Question

Mn(3+) + e --> Mn(2+) E=+1.51V

Br2(l) +2e --> 2Br(-) E=+1.09V

a. Write a balanced overall reaction so that the rxn is spontaneous in the right direction, and calculate Ecell.

b. Caclculate K for the rxn

c. Calculated Ecell fro the conditions: [Mn(3+)]=1.0x10^-2M, [Mn(2+)]=2.0x10^-2M, [Br-]=1.0x10^-3M

Answer 1

a.

balanced reaction

Mn reduces, since it has the higher potential

2Mn+3 + 2Br- --> Br2 + 2 Mn+2

b.

E°cell = Ered - Eox = 1.51 - 1.09 = 0.42 V

then

RT*lnK = nF*E°cell

K = exp((nF)/(RT)*E°cell)

K = exp(2*96500/(8.314*298)*0.42)

K = 1.618*10^14

c.

apply nerst equation

Ecell = E°cell - 0.0592/n*log(Q)

Q = [Mn+2]^2/ [Br-2]^2[Mn+3]^2

Q = (2*10^-2)^2 / ((10^-3)^2)* ((10^-2)^2)) = 0.04

Ecell = E°cell - 0.0592/n*log(Q) = 0.42 - 0.0592/2*log(0.04)

Ecell = 0.4613 V