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.
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