The figure shows a drop of rainwater, $\text{pH} = 5.60$, resting on a clean iron sheet in air at $298\ \text{K}$. Iron is oxidised at one spot under the drop, $\text{Fe} \rightarrow \text{Fe}^{2+} + 2e^-$. The electrons flow through the metal to the edge of the drop, where oxygen from the air is reduced: $\text{O}_2 + 4\text{H}^+ + 4e^- \rightarrow 2\text{H}_2\text{O}$. The partial pressure of oxygen is $0.21\ \text{bar}$, and near the iron $[\text{Fe}^{2+}] = 1.0 \times 10^{-6}\ \text{mol L}^{-1}$. Take the drop as a small cell in which the iron sheet is both electrodes. Use NCERT's values: $E^\circ(\text{O}_2/\text{H}_2\text{O}) = 1.23\ \text{V}$, $E^\circ(\text{Fe}^{2+}/\text{Fe}) = -0.44\ \text{V}$, $E^\circ(\text{Zn}^{2+}/\text{Zn}) = -0.76\ \text{V}$ and $E^\circ(\text{Sn}^{2+}/\text{Sn}) = -0.14\ \text{V}$, with $\dfrac{2.303RT}{F} = 0.059\ \text{V}$. Which of the following statements are correct?

Fig. 1: A cross-section of a drop of rainwater, pH 5.60, on an iron sheet. In the middle of the drop, at the iron surface, iron is oxidised to Fe2+ ions. At the edges of the drop, where air with oxygen at 0.21 bar meets the water, oxygen is reduced to water. Arrows inside the iron show electrons flowing from the middle to the edges.