1. For a galvanic cell consisting of a nickel electrode in contact with a solution of N+i ²(aq )  ions, and a silver electrode in contact with a solution of Ag+ (aq )  i ons:

 

1. Determine the anode and cathode and write the half reactions
2. Calculate the standard cell potential
3. Write the shorthand cell notation

 

1) Anode is Ni

Cathode is Ag

Half reactions:

Anode: Ni²⁺ ==> Ni + 2e^-

Cathode: Ag⁺ + e^- ==> Ag

2) E°_cell = 1.056 V

3) Ni | Ni²⁺||Ag⁺ | Ag

Step-by-step explanation

Use the electrochemical series to determine the reduction potentials of each metal.

Note:

• The metal whose standard reduction potential is greater is easily reduced.
• The metal whose whose standard reduction potential is low is easily oxidized.

1) From the electrochemical series table, the following reduction potentials are obtained:

Ni²⁺ + 2e^- ==> Ni E° = -0.257 V

Ag⁺ + e^- ==> Ag E° = +0.799 V

Since Ni has a lower reduction potential than Ag, Ni is oxidized while Ag is reduced.

Oxidation occurs at the anode while reduction occurs at the cathode.

The half redox reactions become:

Reaction at Anode(Oxidation):

Ni²⁺ ==> Ni + 2e^- E° = +0.257 V

Reaction at Cathode(Reduction):

Ag⁺ + e^- ==> Ag E° = +0.799 V

2) E°_cell = E°_reduction + E°_0xidation = 0.799 V + 0.257 V = 1.056 V

3) Cell notation

When writing the cell notation, list the metal at the cathode on the left and that at the anode on the right.

The single vertical lines(|) indicate a phase separation between the solid metal and its aqueous ions.

the double || line indicate a salt bridge.

Hence, the shorthand cell notation of the given cell is:

Ni | Ni²⁺||Ag⁺ | Ag