Two half-cells are shown below: Fe2+(aq)  +  2e-   ↔  Fe(s) …

Two half-cells 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. Based on the cell voltage you received for question 21, what can you say about this cell?

Consider the following solubility equilibrium:             …

Consider the following solubility equilibrium:                              MgCO3(s)⇄ Mg2+(aq) + CO32-(aq)            The addition of which of the following substances would decrease the solubility of MgCO3?