consider the following electrochemical cell:

Question

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.

Answer 1

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.

Answer 2

What should I multiply each reaction by to make it balanced?

Answer 3

Hg2^2+ + 2e ==> 2Hg^2+ multipy by 5
MnO4^- + 8H^+ + 5e ==> Mn^2+ + 4H2O multiply by 2

10 electrons exchanged then.

Answer 4

How would the final equation turn out? would those 10 electrons cancel out?

Answer 5

so far what I did:
10Hg2^2+ + 10e ==> 10Hg^2+
2MnO4^- + 16H^+ + 10e ==> Mn^2+ + 8H2O

Answer 6

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.

Answer 7

20 Hg??? from where did you get that?

Answer 8

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?