For the electrode process,
`H^(+)+e=1/2H_(2),E_(H_(H)^(+),H_(2))` = x volt
then for `2H^(+)+2e=H_(2),E_(2H^(+),H_(2))` is equal to

Options

A. x volt
B. 2x volt
C. `x/2` volt
D. 0 volt

Answer

To solve the problem, we need to analyze the given electrode processes and their standard reduction potentials.

### Step-by-Step Solution:

1. **Identify the Given Reaction and Its Standard Potential:**
The first reaction is given as:
\[ H^+ + e^- \rightarrow \frac{1}{2} H_2 \]
The standard reduction potential for this reaction is given as \( E(H^+, H_2) = x \) volts.

2. **Understand the Relationship Between Reactions and Their Potentials:**
The second reaction we need to consider is:
\[ 2H^+ + 2e^- \rightarrow H_2 \]
This reaction is essentially the first reaction multiplied by 2.

3. **Apply the Concept of Standard Reduction Potentials:**
It is important to note that when a half-reaction is multiplied by a coefficient, the standard reduction potential (\( E^\circ \)) does not change. This is because the standard reduction potential is an intensive property, meaning it does not depend on the amount of substance involved.

4. **Conclude the Standard Potential for the Second Reaction:**
Since we have multiplied the first reaction by 2, the standard reduction potential for the second reaction remains the same:
\[ E(2H^+, H_2) = x \text{ volts} \]

### Final Answer:
Thus, the standard reduction potential for the reaction \( 2H^+ + 2e^- \rightarrow H_2 \) is equal to \( x \) volts.

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