If the [Zn^2] = 2.0 M and [Al^3+] = 0.50 M, this would be a non-standard cell, U

Question

If the [Zn^2] = 2.0 M and [Al^3+] = 0.50 M, this would be a non-standard cell, Using the relationships Delta G = - nF xi and Delta G degree = -n F xi degree we can come up with an equation to calculate non-standard cell potentials, the Nernst equation. xi = xi degree - RT/nF ln Q where n = # moles of electrons in the balanced half-reactions F = faradays constant = 6,000 C/mol e Q = reaction quotient, [products]^/[reactants] Calculate non-standard cell potential (xi) at 298 K when the [Zn^2+] = 2.0 M and [Al^] = 0.50 M. a) Q _____ b) xi degree = _____ c) n = _____ d) xi = _____ Calculate Delta G and Delta G degree and discuss the difference in the work being done, in terms of [products] and [reactants]. a) Delta G degree = _____ b) Delta G = _____ What is the effect on xi of doubling the size of the aluminum electrode? What is the effect on xi of adding 500 mL of water to the cathode, assuming there was 1 L of the solution to begin with?

Explanation / Answer

the reaction:

Al3+(aq) + 3e– Al(s) –1.66

Zn2+(aq) + 2e– Zn(s) –0.76

Balance

3Zn2+ + 2Al(s) -- > 2Al3+ + 3Zn(s)

c)

therefore, there are 3x2 = 6 electrons of moles being transferred

b)

E º = Ered - Eox = -0.76 -- 1.66 = 0.9 V

now

Q = [Al3+]^2 / [Zn+2 ]^3

a)

Q = (0.5^2)/ (2^3) = 0.03125

d)

E = E º - 0.0592/(n) log(Q)

E = 0.90 - 0.0592/(6) * log(0.03125) = 0.91485 V

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