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Cr(s)| Cr^(3+)||Fe^(2+)|Fe(s)In the abov...

`Cr(s)| Cr^(3+)||Fe^(2+)|Fe(s)`In the above cell, the value of n in the Nernst equation i.e., `E=E^(@) - (0.059)/(n) log_(10)Q` will be :

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To determine the value of \( n \) in the Nernst equation for the given electrochemical cell \( \text{Cr(s)} | \text{Cr}^{3+} || \text{Fe}^{2+} | \text{Fe(s)} \), we need to analyze the oxidation and reduction reactions occurring in the cell. ### Step-by-Step Solution: 1. **Identify the Half Reactions**: - The anode reaction involves the oxidation of chromium: \[ \text{Cr(s)} \rightarrow \text{Cr}^{3+} + 3e^- ...
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