I looked up the reduction potentials.
Ag ==> Ag^+ + e Eo = +0.799
H2O2 + 2H^+ + 2e ==>2H2O Eo = +1.77
You need to confirm therse. My table was small and I may have read the wrong columns.
To make the cell spontaneous, we use H2O2 as written and reverse the Ag couple so
Ecell = +1.77 + (-0.8) = ??
But you don't need to do that for this problem. I put it there to help you understand.
Look at individual Eo values.
For the Ag couple,
E as a reduction is
Eo-(0.0592/n)*log (red/ox) = Eo- (0.0592/n)*log(Ag)/(Ag^+) =
0.799 -(0.0592/n)*log(1/1)= 0.799
(Note: 1 for Ag since that is the elemental state and 1 for Ag^+ from the problem.) The point of this is to show that log 1/1 = 0 so E = Eo; that is, using Ag^+ as 1.0 M does not change EAg couple so it stays as is. You don't need to do the calculation to see that.
Now what happens to the H2O2 couple.
E=Eo-(0.0592/n)*log(red/ox)
E = ........*log[(H2O)^2/(H2O2)(H^+)^2]=
....,
Whatever the value of water is is a constant so let's just call it 1. H2O2 is 2 M from the problem and H^+ is 2 M from the problem. So the denominator is increased by 2*2^2=8
log 1/8 is a negative number, minus number x minus sign from the -0.0592/n gives a + so 1.77 + some number will be a higher positive number.
So when we add a larger number than 1.77 to the same value for Ag, we will get a larger value for Ecell. Again, you don't need to go through the calculations to see that. Since the problem state to "predict" I assume you are NOT supposed to go through the calculation. Check my work. I hope this helps.
A galvanic cell is based on the following half-reactions at 25C:
Ag+ + e- --> Ag
H2O2 + 2H+ + 2e- ---> 2H2O (not H2O2)
Predict whether E_cell is larger or smaller than E°_cell for the following case.
[Ag+]= 1.0M, [H2O2]= 2.0 M, [H+]=2.0M
======================================
To make E_cell smaller than E°, log(Q) has to be negative in the equation E = E° - (0.0591)(log(Q))/n, I think.
But I have no clue which species to use on the calculation!
Please help.
Thanks!
2 answers
Thank you so much! I was trying to work with wrong E_cell
1 answer