If `E_(A^(+2)//A)^(@)=-0.30 V and E_(A^(+3)//A^(+2))^(@)=0.40 V,` the standard
Emf of the reaction: `A+2A^(+3)to3A^(+2)` will be :-

Options

A. 0.30 V
B. 0. 40 V
C. 0. 70 V
D. 0. 10 V

Explanation

To find the standard EMF of the reaction \( A + 2A^{+3} \rightarrow 3A^{+2} \), we will follow these steps:

### Step 1: Identify the half-reactions and their standard electrode potentials.
We have two half-reactions provided:
1. \( A^{+2} + 2e^- \rightarrow A \) with \( E^\circ = -0.30 \, V \) (this is the reduction reaction at the cathode).
2. \( A^{+3} + e^- \rightarrow A^{+2} \) with \( E^\circ = 0.40 \, V \) (this is the oxidation reaction at the anode).

### Step 2: Write the oxidation and reduction reactions.
- **Oxidation (Anode)**:
\[ A \rightarrow A^{+2} + 2e^- \]
The standard potential for this reaction is the negative of the given reduction potential:
\[ E^\circ_{\text{anode}} = +0.30 \, V \]

- **Reduction (Cathode)**:
\[ A^{+3} + e^- \rightarrow A^{+2} \]
The standard potential for this reaction is given as:
\[ E^\circ_{\text{cathode}} = 0.40 \, V \]

### Step 3: Combine the half-reactions to find the overall reaction.
The overall reaction can be written as:
\[ A + 2A^{+3} \rightarrow 3A^{+2} \]

### Step 4: Calculate the standard EMF of the cell.
The standard EMF of the cell (\( E^\circ_{\text{cell}} \)) can be calculated using the formula:
\[ E^\circ_{\text{cell}} = E^\circ_{\text{cathode}} - E^\circ_{\text{anode}} \]

Substituting the values:
\[ E^\circ_{\text{cell}} = 0.40 \, V - (-0.30 \, V) = 0.40 \, V + 0.30 \, V = 0.70 \, V \]

### Final Answer:
The standard EMF of the reaction \( A + 2A^{+3} \rightarrow 3A^{+2} \) is \( 0.70 \, V \).

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