consider the following electrochemical cell:
Pt(s) | Hg2^2+ (0.250M), Hg^2+ (0.800M) || MnO4^-(0.650M), H^+ (1.00M), Mn^2+ (0.375M) | Pt(s)
Using the half-reaction method, write a balanced chemical equation for the electrochemical cell.
As drawn in the cell,
Hg2^2+ + 2e ==> 2Hg^2+
MnO4^- + 8H^+ + 5e ==> Mn^2+ + 4H2O
Multiply each half reaction to make the electrons balance and add the two half reactions to obtain the full equation.
What should I multiply each reaction by to make it balanced?
Hg2^2+ + 2e ==> 2Hg^2+ multipy by 5
MnO4^- + 8H^+ + 5e ==> Mn^2+ + 4H2O multiply by 2
10 electrons exchanged then.
How would the final equation turn out? would those 10 electrons cancel out?
so far what I did:
10Hg2^2+ + 10e ==> 10Hg^2+
2MnO4^- + 16H^+ + 10e ==> Mn^2+ + 8H2O
You multiply the first one by 2 and the second by 5 as Bob P did, then you add the two half rxn together. You didn't do that. Your first equation isn't balanced with Hg. You have 20 Hg on the left and only 10 on the right. Yes, the electrons balance, that's why you multiplied by 5 and by 2 to make the electrons equal so they would cancel.
20 Hg??? from where did you get that?
You wrote this
so far what I did:
10Hg2^2+ + 10e ==> 10Hg^2+
2MnO4^- + 16H^+ + 10e ==> Mn^2+ + 8H2O
The first one should be
10Hg2^2+ + 10e ==> 20Hg^2+
Can you see that NOW you have 20 Hg on the left and 20 on the right?
To write the balanced chemical equation for the electrochemical cell, we need to identify the half-reactions happening at each electrode.
First, let's look at the anode (left side of the cell, marked with "|"). Here, we have the half-reaction:
Pt(s) | Hg2^2+ (0.250M), Hg^2+ (0.800M)
From the given half-cell notation, we can see that the Hg2^2+ and Hg^2+ ions are present. The Pt(s) refers to a platinum electrode.
Hence, the anode half-reaction is the oxidation of Hg(s) to Hg2^2+ ions. This can be written as:
Hg(s) → Hg2^2+ + 2e^-
Next, let's focus on the cathode (right side of the cell, also marked with "|"). The half-reaction is:
MnO4^-(0.650M), H^+ (1.00M), Mn^2+ (0.375M) | Pt(s)
Here, MnO4^- and Mn^2+ ions are present along with H^+ ions, and Pt(s) as the electrode.
We can see that MnO4^- is being reduced to Mn^2+. To balance the equation, we need to add water (H2O) and hydrogen ions (H+) as needed. The balanced cathode half-reaction is:
8H^+ + MnO4^- + 5e^- → Mn^2+ + 4H2O
Finally, we can combine the half-reactions to form the overall balanced equation. To do this, we need to multiply the half-reactions by appropriate factors to ensure that the number of electrons canceled out:
2Hg(s) + MnO4^- + 8H^+ → Hg2^2+ + Mn^2+ + 4H2O
Therefore, the balanced chemical equation for the electrochemical cell is:
2Hg(s) + MnO4^- + 8H^+ → Hg2^2+ + Mn^2+ + 4H2O