Using the standard reduction potentials listed in appendix e, calculate the equilibrium constant for the reaction at 298 k: 3 ce4+ (aq) + bi (s) + h2o (l) → 3 ce3+ (aq) + bio+ (aq) + 2 h+ (aq)

2.9×10^65

Explanation:

E°cell= E°cathode- E°anode

E°cell= 1.61 V- 0.32 V= 1.29V

The anode is oxidized while the cathode is reduced hence Bi/BiO+ is the anode while Ce4+/Ce3+ is the cathode

Recall that

R=8.314JK-1

F= 96500 C

n=3 moles of electrons

T=298K

E°cell= RT/nF ln K

lnK= E°cell × nF/RT

ln K = 1.29 × 3×96500/8.314×298

ln K= 150.7

K= 2.9×10^65

when 25.0 ml of 0.700 mol/l naoh was mixed in a calorimeter with 25.0 ml of 0.700 mol/l hcl, both initially at 20.0 °c, the temperature increased to 22.1 °c. the heat capacity of the calorimeter is 279 j/°c.

the equation for the reaction is:

naoh + hcl → nacl + h₂o

moles of hcl = 0.0250 l hcl ×

0.700

mol hcl

1 l hcl = 0.0175 mol hcl

volume of solution = (25.0 + 25.0) ml = 50.0 ml

mass of solution = 50.0 ml soln ×

1.00

ml soln

= 50.0 g soln

δt=t2-t1

= (22.1 – 20.0) °c = 2.1 °c

the heats involved are

heat from neutralization + heat to warm solution + heat to warm calorimeter = 0

q1+q2+q3=0

nδh+mcδt+cδt=0

0.0175 mol ×

δh

+ 50.0 g × 4.184 j·g⁻¹°c⁻¹ × 2.1 °c + 279 j°c⁻¹ × 2.1 °c = 0

0.0175 mol ×

δh

+ 439.32 j + 585.9 j = 0

0.0175 mol ×

δh

= -1025.22 j

δh=1025.22j0.0175 mol= -58 600 j/mol = -58.6 kj/mol