1.) What is the correct net cell reaction for the electrochemical cell diagramed below?

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.