In the reaction `A^(-n2)+xe^(-)rarrA^(-n1)` . Here, x will be :

Options

A. `n_(1)+n_(2)`
B. `n_(2)-n_(1)`
C. `n_(1)-n_(2)`
D. `n_(1).n_(2)`

Answer

To solve the problem, we need to analyze the given redox reaction:

**Given Reaction:**
\[ A^{-n2} + x e^{-} \rightarrow A^{-n1} \]

### Step 1: Identify the Change in Oxidation State
In this reaction, we see that the species \( A \) is undergoing a change in its oxidation state from \( -n2 \) to \( -n1 \). Since \( n2 \) is greater than \( n1 \), this indicates that \( A \) is gaining electrons, which is characteristic of a reduction process.

### Step 2: Understand the Concept of Reduction
Reduction involves the gain of electrons. The number of electrons gained in this process can be determined by the difference in the oxidation states:
\[ \text{Change in oxidation state} = n2 - n1 \]

### Step 3: Calculate the Number of Electrons (x)
From the above understanding, we can express \( x \) (the number of electrons gained) as:
\[ x = n2 - n1 \]

### Step 4: Conclusion
Thus, in the reaction \( A^{-n2} + x e^{-} \rightarrow A^{-n1} \), the value of \( x \) is equal to the difference in the oxidation states of \( A \):
\[ x = n2 - n1 \]

### Final Answer
Therefore, the value of \( x \) is \( n2 - n1 \).

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