In an oxidation proces for a cell,
`M_(1)rarr M_(1)^(n+)+n e^(-)`,
the other metal `(M_(2))` being univalent showing reduction takes up ...... Electrons to complete redox reaction.

Options

A. (n-1)
B. 1
C. n
D. 2

Explanation

To solve the problem, we need to analyze the redox reaction involving two metals, M1 and M2.

### Step-by-Step Solution:

1. **Identify the Oxidation Reaction**:
The given oxidation reaction is:
\[ M_1 \rightarrow M_1^{(n+)} + n e^- \]
This indicates that metal M1 loses n electrons to form its cation \(M_1^{(n+)}\).

2. **Understand the Reduction Reaction**:
The problem states that metal M2 is univalent, which means it can only lose or gain one electron. In a reduction process, M2 will gain electrons to form its neutral state.

3. **Determine the Number of Electrons M2 Takes**:
Since M2 is univalent, it can only accept one electron to become neutral. However, since the oxidation of M1 involves the loss of n electrons, M2 must take up n electrons to balance the reaction.

4. **Write the Reduction Reaction**:
The reduction reaction for M2 can be written as:
\[ M_2^{(n+)} + n e^- \rightarrow M_2 \]
Here, M2 takes n electrons to become neutral.

5. **Combine the Reactions**:
When we combine the oxidation and reduction reactions, we get:
\[ M_1 + M_2^{(n+)} \rightarrow M_1^{(n+)} + M_2 \]
This shows that the electrons lost by M1 are equal to the electrons gained by M2, confirming that M2 takes up n electrons.

### Final Answer:
M2 takes up **n electrons** to complete the redox reaction.

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