Let's solve the problem step-by-step.
1. Understanding the notation:
- [tex]\( E^0 \)[/tex] denotes the standard electrode potential.
- The standard electrode potential for the [tex]\( 2n^{2+} / 2n \)[/tex] half-reaction is given as [tex]\( E^0 = -0.76 \, \text{V} \)[/tex].
- The standard electrode potential for the [tex]\( \text{H}_3\text{O}^+ / \text{H}_2 \)[/tex] half-reaction is given as [tex]\( E^0 = 0.00 \, \text{V} \)[/tex].
2. Determining the Potential Difference:
- To find the potential difference between these two half-reactions, we need to subtract the electrode potential of the [tex]\( 2n^{2+} / 2n \)[/tex] half-reaction from the electrode potential of the [tex]\( \text{H}_3\text{O}^+ / \text{H}_2 \)[/tex] half-reaction.
3. Performing the Calculation:
- Electrode potential for [tex]\( \text{H}_3\text{O}^+ / \text{H}_2 \)[/tex]: [tex]\( E^0 = 0.00 \, \text{V} \)[/tex]
- Electrode potential for [tex]\( 2n^{2+} / 2n \)[/tex]: [tex]\( E^0 = -0.76 \, \text{V} \)[/tex]
The potential difference is calculated as:
[tex]\[
\text{Potential Difference} = 0.00 \, \text{V} - (-0.76 \, \text{V}) = 0.00 \, \text{V} + 0.76 \, \text{V} = 0.76 \, \text{V}
\][/tex]
4. Conclusion:
- The potential difference between the standard electrode potentials of the [tex]\( \text{H}_3\text{O}^+ / \text{H}_2 \)[/tex] and [tex]\( 2n^{2+} / 2n \)[/tex] half-reactions is [tex]\( 0.76 \, \text{V} \)[/tex].
So, the final answer is:
The potential difference is [tex]\( 0.76 \, \text{V} \)[/tex].
