Nernst

Nernst equation for

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

• E_Cell=E⁰_Cell − 0.0592/6 log [Mg²⁺]³/ [Al³⁺]²

• E_Cell=E⁰_Cell − 0.0592/6 log [Al³⁺]²/ [Mg²⁺]²

• E_Cell=E⁰_Cell +0.0591/3 log[Mg²⁺]³/[Al³⁺]²

• E_Cell=E⁰_Cell −0.0591/3 log[Mg²⁺] /[Al³⁺]

Answer explanation

Tags

NGSS.HS-PS1-5

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

• Increase temperature

• Decrease concentration Ni²⁺

• Increase concentration Ni²⁺

• Choose E^ocell anode and cathode the positive value

Answer explanation

Increasing reactant concentration will decrease the Log value. Thus, E^o_cell will be subtracting a smaller value and E_cell value will be higher.

Caculate the E_cell for the following galvanic cell.

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

E_Ag⁺_{/ Ag} = 1.08V and E_Zn²⁺_{/ Zn} = -0.76V

• 0.32 V

• 0.33 V

• 1.84 V

• 1.85 V

Answer explanation

Tags

NGSS.HS-PS1-4

NGSS.HS-PS3-1

NGSS.HS-PS3-5

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

Pb²⁺ + Cd → Pb + Cd²⁺ .

Given that E^o_cell = 0.277 V,

[Cd²⁺] = 0.02M, and [Pb²⁺] = 0.2M.

• 0.207 V

• 0.255 V

• 0.307 V

• 0.355 V

Answer explanation

E_cell = E^o_cell – (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:

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

= 0.277 – (0.0296)(-1) = 0.3066 Volts

Tags

NGSS.HS-PS3-1

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

(give answer in 3 decimal place)

Answer explanation

E^o_cell = E^o_{cathode -} E^o_anode

= 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.

E_cell = E^o_cell – (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