? sic.cengagenow.com Netflix OWLv2 | Online and lea Use the References to access

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? sic.cengagenow.com Netflix OWLv2 | Online and lea Use the References to access important values if needed for this question. The free energy change for the following reaction at 25 C, when [Zn)-6.59x10 M and [Cd*]- 1.13 M, is 822 kJ: Zn2(6.59x103 M) + Cd)Zn(s)+ Cd1.13 M) AG-822 What is the cell potential for the reaction as written under these conditions? Answer: Would this reaction be spontaneous in the forward or the reverse direction???B Retry Entire Group 2 more group attempts remaining Click to have your response(s) scored by OwL "Note that you will not be able to change your response(s) to this version of the item.

Explanation / Answer

Part a

The two half cell reactions are

Oxidation reaction

Cd = Cd2+ + 2e-

Eox = 0.40 V

Reduction reaction

Zn2+ + 2e- = Zn

Ered = - 0.76 V

Overall cell reaction

Zn2+ + Cd = Cd2+ + Zn

E°cell = Eox + Ered

= 0.40 - 0.76

= - 0.36 V

From the Nernst equation

E = E°cell - (0.0592/2) log [Cd2+/Zn2+]

= - 0.36 - (0.0592/2) log [1.13/6.59*10^-3]

= - 0.36 - 0.066

= - 0.426 V

Part b

Equilibrium constant

K = exp(-G°/RT) = exp (-82.2*1000/8.314*298)

= 3.90 x 10^-15

Reaction Quotient

Q = [Cd2+/Zn2+] = [1.13/6.59*10^-3]

= 171.47

Q > K

The reaction is spontaneous in reverse direction

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