Nernst Equation

Nernst Equation

10th Grade - University

5 Qs

quiz-placeholder

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Nernst Equation

Nernst Equation

Assessment

Quiz

Chemistry

10th Grade - University

Practice Problem

Hard

NGSS
HS-PS1-4, HS-PS3-1, HS-PS1-5

Standards-aligned

Created by

Charles Martinez

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5 questions

Show all answers

1.

MULTIPLE CHOICE QUESTION

5 mins • 1 pt

Nernst equation for

Mg(s) | Mg2+(0.001M) || Al3+(0.001 M) | Al(s)

ECell=E0Cell − 0.0592/6 log [Mg2+]3/ [Al3+]2

ECell=E0Cell − 0.0592/6 log [Al3+]2/ [Mg2+]2

ECell=E0Cell +0.0591/3 log[Mg2+]3/[Al3+]2

ECell=E0Cell −0.0591/3 log[Mg2+] /[Al3+]

Answer explanation

Media Image

2.

MULTIPLE CHOICE QUESTION

5 mins • 1 pt

Media Image

How to increase cell potential, Ecell based on this Nernst equation:

Increase temperature

Decrease concentration Ni2+

Increase concentration Ni2+

Choose Eocell anode and cathode the positive value

Answer explanation

Increasing reactant concentration will decrease the Log value. Thus, Eocell will be subtracting a smaller value and Ecell value will be higher.

Tags

NGSS.HS-PS1-4

NGSS.HS-PS1-5

3.

MULTIPLE CHOICE QUESTION

5 mins • 1 pt

Caculate the Ecell for the following galvanic cell.


Zn(s)|Zn2+ (aq, 1.0M) || Ag+ (aq,1.5M)|Ag(s)


EAg+/ Ag = 1.08V and EZn2+/ Zn = -0.76V

0.32 V

0.33 V

1.84 V

1.85 V

Answer explanation

Media Image

Tags

NGSS.HS-PS1-4

NGSS.HS-PS3-1

4.

MULTIPLE CHOICE QUESTION

5 mins • 1 pt

Calculate the cell potential, Ecell of the electrochemical cell in with reaction

Pb2+ + Cd → Pb + Cd2+ .


Given that Eocell = 0.277 V,

[Cd2+] = 0.02M, and [Pb2+] = 0.2M.

0.207 V

0.255 V

0.307 V

0.355 V

Answer explanation

Ecell = Eocell – (0.0592/n) log10 [Cd2+]/[Pb2+]


Here, two moles of electrons are transferred in the reaction. Therefore, n = 2.


[Cd2+]/[Pb2+] = (0.02M)/(0.2M) = 0.1


The equation can now be rewritten as:

Ecell = 0.277 – (0.0592/2) × log10(0.1)

= 0.277 – (0.0296)(-1) = 0.3066 Volts

5.

FILL IN THE BLANK QUESTION

5 mins • 1 pt

If Eo(Ag+/Ag) = 0.8V, Eo(Cu2+/Cu) = 0.34V, and Cell potential, Ecell(at 25oC) = 0.422V, find the silver ion, Ag+ concentration. Given [Cu2+] = 0.1M

(give answer in 3 decimal place)

Answer explanation

Eocell = Eocathode - Eoanode

= 0.8V – 0.34V = 0.46V


Since the charge on the copper ion is +2 and the charge on the silver ion is +1, the balanced cell reaction is:

2Ag+ + Cu → 2Ag + Cu2+

Since two electrons are transferred in the cell reaction, n = 2.


Ecell = Eocell – (0.0592/2) × log(0.1/[Ag+]2)

0.422= 0.46 -[0.0296 x log(0.1/[Ag+]2)]


log[Ag+] = -1.141

[Ag+] = antilog(-1.141) = 0.0722 M

Tags

NGSS.HS-PS3-1

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