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 balance the chemical equations for the given cell reactions, we need to follow a few steps:

a) Ag(s)|Ag+(aq)||Sn4+(aq), Sn2+(aq)|Pt(s):
Step 1: Write the half-reactions for the oxidation and reduction processes:
Oxidation: Ag(s) → Ag+(aq) + e-
Reduction: Sn4+(aq) + 2e- → Sn2+(aq)

Step 2: Balance the number of electrons transferred by multiplying the half-reactions as necessary. In this case, the reduction reaction needs to be multiplied by 2 to balance the number of electrons:
2Ag(s) → 2Ag+(aq) + 2e-
Sn4+(aq) + 2e- → Sn2+(aq)

Step 3: Combine the half-reactions to obtain the balanced equation for the overall cell reaction:
2Ag(s) + Sn4+(aq) → 2Ag+(aq) + Sn2+(aq)

b) Al(s)|Al3+(aq)||Cu2+(aq)|Cu(s):
Step 1: Write the half-reactions:
Oxidation: Al(s) → Al3+(aq) + 3e-
Reduction: Cu2+(aq) + 2e- → Cu(s)

Step 2: Balance the number of electrons transferred:
2Al(s) → 2Al3+(aq) + 6e-
3Cu2+(aq) + 6e- → 3Cu(s)

Step 3: Combine the half-reactions:
2Al(s) + 3Cu2+(aq) → 2Al3+(aq) + 3Cu(s)

c) Pt(s)|Fe2+(aq), Fe3+(aq)||MnO4-(aq), Mn2+(aq)|Pt(s):
Step 1: Write the half-reactions:
Oxidation: Fe2+(aq) → Fe3+(aq) + e-
Reduction: MnO4-(aq) + 8H+(aq) + 5e- → Mn2+(aq) + 4H2O(l)

Step 2: Balance the number of electrons transferred:
5Fe2+(aq) → 5Fe3+(aq) + 5e-
MnO4-(aq) + 8H+(aq) + 5e- → Mn2+(aq) + 4H2O(l)

Step 3: Combine the half-reactions:
5Fe2+(aq) + MnO4-(aq) + 8H+(aq) → 5Fe3+(aq) + Mn2+(aq) + 4H2O(l)

These are the balanced chemical equations for the given cell reactions.