The standard oxidation potentials, , for the half reactions are as follows :
`Zn rightarrow Zn^(2+) + 2e^(-)` ,` E^(@) = +0.76V`
`Fe rightarrow Fe^(2+)+ 2e^(-), E^(@) = + 0.41 V`
The EMF for the cell reaction,
`Fe^(2+) + Zn rightarrow Zn^(2+) + Fe`

Options

A. `(-0.35V)`
B. (+0.35V)
C. `(+1.17V)`
D. `(1.17V)`

Explanation

To solve the problem, we need to calculate the EMF (Electromotive Force) for the given cell reaction:

**Cell Reaction:**
\[ \text{Fe}^{2+} + \text{Zn} \rightarrow \text{Zn}^{2+} + \text{Fe} \]

**Step 1: Identify the half-reactions and their standard oxidation potentials.**
- The half-reactions given are:
1. \( \text{Zn} \rightarrow \text{Zn}^{2+} + 2e^- \) with \( E^\circ = +0.76 \, \text{V} \) (oxidation potential)
2. \( \text{Fe} \rightarrow \text{Fe}^{2+} + 2e^- \) with \( E^\circ = +0.41 \, \text{V} \) (oxidation potential)

**Step 2: Convert oxidation potentials to reduction potentials.**
- The reduction potential is the negative of the oxidation potential:
- For Zinc:
\[ E^\circ_{\text{Zn}} = -0.76 \, \text{V} \]
- For Iron:
\[ E^\circ_{\text{Fe}} = -0.41 \, \text{V} \]

**Step 3: Determine which species is oxidized and which is reduced.**
- In the reaction, Zn is oxidized (loses electrons) and Fe²⁺ is reduced (gains electrons).
- Therefore:
- Anode (oxidation): Zn
- Cathode (reduction): Fe²⁺

**Step 4: Calculate the EMF of the cell.**
- The EMF (E_cell) can be calculated using the formula:
\[ E_{\text{cell}} = E^\circ_{\text{cathode}} - E^\circ_{\text{anode}} \]
- Substituting the values:
\[ E_{\text{cell}} = (-0.41 \, \text{V}) - (-0.76 \, \text{V}) \]
\[ E_{\text{cell}} = -0.41 + 0.76 \]
\[ E_{\text{cell}} = 0.35 \, \text{V} \]

**Final Answer:**
The EMF for the cell reaction is \( 0.35 \, \text{V} \).

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