Quick questions on Redox II (Topic 13) - Edexcel A-Level Chemistry

The standard hydrogen electrode bubbles $\text{H}_2$ at $100\ \text{kPa}$ over a platinised platinum electrode in $1.00\ \text{mol dm}^{-3}$ $\text{H}^+$. A salt bridge (e.g. saturated $\text{KNO}_3$) completes the circuit while a high-resistance voltmeter measures the potential difference without drawing current.

To get the overall reaction, reverse the half-equation at the negative electrode (so it shows oxidation), balance the electrons between the two half-equations, then add them. The electrons must cancel, which fixes the mole ratio.

Manganate(VII) reacts with iron(II) in a $1:5$ ratio because $\text{MnO}_4^-$ gains $5$ electrons while each $\text{Fe}^{2+}$ loses one: $$\text{MnO}_4^- + 8\text{H}^+ + 5\text{Fe}^{2+} \rightarrow \text{Mn}^{2+} + 4\text{H}_2\text{O} + 5\text{Fe}^{3+}$$Iodine-thiosulfate titrations ($\text{I}_2 + 2\text{S}_2\text{O}_3^{2-} \rightarrow 2\text{I}^- + \text{S}_4\text{O}_6^{2-}$) use starch indicator, with the blue-black colour vanishing at the end point.

Two half-cells have $E^{\ominus}$ values of $+0.77\ \text{V}$ and $-0.76\ \text{V}$. Calculate the cell EMF. [1 mark]

Explain why a reaction with a positive cell EMF might not occur at a noticeable rate. [2 marks]