Zn (s) , Zn(NO 3) 2 (1.0 M) || Ni (s) , Ni(NO 3 ) 2 (1.0 M) (Need help solving t
ID: 971489 • Letter: Z
Question
Zn(s), Zn(NO3)2 (1.0 M) || Ni(s), Ni(NO3)2 (1.0 M)
(Need help solving these questions, so I can solve the rest of the questions like this in my lab. It is greatly appreciated. Thank you)
2. Write the half reactions that occur at the anode and cathode in this electrochemical cell. Annotate which half reaction occurs at the anode and the cathode.
3. Write the overall balance reaction of this electrochemical cell.
4. Calculate E°cell of this electrochemical cell. (include units)
5. Calculate the reaction quotient (Q) of this reaction.
6. Calculate the expected Ecell for this reaction.
Explanation / Answer
Zn(s), Zn(NO3)2 (1.0 M) || Ni(s), Ni(NO3)2 (1.0 M)
2. Write the half reactions that occur at the anode and cathode in this electrochemical cell. Annotate which half reaction occurs at the anode and the cathode.
The anode = electrode in which the OXIDATION takes place, therefore, the reducing agent will get oxidized (lose electrons). The electrons flow from the solid electrode to the voltmeter flow. The solid species will then turn ionic, and go into solution. Since the solid converts to ion in solution, then this electrode will lose mass as time passes by.
The cathode = electrode in which REDUCTION takes place, therefore, the oxidizing agent will get reduced (gain electrons). The electrons are received from the voltmeter ( which come from the anode ). The ions in solution will then form solid over the cathode. Since there is solid formation, there will be an increase in the mass of this electrode.
Electrons, therefore, flow from oxidation cell to reduction cell; that is, from the anode to cathode.
Note that, as expected, there will be a charge overload if we kept this cell running as it is. Therefore, we must add a salt bridge, which allows ionic exchange, i.e. the charges will balance each other, positive ions and negative ions will flow in order to balance the cell’s charge.
Zn(s), Zn(NO3)2 (1.0 M) || Ni(s), Ni(NO3)2 (1.0 M)
Zn2+ + 2 e Zn(s) 0.7618
Ni2+ + 2 e Ni(s) 0.25
From the electrode potentias, the highest poential reduces
this is Nickel
Ni2+ + 2 e Ni(s) 0.25 --> cathode
Zn(s)Zn2+ + 2 e E = 0.7618 --> Ande
3. Write the overall balance reaction of this electrochemical cell.
Add all
Ni2+ + 2 e Ni(s)
Zn(s)Zn2+ + 2 e
Zn(s) + Ni2+ + 2 e Ni(s) + Zn2+ + 2 e
cancel common terms
Zn(s) + Ni2+ Ni(s) + Zn2+
this is the overall reaction; NO3- in remains as spectator ion
4. Calculate E°cell of this electrochemical cell. (include units)
E°Cell = Ecathode + Eanode
E°Cell = -0.25 + 0.7618 = 0.5118V
5. Calculate the reaction quotient (Q) of this reaction.
from
Zn(s) + Ni2+ Ni(s) + Zn2+
Q = [Zn+2]/[Ni2+]
since 1M is in both reacitons
Q = 1/1 = 1
6. Calculate the expected Ecell for this reaction.
When the cell is NOT under standard conditions, i.e. 1M of each reactants at T = 25°C and P = 1 atm; then we must use Nernst Equation.
The equation relates E°cell, number of electrons transferred, charge of 1 mol of electron to Faraday and finally, the Quotient retio between products/reactants
The Nernst Equation:
Ecell = E0cell - (RT/nF) x lnQ
In which:
Ecell = non-standard value
E° or E0cell or E°cell or EMF = Standard EMF: standard cell potential
R is the gas constant (8.3145 J/mol-K)
T is the absolute temperature = 298 K
n is the number of moles of electrons transferred by the cell's reaction
F is Faraday's constant = 96485.337 C/mol or typically 96500 C/mol
Q is the reaction quotient, where
Q = [C]^c * [D]^d / [A]^a*[B]^b
pure solids and pure liquids are not included. Also note that if we use partial pressure (for gases)
Q = P-A^a / (P-B)^b
substitute in Nernst Equation:
Ecell = E° - (RT/nF) x lnQ
Ecell = 0.5118V - (8.314*298)/(2*96500) * ln(1)
Ecell = 0.5118V
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