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)