What is the standard e.m.f of the cell containing `Sn^(2+)|Sn` and `Br_2 |Br^(-)` electrodes ?
`E^(Theta) (Sn^(2+) |Sn) = -0.14 V ,E^(Theta) (Br_2 |Br^-) = 1.08 V)`

Explanation

To find the standard e.m.f (electromotive force) of the cell containing the electrodes \( \text{Sn}^{2+} | \text{Sn} \) and \( \text{Br}_2 | \text{Br}^- \), we will follow these steps:

### Step 1: Identify the reduction potentials
We are given the standard reduction potentials:
- For \( \text{Sn}^{2+} + 2e^- \rightarrow \text{Sn} \), \( E^\circ = -0.14 \, \text{V} \)
- For \( \text{Br}_2 + 2e^- \rightarrow 2\text{Br}^- \), \( E^\circ = 1.08 \, \text{V} \)

### Step 2: Determine which half-reaction will undergo reduction and which will undergo oxidation
The half-reaction with the higher reduction potential will occur at the cathode (reduction), while the one with the lower reduction potential will occur at the anode (oxidation).
- \( \text{Br}_2 \) has a higher reduction potential (1.08 V) compared to \( \text{Sn}^{2+} \) (-0.14 V), so \( \text{Br}_2 \) will be reduced, and \( \text{Sn} \) will be oxidized.

### Step 3: Write the half-reactions
- Reduction half-reaction (cathode):
\[ \text{Br}_2 + 2e^- \rightarrow 2\text{Br}^- \]
- Oxidation half-reaction (anode):
\[ \text{Sn} \rightarrow \text{Sn}^{2+} + 2e^- \]

### Step 4: Calculate the oxidation potential for the anode
The oxidation potential is the negative of the reduction potential:
\[ E^\circ_{\text{oxidation}} = -E^\circ_{\text{reduction}}(\text{Sn}^{2+}/\text{Sn}) = -(-0.14 \, \text{V}) = 0.14 \, \text{V} \]

### Step 5: Calculate the standard e.m.f of the cell
The standard e.m.f (E°cell) is calculated by adding the reduction potential of the cathode to the oxidation potential of the anode:
\[ E^\circ_{\text{cell}} = E^\circ_{\text{cathode}} + E^\circ_{\text{anode}} = 1.08 \, \text{V} + 0.14 \, \text{V} = 1.22 \, \text{V} \]

### Final Answer
The standard e.m.f of the cell is:
\[ \boxed{1.22 \, \text{V}} \]

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