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16. Calculate AG in wate agin) (2 pts) and (2 pts) and (3 points) points for the following reaction at 290 K. Mg(s) + Fe(aq)
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Q16. Given : Standard cell potential, Eocell = 1.84 V

\small \DeltaGo = -(n) * (F) * (Eocell)

where

n = number of electrons transferred in reaction = 2

F = Faraday's constant = 96485 J/V-mol

Substituting the values,

\small \DeltaGo = -(2) * (96485 J/V-mol) * (1.84 V)

\small \DeltaGo = -355064.8 J

ln(K) = -\small \DeltaGo / (R) * (T)

where

R = constant = 8.314 J/mol-K

T = temperature = 298 K

ln(K) = -(355064.8 J) / (8.314 J/mol-K) * (298 K)

ln(K) = 143.3

K = e143.3

K = 1.74 x 1062

Q17. The balanced reduction half reaction is : MnO4- (aq) + 4 H+ (aq) + 3 e-\small \rightarrow MnO2 (aq) + 2 H2O (l)

The balanced oxidation half reaction is : Br- (aq) + 3 H2O (l) \small \rightarrow BrO3- (aq) + 6 H+ (aq) + 6 e-

The overall balanced equation is : 2 MnO4- (aq) + Br- (aq) + 2 H+ (aq) \small \rightarrow 2 MnO2 (aq) + BrO3- (aq) + H2O (l)

Q18. (a) The balanced chemical equation is : 2 Al (s) + 3 Ni2+ (aq) \small \rightarrow 2 Al3+ (aq) + 3 Ni (s)

(b) 6 (six) moles of electrons were transferred

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