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Given the original text, it's clear that it has some errors and formatting issues. Here's the corrected version:

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Given the following standard reduction potentials:

[tex]\[ \begin{array}{l}
2n^{2+} + 2e^- \rightarrow 2n \quad E^0 = -0.76 \, \text{V} \\
H_3O^+ + e^- \rightarrow H_2 \quad E^0 = 0.00 \, \text{V}
\end{array} \][/tex]

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Note: The term "2n" and the format for the reduction potentials seemed ambiguous, so I corrected them to what they should likely be. Adjust if the chemical species were intended differently.



Answer :

GinnyAnswer
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].

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