Question

Working on an electrochemistry lab and I need to write the following net ionic equations based on the half reactions

Mg + AlCl3; Al + Pb(NO3)2; AgNO3 + Pb; Cu + MgSO4

I am aware of the half reactions for each of the solids, which I will include, but I'm confused about writing the half reactions for the compounds

Mg(aq) + 2e- --> Mg(s)

Al(aq) + 3e- --> Al(s)

Cu2+ (aq) + e- --> Cu+ (aq)

Pb2+ (aq) + 2e- --> Pb(s)

Answer 1

a)

Mg(0)   --------Mg+2 + 2e-

AlCl3 + 3e- ------ Al(O)

I have to have the same numbers of e that reduces than oxide, so I multiply by 2 and

3Mg(0)   -------- 3Mg+2 + 6e-

2AlCl3 + 6e- ------ 2Al(O)

The sum of both reactions gets you the total net reaction

3Mg + 2AlCl3 ----- 3MgCl2 + 2Al

b)

Al(0)   -------- Al+3 + 3e-

Pb(NO3)2 + 2e- ------ Pb(0) + 2NO3-

I have to have the same numbers of e that reduces than oxide, so I multiply by 2 and

2Al(0)   -------- 2Al+3 + 6e-

3Pb(NO3)2 + 6e- ------ 3Pb(O) + 6NO3-

The sum of both reactions gets you the total net reaction

2AL + 3Pb(NO3)2 ------ 2Al(NO3)3 + 3Pb

c)

AgNO3 +1e- -------- Ag(O)

Pb(0) + 2NO3- ---------- Pb(NO3)2 + 2e-

I have to have the same numbers of e that reduces than oxide, so I multiply by 2 the reaction of Ag

2AgNO3 +2e- -------- 2Ag

Pb(O) + 2NO3- ----------- Pb(NO3)2 + 2e-

The sum of both reactions gets you the total net reaction

Pb + 2AgNO3 --- Pb(NO3)2 + 2Ag

d)

MgSO4 + 2e-    -------- Mg+2 + SO4^-2

Cu(0) ---------- Cu +2 + 2e-

I have to have the same numbers of e that reduces than oxide, the sum of both reactions gets you the total net reaction

Cu + MgSO4 ----- CuSO4 + Mg