Question

Use standard reduction potentials to calculate the equilibrium constant for the reaction: 2Cu-+ (aq) + Cd(s) +2Cu+ (aq) + Cd + (aq) Hint: Carry at least 5 significant figures during intermediate calculations to avoid round off error when taking the antilogarithm Equilibrium constant AGº for this reaction would be than zero Submit Answer Retry Entire Group 7 more group attempts remaining

Answer 1

Cd(s) ----------------> Cd^2+ (aq) + 2e^-              E0   = 0.40v

2Cu^2+ (aq) + 2e^- ------> 2Cu(s)                       E0   = 0.16v

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Cd(s) + 2Cu^2+(aq) ------. Cd^2+ (aq) + 2Cu(s)     E0cell = 0.56v

n = 2

$\Delta$G⁰    = -nE0cell*F

             = -2*0.56*96500

              = -108080J

$\Delta$G⁰      = -RTlnK

-108080   = -8.314*298lnK

lnK   = -108080/(-8.314*298)

lnK   = 43.62

K     = 8.8*10^18 >>>answer

$\Delta$G⁰ for this reaction would be less than zero

b.

   Hg(l) ------------> Hg^2+ (aq) + 2e^-             E0   = -0.85g

2Cu^2+ (aq) + 2e^- ------> 2Cu(s)                  E0   = 0.16v

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Hg(l) + 2Cu^2+(aq) -----> Hg^2+ (aq) + 2Cu(s)     E0cell = -0.69v

n = 2

$\Delta$G⁰    = -nE0cell*F

             = -2*-0.69*96500

              = 133170J

$\Delta$G⁰      = -RTlnK

133170 = -8.314*298lnK

lnK   = 133170/(-8.314*298)

lnK   = -53.75

K     = 4.5*10^-24 >>>answer

$\Delta$G⁰ for this reaction would be greater than zero