A galvanic cell with e o cell = 0.30 v can be constructed using an iron electrode in a 1.0 m fe(no3)2 solution, and either a tin electrode in a 1.0 m sn(no3)2 solution, or a chromium electrode in a 1.0 m cr(no3)3 solution even though sn2+/sn and cr3+/cr have different reduction potentials. give the overall balanced reaction for fe-sn cell. do not include the states of matter.
To find the overall balanced reaction for an Fe-Sn galvanic cell, you'll need to determine the individual half-reactions first and then combine them.
1. Identify the half-reaction for the oxidation process (where Fe is being oxidized) and the half-reaction for the reduction process (where Sn is being reduced).
The half-reaction for the oxidation process involving Fe can be determined using the reduction potential values (E°) for Fe(II) --> Fe(III) and Sn(II) --> Sn.
From the reduction potentials table:
- E° for Fe(II) --> Fe(III) = +0.77 V
- E° for Sn(II) --> Sn = -0.14 V
Since oxidation occurs at the anode (negative electrode), the reaction with the more positive reduction potential will be the oxidation half-reaction. Therefore, Fe(II) is oxidized to Fe(III) producing Fe(III) ions. The half-reaction for Fe is:
Fe(II) --> Fe(III) + e⁻
2. Determine the half-reaction for the reduction process involving Sn.
Since reduction occurs at the cathode (positive electrode), the reaction with the more negative reduction potential will be the reduction half-reaction. Therefore, Sn(II) is reduced to Sn producing Sn(II) ions. The half-reaction for Sn is:
Sn(II) + 2e⁻ --> Sn
3. Balance the half-reactions.
To balance the half-reactions, you need to ensure that the number of electrons transferred in each half-reaction is equal. In this case, multiply the oxidation half-reaction by 2 to balance the number of electrons:
2 Fe(II) --> 2 Fe(III) + 2e⁻
Now the number of electrons transferred in each half-reaction is the same.
4. Combine the half-reactions to obtain the overall balanced reaction.
The overall balanced reaction can be obtained by adding the two half-reactions together:
2 Fe(II) + Sn(II) + 2e⁻ --> 2 Fe(III) + Sn
Therefore, the overall balanced reaction for the Fe-Sn galvanic cell is:
2 Fe(II) + Sn(II) --> 2 Fe(III) + Sn
To find the overall balanced reaction for the Fe-Sn cell, we first need to write the individual half-reactions for each electrode:
At the anode (oxidation half-reaction):
Fe(s) → Fe2+(aq) + 2e-
At the cathode (reduction half-reaction):
Sn2+(aq) + 2e- → Sn(s)
Now, we need to balance the number of electrons transferred in the two half-reactions so that when they are combined, they cancel out:
2 Fe(s) → 2 Fe2+(aq) + 4e-
Sn2+(aq) + 4e- → Sn(s)
By multiplying the reduction half-reaction by 2, we can balance the number of electrons. Now, we can combine the two half-reactions to get the overall balanced reaction:
2 Fe(s) + Sn2+(aq) → 2 Fe2+(aq) + Sn(s)
Therefore, the overall balanced reaction for the Fe-Sn cell is:
2 Fe(s) + Sn2+(aq) → 2 Fe2+(aq) + Sn(s)