What is the value of Q and what is the concentration of the Ni2+ ions?

Question

You have a battery composed of the following reaction:

Ni(s) + 2 Ag+(aq) → Ni2+(aq) + 2 Ag(s), Ecellº=1.03 V, n = 2 e- 

If you want to make a cell with an operating potential of 1.05 V, what is the value of Q for this reaction? 

Use the Nernst Equation.

Give your answer to 3 decimal places.

You're using the same battery as above:

Ni(s) + 2 Ag⁺(aq) → Ni²⁺(aq) + 2 Ag(s)

 

To get your desired E_cell, you've determined that Q must be 0.818. If the concentration of silver ions is 1.44 M, what should the concentration of Ni²⁺ ions be? 

Give your answer to 2 decimal places.

Answer 1

Nernst equation

Ecell = Eºcell - RT/nF ln Q

Again, as in your previous question, no temperature is given so we will assume 298K and reduce the Nernst to

Ecell = Eºcell - 0.0592/n log Q

Ecell = 1.05 V

Eºcell = 1.03 V

n = 2

Q = [Ni²⁺] / [Ag⁺]² = ?

1.05 = 1.03 - 0.0296 log Q

log Q = -0.675

Q = 0.211

(be sure the check the math)

If Q = 0.818

Q = [Ni²⁺] / [Ag⁺]²

0.818 = [Ni²⁺] / (1.44)² = [Ni²⁺] / 2.0736

1.70 = [Ni²⁺]

(be sure the check the math)