Consider the following half-cell reactions: Zn2+(aq)  +  2e-  ↔  Zn(s)  Ecell = -0.76 V Cu2+(aq)  +  2e-  ↔  Cu(s)  Ecell = +0.34 V If a cell is built by combining above two half-cells, what will be its cell potential?  Hint: The half-cell with the highest positive potential gets reduced: Ecell= E(cathode) – E(anode)

Which statement is true regarding a half-cell potential of a reaction

Which of the following metal ion gives the highest acidity in water?

In the diagram shown, if A was Cu metal and B was Mg metal, what would be the cell potential of this electrochemical cell? Use: Cu2+(aq)  +  2e-  = Cu(s)       Ecell = +0.34 V Mg2+(aq)  +  2e-  = Mg(s)     Ecell= -2.36 V Please open the file Question 20.docx

A 70-year old female may need to cash in her contract in the future. The Funeral Practitioner should suggest what type of contract?

What are those appropriate and helpful acts of counseling that come after the funeral?

What statement is true for the following half cell reaction F2(g)  +   2e-  ↔  2F-(aq)  Ecell = + 2.87 V

A process in which a very heavy nucleus splits into more stable nuclei of intermediate mass is called

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

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