Question

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) + Mg(aq) + Fe(s) - 1.84 V AGE ANT 8314 K-mol F-96,485 )/V-mol K R 17. Balance the following redox reaction in acidic aqueous solution. Mno. (a) + Br" (a) → MnO2 (s) + Broj (na) 18. Consider the following voltaic cell designation: Al(s) AP+(aq) || Ni+(aq) Ni(s) a. (3 points): Write a balanced chemical equation for the overall reaction in this voltaic cell. C. (2 points): How many moles of electrons were transferred during the overall reaction?

Answer 1

Q16. Given : Standard cell potential, E^o_cell = 1.84 V

$\small \Delta$G^o = -(n) * (F) * (E^o_cell)

where

n = number of electrons transferred in reaction = 2

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

Substituting the values,

$\small \Delta$G^o = -(2) * (96485 J/V-mol) * (1.84 V)

$\small \Delta$G^o = -355064.8 J

ln(K) = -$\small \Delta$G^o / (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 = e^143.3

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