Two half-cell reactions are shown below: Fe2+(aq) + 2e- …
Two half-cell reactions are shown below: Fe2+(aq) + 2e- ↔ Fe(s) Ecell = – 0.44 V Cu2+(aq) + 2e- ↔ Cu(s) Ecell = +0.34 V Hint: The half-cell with the highest positive potential gets reduced. the other half-cell must get oxidized. Ecell = E(cathode) – E(anode) Calculate the cell voltage of a cell built by combining above two half-cells
Read DetailsA titration was carried out to find the concentration of an…
A titration was carried out to find the concentration of an unknown base (NaOH) with a known concentration of HCl. Concentration of HCl =0.35 M Volume of HCl dispensed from the buret = 24.35 mL Volume of NaOH added into the beaker = 36.44 mL Calculate the number of moles of NaOH in 36.44 mL
Read Details