*These notes were taken from LG 5.3
LG:
file:///C:/Users/Valeska/Desktop/Valeska's%20School%20Files/PSHS%20Grade%2010/Chemistry%202/4th%20Quarter/Learning%20Guides/SLG%20CHEM2%20LG%205.3%20Cell%20Potentials.pdf
Given below are statements about galvanic cells. To check your understanding of galvanic cells, write TRUE if the statement is correct and FALSE if the statement is incorrect.
STATEMENT | ANSWERS |
---|---|
1. Oxidation occurs in the anode. | |
2. The anode carries a negative charge. | |
3. The oxidation number of chemical species decrease at the cathode. | |
4. The standard reduction potentials are measured at room temperature. | |
5. Electrons flow from the anode to the cathode through the salt bridge. | |
6. The electrode potentials are measured relative to the standard hydrogen electrode. | |
7. Eocell are written as reduction half-reactions. | |
8. The anode has the more positive the Eo half-cell. | |
9. For the oxidation half-reaction, the value of the Eo half-cell is reversed. | |
10. The more positive the Eocell, the stronger the oxidizing agent. |
Write the cell notation for the reaction:
2Al(s) + 3Pb2+(aq) → 2Al3+(aq) + 3Pb(s)
Given the cell notation, write the anode half-reaction, the cathode half-reaction and the overall balanced equation.
Pt(s) | H2(g) | H+(aq) || Ag+(aq) | Ag(s)
Oxidation occurs in the anode.
✅ TRUE
In a galvanic cell, oxidation always occurs at the anode and reduction at the cathode.
The anode carries a negative charge.
✅ TRUE
In galvanic cells (spontaneous reactions), the anode is negative because it supplies electrons.
The oxidation number of chemical species decrease at the cathode.
✅ TRUE
Reduction (gain of electrons) happens at the cathode, which lowers the oxidation number.
The standard reduction potentials are measured at room temperature.
✅ TRUE
Standard conditions for electrochemical measurements are 25°C (298 K), 1 atm, and 1 M solutions.
Electrons flow from the anode to the cathode through the salt bridge.
❌ FALSE
Electrons flow through the external wire, not the salt bridge. The salt bridge allows ion flow to maintain charge balance.
The electrode potentials are measured relative to the standard hydrogen electrode.
✅ TRUE
All standard electrode potentials are referenced against the standard hydrogen electrode (SHE), which is assigned 0.00 V.
E°cell are written as reduction half-reactions.
✅ TRUE
Standard electrode potentials (E°) are tabulated for reduction half-reactions.
The anode has the more positive the E° half-cell.
❌ FALSE
In a galvanic cell, the cathode has the more positive E° value; electrons flow from lower to higher potential.
For the oxidation half-reaction, the value of the E° half-cell is reversed.
✅ TRUE
When using standard reduction potentials to calculate E°cell, reverse the sign of E° for oxidation.
The more positive the E°cell, the stronger the oxidizing agent.
✅ TRUE
A more positive E° indicates a greater tendency to gain electrons, making it a stronger oxidizing agent.
2Al(s)+3Pb(aq)2+→2Al(aq)3++3Pb(s)2Al_{(s)} + 3Pb^{2+}_{(aq)} \rightarrow 2Al^{3+}_{(aq)} + 3Pb_{(s)}
Step-by-step:
Anode (oxidation): Al(s)→Al(aq)3++3e−Al_{(s)} \rightarrow Al^{3+}_{(aq)} + 3e^-
Cathode (reduction): Pb(aq)2++2e−→Pb(s)Pb^{2+}_{(aq)} + 2e^- \rightarrow Pb_{(s)}
Cell notation format:
Anode ∣ Anode solution ∣∣ Cathode solution ∣ Cathode\text{Anode} \ | \ \text{Anode solution} \ || \ \text{Cathode solution} \ | \ \text{Cathode}
Answer:
Al(s)∣Al(aq)3+ ∣∣ Pb(aq)2+∣Pb(s)Al_{(s)} | Al^{3+}_{(aq)} \ || \ Pb^{2+}_{(aq)} | Pb_{(s)}
Cell notation:
Pt(s)∣H2(g)∣H(aq)+ ∣∣ Ag(aq)+∣Ag(s)Pt_{(s)} | H_2(g) | H^+_{(aq)} \ || \ Ag^+_{(aq)} | Ag_{(s)}
Interpretation:
Left side (anode): H2(g)→2H+(aq)+2e−H_2(g) \rightarrow 2H^+(aq) + 2e^- (oxidation)
Right side (cathode): Ag+(aq)+e−→Ag(s)Ag^+(aq) + e^- \rightarrow Ag(s) (reduction)
Balance the reaction: Multiply Ag⁺ reaction by 2 to match electrons
Overall balanced equation:
H2(g)+2Ag+(aq)→2H+(aq)+2Ag(s)H_2(g) + 2Ag^+(aq) \rightarrow 2H^+(aq) + 2Ag(s)
Answers:
Anode half-reaction:
H2(g)→2H+(aq)+2e−H_2(g) \rightarrow 2H^+(aq) + 2e^-
Cathode half-reaction:
2Ag+(aq)+2e−→2Ag(s)2Ag^+(aq) + 2e^- \rightarrow 2Ag(s)
Overall equation:
H2(g)+2Ag+(aq)→2H+(aq)+2Ag(s)H_2(g) + 2Ag^+(aq) \rightarrow 2H^+(aq) + 2Ag(s)