Design an electrochemical cell using Pb(s) and Mn(s) and their solutions to answer the following questions.
I just want to see if my answers are correct. I drew the cell already. Thanks
1. Give the line notation for this electrochemical cell.
Mn(s) ∣ Mn2+(aq) ∥ Pb2+ ∣ Pb(s)
2. Calculate the net voltage of the cell.
E0 net(cell) = E0 ox + E0 red
= (1.18 V) + (-0.13 V)
= +1.05 V
3. State the anode and cathode.
The anode is Mn and the cathode is Pb.
4. Write the half reactions taking place at each electrode and indicate whether it is a n oxidation or reduction reaction.
Mn(s)--> Mn2+(aq) +2e- is the oxidation reaction.
Pb 2+(aq) + 2 e- -->Pb(s) is the reduction reaction.
5. Write the balanced equation for the net reaction.
2Mn(s) + 2Pb2+(aq) 2Mn2+ +2Pb(s)
6. Which electrode gains mass? Which loses mass?
Pb gains mass Mn losses mass.
7. State the direction of electron flow. Be specific.
The direction of electron flow is from Mn anode to Pb cathode.
I didn't look up the voltages for #2 but everything else looks ok except for #5. The coefficient of 2 is not needed.
that is correct
Your answers are correct. Well done! Here is a summary:
1. Line notation: Mn(s) ∣ Mn2+(aq) ∥ Pb2+ ∣ Pb(s)
2. Net voltage of the cell: +1.05 V
3. Anode: Mn
Cathode: Pb
4. Half reactions:
Oxidation (anode): Mn(s) → Mn2+(aq) + 2e- (oxidation)
Reduction (cathode): Pb2+(aq) + 2e- → Pb(s) (reduction)
5. Balanced net reaction: 2Mn(s) + 2Pb2+(aq) → 2Mn2+(aq) + 2Pb(s)
6. Electrode mass changes:
The Pb electrode gains mass.
The Mn electrode loses mass.
7. Electron flow: Electrons flow from the Mn anode to the Pb cathode.
You did a great job!
Your answers are all correct. Well done!
1. The line notation for the electrochemical cell is: Mn(s) | Mn2+(aq) || Pb2+ | Pb(s)
The vertical line (|) represents a phase boundary, the double vertical line (||) represents a salt bridge or porous barrier separating the two half-cells.
2. To calculate the net voltage of the cell, you add the reduction potentials of the half-reactions:
E0 net(cell) = E0 red(cathode) - E0 ox(anode)
= (-0.13 V) - (-1.18 V)
= +1.05 V
3. The anode is where oxidation occurs, so Mn(s) is the anode.
The cathode is where reduction occurs, so Pb(s) is the cathode.
4. The half-reaction at the anode (Mn(s)) is an oxidation reaction:
Mn(s) --> Mn2+(aq) + 2e-
The half-reaction at the cathode (Pb2+(aq)) is a reduction reaction:
Pb2+(aq) + 2e- --> Pb(s)
5. The balanced equation for the net reaction is:
2Mn(s) + 2Pb2+(aq) → 2Mn2+(aq) + 2Pb(s)
6. The electrode where reduction occurs (cathode) gains mass, so Pb(s) gains mass.
The electrode where oxidation occurs (anode) loses mass, so Mn(s) loses mass.
7. The direction of electron flow is from the anode (Mn) to the cathode (Pb). So, electrons flow from Mn(s) to Pb(s) through the external circuit.