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

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.

possibly, yes.

explanation:

the amount of sugar you take in is dangerously high, risk of heart disease, high risk of diabetes, and etc.

the ideal gas law equation is as follows

pv = nrt

where p - pressure

v - volume

n - number of moles

r - universal gas constant

t - temperature

so if the pressure, volume and temperature are already known

we are left with n and r

since r is the universal gas constant that has a known fixed value then r too is known

so we are left with 'n'

once we know temperature volume and pressure

we can find the number of moles of gas present using the ideal gas law equation

this is not my own answer, this is an answer for the same question by the user blahblahmali. i do not claim ownership of this account.