Question

1) Assuming a temperature of 25 degree celsius, and a Cu2+ concentration of 0.11 M, perform the first calculation in the experiment (Eo fir Cu = 0.340 V)

2) Calculate the value of Eo cell for a galvanic cell based on the following half-reaction (See picture attached below)

Question 1 0.4 / 2 points Assuming a temperature of 25 °C, and a Cu2+ concentration of 0.11 M, perform the first calculation in the Calculations section of this experiment (E for Cu = 0.340 V). Answer: 0.331 V Question 2 0/2 points Calculate the value of Ecell for a galvanic cell based on the following half reactions. Be sure to report your answer with the correct significant figures and units. Include a space between the value and unit. Ni2+(aq) + 2e - Ni(s) Cr3+ (aq) + 3e - Cr(s) E (V) -0.257 -0.744 Answer: 0.484 v

Answer 1

1- The given cell has Cu+2 with concentration =  0.11 M

Now Cu+2 can be collected from Cu from the reaction-

Cu ------------> Cu+2 + 2e

Now in this reaction, Cu has lost 2 electrons to get converted to Cu+2. Thus Oxidation (i.e loss of electron) take place here.

According to Nernst equation, the E_cell for an Oxidation half cell is

E_cell = E⁰_cell - 0.0591/n log [Oxd] where n = number of electrons involved

Now putting the given values-

E_cell = E⁰_cell - 0.0591/n log [Oxd]

E_cell = 0.340 V - 0.0591/2 * log [0.11]

= 0.340 V - 0.0591/2 * (-0.958)

= 0.340 V + 0.028

= 0.368‬ V

2-

Here the given two half cells are-

Ni+2 + 2e ----------> Ni E⁰ = -0.257 V

Cr+3 + 3e ---------> Cr   E⁰ = -0.744 V

Now the rule says, the half cell where the E⁰red values is more is the actual reduction half cell and the other is the oxidation half cell.

Thus here the

Oxidation half cell Cr ---------> Cr+3 + 3e E⁰ = -0.744 V

Reduction half cell Ni+2 + 2e ----------> Ni   E⁰ = -0.257 V

Now to balance the electrons, we have to multiply 2 in eqn 1 and 3 in eqn 2. So that

Oxidation half cell 2Cr ---------> 2Cr⁺³ + 6e E⁰ = -0.744 * 2 = - 1.488 V

Reduction half cell 3Ni⁺² + 6e ----------> 3Ni E⁰ = -0.257 * 3 = - 0.771 V

Overall reaction   2Cr + 3Ni⁺² -------------> 2Cr⁺3 + 3Ni E⁰cell = E⁰red - E⁰oxd

= - 0.771 V - (- 1.488 V)

= 0.717‬ V