This question is about electrode potentials and electrochemical cells.
Table 1 shows some electrode potentials.
Table 1
| Half-equation | E⊖/VE^\ominus / \text{V}E⊖/V |
|---|---|
| Ni2+(aq)+2e−→Ni(s)\text{Ni}^{2+}(\text{aq}) + 2\text{e}^- \rightarrow \text{Ni}(\text{s})Ni2+(aq)+2e−→Ni(s) | −0.25-0.25−0.25 |
| H+(aq)+e−→12H2(g)\text{H}^+(\text{aq}) + \text{e}^- \rightarrow \frac{1}{2} \text{H}_2(\text{g})H+(aq)+e−→21H2(g) | 0.000.000.00 |
| [Co(NH3)6]3+(aq)+e−→[Co(NH3)6]2+(aq)[\text{Co}(\text{NH}_3)_6]^{3+}(\text{aq}) + \text{e}^- \rightarrow [\text{Co}(\text{NH}_3)_6]^{2+}(\text{aq})[Co(NH3)6]3+(aq)+e−→[Co(NH3)6]2+(aq) | +0.11+0.11+0.11 |
| Cr2O72−(aq)+14H+(aq)+6e−→2Cr3+(aq)+7H2O(l)\text{Cr}_2\text{O}_7^{2-}(\text{aq}) + 14\text{H}^+(\text{aq}) + 6\text{e}^- \rightarrow 2\text{Cr}^{3+}(\text{aq}) + 7\text{H}_2\text{O}(\text{l})Cr2O72−(aq)+14H+(aq)+6e−→2Cr3+(aq)+7H2O(l) | +1.33+1.33+1.33 |
| [Co(H2O)6]3+(aq)+e−→[Co(H2O)6]2+(aq)[\text{Co}(\text{H}_2\text{O})_6]^{3+}(\text{aq}) + \text{e}^- \rightarrow [\text{Co}(\text{H}_2\text{O})_6]^{2+}(\text{aq})[Co(H2O)6]3+(aq)+e−→[Co(H2O)6]2+(aq) | +1.82+1.82+1.82 |
Define the term standard electrode potential.
State two standard conditions that must be maintained in a standard hydrogen electrode for it to have E⊖=0.00 VE^\ominus = 0.00 \text{ V}E⊖=0.00 V.
Identify the weakest reducing agent in Table 1.
Use half-equations from Table 1 to deduce a balanced chemical equation for the reduction of Cr2O72−\text{Cr}_2\text{O}_7^{2-}Cr2O72− to form Cr3+\text{Cr}^{3+}Cr3+ in acidic solution by nickel metal (Ni\text{Ni}Ni).
Use data from Table 1 to explain why [Co(H2O)6]3+(aq)[\text{Co}(\text{H}_2\text{O})_6]^{3+}(\text{aq})[Co(H2O)6]3+(aq) will undergo a redox reaction with [Co(NH3)6]2+(aq)[\text{Co}(\text{NH}_3)_6]^{2+}(\text{aq})[Co(NH3)6]2+(aq).
Give an equation for this reaction.
Suggest why the two cobalt complex systems in Table 1 have different electrode potentials.
