Question

What are the balanced reaction equations for:

1.0 M lead (II) nitrate and Zn

0.1 M lead (II) nitrate and Zn

1.0 M copper nitrate and Mg

1.0 M potassium nitrate and Mg

1.0 M copper nitrate and Pb

1.0 M copper nitrate and Zn

0.1 M iron sulfate and Mg

0.1 M copper nitrate and Zn

Answer 1

This is done via activity of better yet, with reductino potentials:

Zn2+ + 2 e− ⇌ Zn(s) −0.7618;

Pb2+ + 2 e− ⇌ Pb(s) −0.126;

Cu2+ + 2 e− ⇌ Cu(s) +0.337;

Mg2+ + 2 e− ⇌ Mg(s) −2.372;

Fe2+ + 2 e− ⇌ Fe(s) −0.44

then

1.0 M lead (II) nitrate and Zn --> Pb(NO3)2 + Zn(s) --> Zn(NO3)2 + Pb(s); since Pb > Zn potential

0.1 M lead (II) nitrate and Zn --> Pb(NO3)2 + Zn(s) --> Zn(NO3)2 + Pb(s); since Pb > Zn potential

1.0 M copper nitrate and Mg --> Cu(NO3) + Mg(s) --> Cu(s) + Mg(NO3)2, since Mg < Cu, reduciton potential

1.0 M potassium nitrate and Mg --> no reaction, K+ remains aquoeus

1.0 M copper nitrate and Pb --> Cu(NO3)2 + Pb --> Cu(s) + Pb(NO3)2, since CU > Pb

1.0 M copper nitrate and Zn --> Cu(NO3)2 + Zn -> Zn(NO3)2 + Cu(s) , since CU > Zn

0.1 M iron sulfate and Mg --> FeSO4 + Mg --> MgSO4 + Fe+2

0.1 M copper nitrate and Zn --> Cu(NO3)2 + Zn --> Zn(NO3)2 + Cu(s) since Cu > Zn