Table 1 shows some electrode half-equations and their standard electrode potentials.
Table 1
| Electrode half-equation | EθE^\thetaEθ / V |
|---|---|
| F2(g)+2e−→2F−(aq)\text{F}_2\text{(g)} + 2\text{e}^- \rightarrow 2\text{F}^-\text{(aq)}F2(g)+2e−→2F−(aq) | +2.87+2.87+2.87 |
| MnO4−(aq)+8H+(aq)+5e−→Mn2+(aq)+4H2O(l)\text{MnO}_4^-\text{(aq)} + 8\text{H}^+\text{(aq)} + 5\text{e}^- \rightarrow \text{Mn}^{2+}\text{(aq)} + 4\text{H}_2\text{O}\text{(l)}MnO4−(aq)+8H+(aq)+5e−→Mn2+(aq)+4H2O(l) | +1.51+1.51+1.51 |
| NO3−(aq)+4H+(aq)+3e−→NO(g)+2H2O(l)\text{NO}_3^-\text{(aq)} + 4\text{H}^+\text{(aq)} + 3\text{e}^- \rightarrow \text{NO}\text{(g)} + 2\text{H}_2\text{O}\text{(l)}NO3−(aq)+4H+(aq)+3e−→NO(g)+2H2O(l) | +0.96+0.96+0.96 |
| Cu2+(aq)+2e−→Cu(s)\text{Cu}^{2+}\text{(aq)} + 2\text{e}^- \rightarrow \text{Cu}\text{(s)}Cu2+(aq)+2e−→Cu(s) | +0.34+0.34+0.34 |
| S4O62−(aq)+2e−→2S2O32−(aq)\text{S}_4\text{O}_6^{2-}\text{(aq)} + 2\text{e}^- \rightarrow 2\text{S}_2\text{O}_3^{2-}\text{(aq)}S4O62−(aq)+2e−→2S2O32−(aq) | +0.08+0.08+0.08 |
| 2H+(aq)+2e−→H2(g)2\text{H}^+\text{(aq)} + 2\text{e}^- \rightarrow \text{H}_2\text{(g)}2H+(aq)+2e−→H2(g) | 0.000.000.00 |
| Cr3+(aq)+e−→Cr2+(aq)\text{Cr}^{3+}\text{(aq)} + \text{e}^- \rightarrow \text{Cr}^{2+}\text{(aq)}Cr3+(aq)+e−→Cr2+(aq) | −0.41-0.41−0.41 |
Deduce the oxidation state of sulfur in S4O62−\text{S}_4\text{O}_6^{2-}S4O62− and in S2O32−\text{S}_2\text{O}_3^{2-}S2O32−.
State the weakest reducing agent in Table 1.
Write the conventional representation of the cell that has an EMF of +0.75 V+0.75\text{ V}+0.75 V.
Use data from Table 1 to identify an acid that will oxidise copper metal. Explain your choice, suggest a possible overall equation for the reaction, and calculate the EMF of the cell that has the same overall reaction.
