For the given reactions
`Sn^(2+)+2e^(-) rarr Sn`
`Sn^(4+)+4e^(-) rarr Sn`
the electrode potentials are, `E^@""_(Sn^(2+)//Sn)=-0.140 V` and `E^@_(Sn^(4+)//Sn)=0.010V`. The magnitude of standard electrode potential for `Sn^(4+)//Sn^(2+) i.e E^@""_(Sn^(4+)//Sn^(2+))` is _________ `xx 10^(-2)V` (Nearest integer)

Explanation

To solve the problem, we need to determine the standard electrode potential for the half-reaction \( \text{Sn}^{4+} + 2e^- \rightarrow \text{Sn}^{2+} \). We will use the given standard electrode potentials for the reactions involving \( \text{Sn}^{2+} \) and \( \text{Sn}^{4+} \).

### Step-by-Step Solution:

1. **Identify the Given Reactions and Their Potentials:**
- The first reaction is:
\[ \text{Sn}^{2+} + 2e^- \rightarrow \text{Sn} \quad (E^\circ = -0.140 \, \text{V}) \]
- The second reaction is:
\[ \text{Sn}^{4+} + 4e^- \rightarrow \text{Sn} \quad (E^\circ = 0.010 \, \text{V}) \]

2. **Write the Desired Reaction:**
- We want to find the potential for the reaction:
\[ \text{Sn}^{4+} + 2e^- \rightarrow \text{Sn}^{2+} \]
- To obtain this reaction, we can manipulate the given reactions.

3. **Reverse the First Reaction:**
- Reverse the first reaction to express it in terms of \( \text{Sn}^{2+} \):
\[ \text{Sn} \rightarrow \text{Sn}^{2+} + 2e^- \quad (E^\circ = +0.140 \, \text{V}) \]

4. **Combine the Reactions:**
- Now, we can add the reversed first reaction to the second reaction:
\[ \text{Sn}^{4+} + 4e^- \rightarrow \text{Sn} \quad (E^\circ = 0.010 \, \text{V}) \]
\[ \text{Sn} \rightarrow \text{Sn}^{2+} + 2e^- \quad (E^\circ = +0.140 \, \text{V}) \]
- The combined reaction will be:
\[ \text{Sn}^{4+} + 2e^- \rightarrow \text{Sn}^{2+} \]

5. **Calculate the Standard Electrode Potential:**
- The overall standard electrode potential \( E^\circ \) for the combined reaction can be calculated using the formula:
\[ E^\circ = E^\circ_{\text{reduction}} + E^\circ_{\text{oxidation}} \]
- Here, \( E^\circ_{\text{reduction}} = 0.010 \, \text{V} \) and \( E^\circ_{\text{oxidation}} = +0.140 \, \text{V} \).
- Therefore:
\[ E^\circ = 0.010 \, \text{V} + 0.140 \, \text{V} = 0.150 \, \text{V} \]

6. **Convert to Required Format:**
- The problem asks for the magnitude in the form \( xx \times 10^{-2} \, \text{V} \):
\[ 0.150 \, \text{V} = 15.0 \times 10^{-2} \, \text{V} \]
- The nearest integer for \( xx \) is 15.

### Final Answer:
The magnitude of the standard electrode potential for \( \text{Sn}^{4+} // \text{Sn}^{2+} \) is \( 15 \times 10^{-2} \, \text{V} \).

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