Consider the following reactions: 3 Ni 2+(aq) + 2Cr(OH)3 (s) + 10 OH –(aq) → 3 Ni (s) + 2 CrO4 –2 (aq) + 8 H2O (l)

A) What is the value for n for this reactions?

B) Calculate the standard Gibbs Free-Energy

C) How long does it take a current of 5 A to deposit 95 g of nickel?

For the reaction

3 Ni2+(aq) + 2 Cr(OH)3(s) + 10 OH- (aq)  →  3 Ni(s) + 2 CrO42-(aq) + 8 H2O(l) G = +87 kJ/mol

Given:

(Ni2+(aq) + 2 e- →  Ni(s)) x 3 Ered = -0.28 V

(CrO42-(aq) + 4 H2O(l) + 3 e- →  Cr(OH)3(s) + 5 OH-(aq)) x 2

3 Ni2+(aq) + 6 e- →  3 Ni(s)

2 CrO42-(aq) + 8 H2O(l) + 6 e- →  2 Cr(OH)3(s) + 10 OH-(aq)

true

hope that

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