Write a balanced half-reaction for the reduction of aqueous nitrous acid hno2 to gaseous nitric oxide no in basic aqueous solution. be sure to add physical state symbols where appropriate.

The unbalanced chemical reaction of the transformation of aqueous nitrous acid to gaseous nitric oxide is as follows:

HNO₂ (aq) → NO (g)

In the above-mentioned equation, balancing is done by adding a OH,

HNO₂ (aq) → NO (g) + OH⁻ (aq)

Now, the balancing of the charge is done by adding the electrons on the left side,

HNO₂ (aq) + e⁻ → NO (g) + OH⁻ (aq)

Therefore, the balanced half-reaction in the basic medium is as follows:

HNO₂ (aq) + e⁻ = NO (g) + OH⁻ (aq)

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