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crimeas [40]
2 years ago
14

Consider two electrochemical reactions: A and B. Reaction A results in the transfer of 2 mol of electrons per mole of reactant a

nd generates a current of 5 Amps on an electrode 2 cm2 in area. Reaction B results in the transfer of 3 mol of electrons per mole of reactant and generates a current of 15 Amps on an electrode having 5 cm2 in area. What are the net reaction rates for reactions A and B (in mol reactant per square centimeter per second)
Chemistry
1 answer:
soldier1979 [14.2K]2 years ago
3 0

Answer:

See explaination

Explanation:

The rate of the electrochemical reaction per unit area is given by

r=\frac{i}{nFA}

where r is the Reaction Rate (in mol s-1 cm-2)

i is the Current Generated (in A)

n is the Number of moles of electrons per mole of reactant

F is Faraday's Constant= 96,500 C mol-1

A is the Electrode Area (in cm2)

(Unit of current should be expressed in terms of Columb per second, C s-1 to obtain reaction rate in mol s-1 cm-2)

Net Reaction Rate for Reaction A:

Given: Number of moles of A, n=2 moles

Current generated, i=5 A=5 C s-1

Electrode Area, A= 2cm2

Thus reaction rate is

\mathbf{r=\frac{5 C s^{-1}}{2*96500 C mol^{-1} *2 cm^{2}}= 1.3*10^{-5} mol s^{-1} cm^{-2}}

Net Reaction Rate for Reaction B:

Given: Number of moles of B, n=3 moles

Current generated, i=15 A=15 C s-1

Electrode Area, A= 5cm2

Thus reaction rate is

\mathbf{r=\frac{15 C s^{-1}}{3*96500 C mol^{-1}*5cm^{2}}= 1.036*10^{-5} mol s^{-1} cm^{-2}}

Reaction A has higher net reaction rate.

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A 25.0-mL sample of 0.150 M hydrazoic acid, HN3, is titrated with a 0.150 M NaOH solution. What is the pH after 13.3 mL of base
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pH ≅ 4.80

Explanation:

Given that:

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number of moles of HN₃ =  0.00375  mol

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the volume of NaOH = 13.3 mL = 0.0133

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