From the fllowing half-cell reactions and their standard potentials ,what is the smallest possible standard e.m.f for spontaneous reactions?
`PO_4^(3-)(aq)+2H_2O(l)+2e^(-)toHPO_3^(2-)+3OH^(-)(aq), E^(@)=-1.05V`
`PbO_2(s)+H_2O(l)+2e^(-)toPbO(s)+2OH^(-)(aq),E^(@)=+0.28 V`

Answer

To determine the smallest possible standard e.m.f for spontaneous reactions from the given half-cell reactions and their standard potentials, we can follow these steps:

### Step 1: Identify the half-cell reactions and their standard potentials
We have two half-cell reactions:
1. \( \text{PO}_4^{3-}(aq) + 2H_2O(l) + 2e^- \rightarrow \text{HPO}_3^{2-} + 3OH^-(aq), \quad E^\circ = -1.05 \, V \)
2. \( \text{PbO}_2(s) + H_2O(l) + 2e^- \rightarrow \text{PbO}(s) + 2OH^-(aq), \quad E^\circ = +0.28 \, V \)

### Step 2: Determine the cathode and anode reactions
For a spontaneous reaction, we need to identify which half-reaction will occur at the cathode (reduction) and which will occur at the anode (oxidation).

- The half-reaction with a more positive standard potential will act as the cathode. In this case, the second reaction (lead) has a higher potential (+0.28 V) and will be the cathode.
- The first reaction (phosphate) will be reversed to act as the anode, which means we will change the sign of its standard potential.

### Step 3: Reverse the anode reaction and change its potential
When we reverse the first half-reaction, the new reaction will be:
\[ \text{HPO}_3^{2-} + 3OH^-(aq) \rightarrow \text{PO}_4^{3-}(aq) + 2H_2O(l) + 2e^- \]
The standard potential for this reversed reaction becomes:
\[ E^\circ = +1.05 \, V \]

### Step 4: Calculate the standard e.m.f (E°cell)
The standard e.m.f of the cell can be calculated using the formula:
\[ E^\circ_{cell} = E^\circ_{cathode} - E^\circ_{anode} \]
Substituting the values:
- \( E^\circ_{cathode} = +0.28 \, V \)
- \( E^\circ_{anode} = -1.05 \, V \) (since we reversed it, we take it as +1.05 V)

Thus, we have:
\[ E^\circ_{cell} = 0.28 \, V - (-1.05 \, V) \]
\[ E^\circ_{cell} = 0.28 \, V + 1.05 \, V = 1.33 \, V \]

### Conclusion
The smallest possible standard e.m.f for spontaneous reactions is:
\[ \boxed{1.33 \, V} \]