For the reactions that occurred spontaneously in the individual tables, balance these reactions assuming they are in an acidic solution.
Reaction 1:
Cu
I^-
Ag
Br^-
Reaction 2:
Zn
Pb
Cu
Ag
Reaction 3:
Cl2
Br2
I2
Reaction 1
I 2(aq) Cu2 + (aq) Ag + (aq) Br2 (aq)
Reaction 2
Ag + (aq) Pb2 + (aq) Cu2 + (aq) Zn2 +(aq)
Reaction 3
Br - (aq) Cl - (aq) I - (aq)
Your notation is still somewhat confusing to me but here is what you do for one of the reactions. The others are done the same way.
For the copper, Ag^+ reaction:
Cu ==> Cu^+2 + 2e
Ag^+ + e ==> Ag
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The electrons must be kept equal; therefore, multiply the first equation by 1 and the second equation by 2, then add them.
Cu + 2Ag^+ ==> Cu^+2 + 2Ag.
The others are done the same way.
To balance the reactions in an acidic solution, you need to follow these steps:
Step 1: Identify the unbalanced species and write down their chemical formulas.
In Reaction 1:
Unbalanced species: I2, Cu2+, Ag+, Br2
In Reaction 2:
Unbalanced species: Ag+, Pb2+, Cu2+, Zn2+
In Reaction 3:
Unbalanced species: Br-, Cl-, I-
Step 2: Balance the elements that are not involved in redox reactions.
In Reaction 1:
There are no additional elements to balance.
In Reaction 2:
There are no additional elements to balance.
In Reaction 3:
There are no additional elements to balance.
Step 3: Balance the reduction half-reactions.
In Reaction 1:
Start with the reduction half-reaction involving the easiest species to balance.
Ag+ + e- -> Ag (Silver reduction)
Since there is one Ag on the right side, you place a coefficient of 1 in front of the Ag+ and the e-.
Ag+ + e- -> Ag (Silver reduction) [Balanced]
Now, balance the reduction half-reaction for Cu2+ and I2.
Cu2+ + 2e- -> Cu (Copper reduction)
I2 + 2e- -> 2I- (Iodine reduction)
Since there are two I- ions on the right side, place a 2 in front of the I2:
Cu2+ + 2e- -> Cu (Copper reduction) [Balanced]
I2 + 2e- -> 2I- (Iodine reduction) [Balanced]
Step 4: Balance the oxidation half-reactions.
In Reaction 1:
Start with the oxidation half-reaction involving the easiest species to balance.
Br2 -> 2Br- (Bromine oxidation)
Since there are two Br- ions on the right side, place a 2 in front of the Br2:
2Br2 -> 4Br- (Bromine oxidation) [Balanced]
Now, balance the oxidation half-reaction for Cu2+.
Cu2+ -> Cu (Copper oxidation)
Since there is one Cu on each side, there is no need to place a coefficient.
Cu2+ -> Cu (Copper oxidation) [Balanced]
Step 5: Combine the balanced half-reactions and make sure the number of electrons transferred is the same for both half-reactions.
To do this, multiply the reduction half-reactions by appropriate factors to ensure equal numbers of electrons.
In Reaction 1, multiply the iodine reduction half-reaction by 2 to balance the electrons.
2Cu2+ + 4e- -> 2Cu (Copper reduction)
2I2 + 4e- -> 4I- (Iodine reduction)
Now, combine the balanced oxidation and reduction half-reactions:
2I2 + 2Cu2+ -> 4I- + 2Cu (Balanced)
Finally, sum up the species on both sides of the equation:
2I2 + 2Cu2+ + 4Ag+ + 2Br2 -> 4I- + 2Cu + 4Ag+ + 2Br2 (Balanced)
Repeat the same steps for Reaction 2 and Reaction 3 to balance them in an acidic solution.