The silver half-reaction will occur at the cathode, while the copper half-reaction will occur at the anode in the cell.
The ionic equation for copper(II) nitrate solution (Cu(NO3)2) is: Cu2+(aq) + 2NO3-(aq) -> Cu(NO3)2(aq)
-2.71v
Molecular equation: Cu(NO3)2(aq) + 2LiOH(aq) -> Cu(OH)2(s) + 2LiNO3(aq) Full ionic equation: Cu2+(aq) + 2NO3-(aq) + 2Li+(aq) + 2OH-(aq) -> Cu(OH)2(s) + 2Li+(aq) + 2NO3-(aq) Net ionic equation: Cu2+(aq) + 2OH-(aq) -> Cu(OH)2(s)
-2.71v
In solution, CuC2H3O2 would dissociate into Cu2+ and C2H3O2- ions. The phases involved would be solid CuC2H3O2 dissociating into aqueous Cu2+ and C2H3O2- ions.
-2.71v
The complete ionic equation is Fe(s) + Cu^2+(aq) + SO4^2-(aq) -> Fe^2+(aq) + Cu(s) + SO4^2-(aq). The net ionic equation is Fe(s) + Cu^2+(aq) -> Fe^2+(aq) + Cu(s).
Cr(s) Cr3+(aq) Ag+(aq) Ag(s) Cu(s) Cu2+(aq) Sn4+(aq), Sn2+(aq) Pt(s)
Cu2+(aq) + 2e- → Cu(s) and Fe(s) → Fe2+(aq) + 2e-