I am on Step 4 on pg. 8 of the Galvanic cells activity. I used E(nonstandard) =

Question

I am on Step 4 on pg. 8 of the Galvanic cells activity. I used E(nonstandard) = Ecell (standard) - 0.0592/2 log Q

For the E cell nonstandard I used the voltage found when the solution was diluted in either the anode or the cathode and for the Ecell standard I used the voltage found when the cathode and anode were both undiluted and at 1M concentration.

Reaction 1: Sn(s) + Cu2+(aq) --> Sn2+(aq) + Cu(s) 0.48V UNDILUTED VOLTAGE

Reaction 2: Zn(s) + Sn2+(aq) --> Zn2+(aq) + Sn(s) 0.62V UNDILUTED VOLTAGE

Reaction 1: diluted voltage = 0.51V and Reaction 2 diluted voltage = 0.62V

Reaction 1: Sn(s) + Cu2+(aq) --> Sn2+(aq) + Cu(s)

Reaction 2: Zn(s) + Sn2+(aq) --> Zn2+(aq) + Sn(s)   Both of these reactions have an n=2.

This is where my problem arises.. I put it into my solve button as exactly this for reaction 1:

solve(.51=.48-0.0592/2*Log(x/1) solve for x. x=0.0969

For reaction 2 it looks similar but the voltages are different

solve(.59=.62-0.0592/2logQ1./x) , x x= 0.0969

I don't understand how I am getting the same concentration if the voltages are different!

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Explanation / Answer

ANSWER =

DIFFERENCE BETWEEN THEM CONSIDER ,NOT QUANTITY, SO IF SEE DEEFERENCE= 0.51-0.48 =0.03 AND

ANOTHER ONE IS 0.59-0.62=-0.03 NEGATIVE SIGN LAPS WHEN COCENTRATION FACTOR BECOME REVERSE AS Q USE =1/X ,NOT X/1 SO OVERALL IT BECOME SAME. NO PROBLEM

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