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Chemistry

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!

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Andrew

  1. 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.

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    DrBob222

  2. Thank you so much! I was trying to work with wrong E_cell

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    Andrew

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