OK. Recall that, according to the activity series, an elemental metal will react with (be oxidized by) the ions of any species below it. We figure out what the elemental species is, what metal ion we have, and look at the activity series...
(a)Mn(s) + NiCl2(aq) -->
We have elemental Mn and Ni2+(aq); on the activity series, Mn is above nickel. Therefore, Mn will be oxidized to the +2 ion (look on the activity series) and Ni2+(aq) will be reduced to elemental Ni:
Mn(s) + Ni2+(aq) --> Mn2+(aq) + Ni(s)
This is a prime example of a single displacement reaction!
(b) Cu(s) + Cr(CH3COO)3(aq) -->
Here, we have elemental copper and a solution of Cr+3 ions. Copper is pretty low on the activity series (try to remember this!) - it is lower than Cr, so there will not be a reaction here.
(c) Cr(s) + NiSO4(aq) -->
We have elemental chromium and aqueous nickel (II) ions. Chromium is above Ni on the acvitity series, and Cr tends to form a +3 ion (look on the activity series), so in net ionic form,
2Cr(s) + 3Ni2+(aq) --> 2Cr+3 + 3Ni(s)
Hint here: do a single displacement to form chromium (III) sulfate, balance the equation, and put in net ionic form.
(d) Pt(s) + HBr(aq) -->
Pt is very low on the activity series - it is below hydrogen and therefore won't react with hydrogen ions (aka H+, or acid!!) No reaction here.
H2(g) + CuCl2 -->
We have elemental hydrogen and aqueous copper (II) ions. Copper is below hydrogen in the activity series, and since any elemental species will react with the ions of species below, we have
H2(g) + Cu2+ --> 2H+(aq) + Cu(s)
Doug Chapman firstname.lastname@example.org 6/23/08