Question
Explain why $\text{E}^0_\frac{\text{Fe}^{3+}}{\text{Fe}^{2+}}$ and $\text{C}^0_\frac{\text{Cu}^{2+}}{\text{Cu}}$ form a redox of couple?
✓
Answer
$\text{E}^0_\text{cell}=\text{E}^0_\frac{\text{Fe}^{3+}}{\text{Fe}^{2+}}-\text{E}^0_\frac{\text{Cu}^{2+}}{\text{Cu}}$
$=+0.77\text{V}-0.34\text{V}$
$=0.43\text{V}$
Since $\text{E}^0_\text{cell}$ is +ve, therefore, it will act as redox couple.
$2\text{Fe}^{3+}+\text{Cu}\xrightarrow{ \ \ \ \ \ \ }\text{Cu}^++2\text{Fe}^{2+}$
$=+0.77\text{V}-0.34\text{V}$
$=0.43\text{V}$
Since $\text{E}^0_\text{cell}$ is +ve, therefore, it will act as redox couple.
$2\text{Fe}^{3+}+\text{Cu}\xrightarrow{ \ \ \ \ \ \ }\text{Cu}^++2\text{Fe}^{2+}$
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