Consider the following reaction at 298K. 2 H+(aq) + 2 Cr2+(aq) =H2(g) + 2 Cr3+(aq)
Which of the following statements are correct?
a) K > 1
b) ΔGo < 0
c) Eocell < 0
d) n = 2 mol electrons

The question is incomplete, the complete question is;

Consider the following reaction at 298K.

2 H+(aq) + 2 Cr2+(aq) =H2(g) + 2 Cr3+(aq)

Which of the following statements are correct?

Choose all that apply.

K > 1

ΔGo < 0

Eocell < 0

n = 2 mol electrons

The reaction is reactant-favored.

K > 1

ΔGo < 0

n = 2 mol electrons

Explanation:

If we look at the reaction; 2 H+(aq) + 2 Cr2+(aq) > H2(g) + 2 Cr3+(aq), we will notice that the reaction is a spontaneous reaction because E°cell= 0-(-0.41) = 0.41 V

If the reaction is a spontaneous electrochemical process, we expect it to be product favoured with K>1. Also, a spontaneous reaction must have ∆G<0.

Lastly, we can see that two electrons were transferred according to the balanced equation for the electrochemical reaction.

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

answer: they produce more slow than unicellular organisms.