Write a balanced chemical equation for the overall cell reaction represented as:

a) Ag(s)|Ag^+(aq)||Sn^4+(aq), Sn^2+(aq)|Pt(s)
b) Al(s)|Al^3+(aq)||Cu^2+(aq)|Cu(s)
c) Pt(s)|Fe^2+(aq), Fe^3+(aq)||MnO4^-(aq),Mn^2+(aq)|Pt(s)

2Ag(s) + Sn^+4(aq) ==> 2Ag^+(aq) + Sn^+2(aq)

Don't listen to me. I'm wrong in every post I comment on. :|

Write a balanced chemical equation for the overall cell reaction represented as:

a) Ag(s)|Ag^+(aq)||Sn^4+(aq), Sn^2+(aq)|Pt(s)
b) Al(s)|Al^3+(aq)||Cu^2+(aq)|Cu(s)
c) Pt(s)|Fe^2+(aq), Fe^3+(aq)||MnO4^-(aq),Mn^2+(aq)|Pt(s)

To write a balanced chemical equation for the given cell reactions, we need to break down each half-cell reaction and balance the equation for each one separately.

a) Ag(s)|Ag+(aq)||Sn4+(aq), Sn2+(aq)|Pt(s)

In the first half-cell, Ag is being oxidized to Ag+ ions. The balanced equation for this oxidation half-reaction is:

Ag(s) -> Ag+(aq) + e-

In the second half-cell, Sn4+ is being reduced to Sn2+. The balanced equation for this reduction half-reaction is:

Sn4+(aq) + 2e- -> Sn2+(aq)

To balance the overall cell reaction, we need to multiply the half-reactions by appropriate coefficients to ensure that the number of electrons gained and lost is the same. In this case, we need to multiply the oxidation half-reaction by 2 and the reduction half-reaction by 1:

2Ag(s) -> 2Ag+(aq) + 2e-
Sn4+(aq) + 2e- -> Sn2+(aq)

The balanced overall cell reaction for part (a) is:

2Ag(s) + Sn4+(aq) -> 2Ag+(aq) + Sn2+(aq)

b) Al(s)|Al3+(aq)||Cu2+(aq)|Cu(s)

In the first half-cell, Al is being oxidized to Al3+ ions. The balanced equation for this oxidation half-reaction is:

Al(s) -> Al3+(aq) + 3e-

In the second half-cell, Cu2+ is being reduced to Cu. The balanced equation for this reduction half-reaction is:

Cu2+(aq) + 2e- -> Cu(s)

To balance the overall cell reaction, we need to multiply the oxidation half-reaction by 2 and the reduction half-reaction by 3:

2Al(s) -> 2Al3+(aq) + 6e-
3Cu2+(aq) + 6e- -> 3Cu(s)

The balanced overall cell reaction for part (b) is:

2Al(s) + 3Cu2+(aq) -> 2Al3+(aq) + 3Cu(s)

c) Pt(s)|Fe2+(aq), Fe3+(aq)||MnO4-(aq),Mn2+(aq)|Pt(s)

In the first half-cell, Fe2+ and Fe3+ ions are both involved in the reaction. Let's consider the reduction half-reaction for Fe2+:

Fe2+(aq) -> Fe3+(aq) + e-

In the second half-cell, MnO4- is reduced to Mn2+. The balanced equation for this reduction half-reaction is:

8H+(aq) + MnO4-(aq) + 5e- -> Mn2+(aq) + 4H2O(l)

Since both Fe2+ and Fe3+ are participating in the reaction, we need to combine the half-reactions in such a way that the number of electrons gained and lost is the same. To do this, we can multiply the reduction half-reaction for Fe2+ by 5, and the reduction half-reaction for MnO4- by 1:

5Fe2+(aq) -> 5Fe3+(aq) + 5e-
8H+(aq) + MnO4-(aq) + 5e- -> Mn2+(aq) + 4H2O(l)

Now we can write the balanced overall cell reaction for part (c), including both oxidation and reduction half-reactions:

5Fe2+(aq) + 8H+(aq) + MnO4-(aq) -> 5Fe3+(aq) + Mn2+(aq) + 4H2O(l)