1.) What is the correct net cell reaction for the electrochemical cell diagramed below?
Ni(s) ∣ Ni2+(aq) || Ag+(aq) ∣ Ag(s)
a.) Ag+(aq) + Ni(s) ---> Ag(s) + Ni2+(aq)
b.) Ag(s) + Ni2+(aq) ---> Ag+(aq) + Ni(s)
c.) 2Ag+(aq) + Ni(s) ---> 2 Ag(s) + Ni2+(aq)
d.) Ag(s) + 2 Ni2+(aq) ---> Ag+(aq) + 2 Ni(s)
e.) Ag+(aq) + 2 Ni(s) ---> Ag(s) + 2 Ni2+(aq)
Look up the reductions potentials and see which half cell reacts.
To determine the correct net cell reaction for the electrochemical cell, we need to identify the half-reactions occurring at each electrode and check for any necessary balancing.
In this case, looking at the given cell diagram:
Ni(s) ∣ Ni2+(aq) || Ag+(aq) ∣ Ag(s)
The left half-cell consists of a solid nickel electrode (Ni(s)) and a nickel(II) ion solution (Ni2+(aq)). The right half-cell consists of a silver ion solution (Ag+(aq)) and a solid silver electrode (Ag(s)).
To find the net cell reaction, we can start by writing the oxidation half-reaction and the reduction half-reaction for each electrode.
At the left electrode (anode), nickel undergoes oxidation:
Ni(s) → Ni2+(aq) + 2e-
At the right electrode (cathode), silver undergoes reduction:
Ag+(aq) + e- → Ag(s)
Notice that the number of electrons involved in each half-reaction is not equal, which means we need to balance the number of electrons transferred. To do this, we multiply the oxidation half-reaction by 2:
2Ni(s) → 2Ni2+(aq) + 4e-
Now we can combine the half-reactions and cancel out the electrons to obtain the net cell reaction:
2Ni(s) + 4Ag+(aq) → 2Ni2+(aq) + 4Ag(s)
However, we can further simplify this equation by dividing all the coefficients by 2:
Ni(s) + 2Ag+(aq) → Ni2+(aq) + 2Ag(s)
This matches option d.) Ag(s) + 2 Ni2+(aq) ⟶ Ag+(aq) + 2 Ni(s), which is the correct net cell reaction for the given electrochemical cell.