Chemistry, 21.04.2020 22:35 tiniecisneros28

A concentration cell based on the following half reaction at 289 K has initial concentrations of 1.39 M , 0.312 M , and a potential of 0.037206 V at these conditions. After 9.1 hours, the new potential of the cell is found to be 0.0095376 V. What is the concentration of at the cathode at this new potential

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Answer from: ariellake8551

The concentration at the cathode is A_1 = 0.8183M

Explanation:

From the question we are told that

The initial concentration is $1.39 M \ for \ Zn^{2+} \ at \ anode \ , 0.312 M \ for \ Zn^{2+} \ at \ cathode$

The reaction is

At Cathode $Zn^{2+} + 2 e^- ------ Zn$

At anode $Zn ----- Zn ^{2+} + 2e^{-}$

The initial potential is $+ 0.037206V \ for \ Cathode \ and \ + 0.037206V \ for \ anode$

The complete reaction is

$Zn^{2+} + Zn_{(s)} ---- Zn^{2+} + Zn_{(s)}$

(Cathode) ----------------- (anode)

Looking the reaction above we can see that after the reaction $Zn^{2+}$ at cathode loss some concentration and $Zn^{2+}$ will gain some concentration

Let say the amount of concentration lost/gained is = z M

The after the reaction the concentration of $Zn^{2+}$ at the cathode would be

A_1 = 1.39 -z

While the concentration at the anode would be

A_2 = 0.312 +z

At initial the condition the net potential of the cell is

$E^i_{net} = 0V$

After time t = 9.1 hrs = 9.1 *3600 = 32760s

The potential of the cell is

$E_{net} = 0.0095376V$

Generally this potential is defined by Nernst equation as

$E_{cell} = E^i_{cell}- \frac{RT}{nF} ln[\frac{[A_2]}{[A_2]} ]$

Where R is the gas constant with a value of $R = 8.314 J/mol \cdot K$

T is the temperature given as T = 289K

n is the number of mole of electron which $= 2 mol \ e^-$

F is the farad constant with a value of $= 96500 C /mol\ e^-$

Substituting values

$0.0095376 =0.0 -\frac{[8.314][288]}{[2] [9600]} ln [\frac{(0.312 +z)}{(1.39 -z)} ]$

$0.0095376 = 0.12406383 \ ln [\frac{( 0.312 + z)}{[1.39 -z]} ]$

$\frac{0.0095376 }{ 0.12406383} = \ ln [\frac{( 0.312 + z)}{[1.39 -z]} ]$

$0.0768766 = \ ln [\frac{( 0.312 + z)}{[1.39 -z]} ]$

Taking exponent of both sides

$e^{0.0768766 }= [\frac{( 0.312 + z)}{[1.39 -z]} ]$

$1.0799 = [\frac{( 0.312 + z)}{[1.39 -z]} ]$

$[1.39 -z ]1.0799 = 0.312 + z \\\\1.501 - 1.0799z = 0.312 +z\\\\1.501-0.312 = 1.0799z + z\\\\1.1891 =2.0799z\\\\ z =\frac{1.1891}{2.0799}\\\\ z =0.5717$

The concentration at the cathode is A_1 = 1.39 -0.5717

A_1 = 0.8183M

Answer from: Quest

solute : a solute is a substance that is dissolved in another substance is known as a solvent. it is mixed with a special mixture know as homogeneous mixture.

solvent: a solvent is a substance that dissolved a solute, which results in a solution is known as solvent.

Answer from: Quest

answer; 4 electrons;

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A concentration cell based on the following half reaction at 289 K has initial concentrations of 1.3...

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