B. The following reaction takes place in a basic solution. (7 points) MnO4–(aq) + NO2–(aq) MnO2(s) + NO3–(aq)
The half-reactions (balanced only for atoms) are the following:
MnO4– + 2H2O MnO2 + 4OH–
NO2– + 2OH– NO3– + H2O
Use the following steps to finish balancing the equation.
i. Balance each half-reaction for charge. (2 points)
ii. Multiply each half-reaction by the correct number in order to balance charges for the two half-reactions. (2 points)
iii. Add the equations and simplify to get a balanced equation. (2 points)
iv. How can you tell from this equation that the reaction occurred in a basic solution? (1 point)

The final equation has hydroxide ions which indicate that the reaction has occurred in a basic medium.

Explanation:

Redox reaction is defined as the reaction in which oxidation and reduction take place simultaneously.

The oxidation reaction is defined as the reaction in which a chemical species loses electrons in a chemical reaction. It occurs when the oxidation number of a species increases.

A reduction reaction is defined as the reaction in which a chemical species gains electrons in a chemical reaction. It occurs when the oxidation number of a species decreases.

The given redox reaction follows:

To balance the given redox reaction in basic medium, there are few steps to be followed:

Oxidation half-reaction:

Reduction half-reaction:

Oxidation half-reaction:         ( × 3)

Reduction half-reaction:             ( × 2)

The half-reactions now become:

Oxidation half-reaction:

Reduction half-reaction:

Overall redox reaction:

As we can see that in the overall redox reaction, hydroxide ions are released in the solution. Thus, making it a basic solution

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

the ratio of the conjugate base to the acid required to attain the desired ph.