# calculate the E cell for this reaction 3Cu(s) + 8H+(aq) + 2NO3-(aq) ---> 2NO(g) + 3Cu2+(aq) + 4H20(l)?

calculate the E cell for this reaction 3Cu(s) + 8H+(aq) + 2NO3-(aq) ---> 2NO(g) + 3Cu2+(aq) + 4H20(l)?

And calculate E at 298K when [H+]=0.100M, [NO3-]=0.0250M [Cu2+]=0.0375M & the partial pressure of NO= 0.00150 atm

the E cell for this reaction is +0.62V and the E at 298K is +0.61V but I do not know how they got those answers.

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Half reactions are:

Oxidation: Cu ----> Cu2+ + 2e Eo = -0.34V

Reduction: NO3- + 4H+ + 3e ---> NO + 2H2O Eo = +0.96V

Add the two voltages (do not multiply this number by the coefficients you use to cancel the number of electrons out) and you get the +0.62V for the Eo of the cell.

The Nernst equation will take this form:

E = Eo - 0.0592/n * log {(PNO)^2 * [Cu2+]^3 / [NO3-]^2 * [H+]^8}

n will equal 6 here, not 8, as that will be the number of electrons that will be cancelled out when you add the oxidation and reduction reactions together (multiplied by the correct number to cancel those electrons out).

The concentration's portion simplifies, once you plug all of the numbers in, to about 18.94, whose log is 1.278. Multiply that by 0.0592 and divide by 6 and you get 0.01, so only 0.01 V will be lost from the standard potential for that reaction at those concentrations.

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use the Nernst Equation.IF u dont know the equation reffer a book for the equation.

I got a perfect answer of 0.61 V.

In the equation

n=8

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