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.

pure water does not conduct electricity very well. however, when certain substances are dissolved in water, the solution does conduct electricity. you can make a simple device that shows how well a solution conducts electricity. this device uses a flashlight bulb to indicate how well the solution conducts electricity. the better the solution conducts electricity, the brighter the bulb will glow.

explanation:

i think it's c. i'm not sure but i think so.