From the following half cells a galvanic cell is made,
`A^(2+)+2e=A" "..." "E_(1)^(0)=0.8V`
`B=B^(3+)+3e" "..." "E_(2)^(0)=-0.3V`
`E^(0)` cell is

Options

A. `E_(1)^(0)-E_(2)^(0)`
B. `E_(1)^(0)+E_(2)^(0)`
C. `3E_(1)^(0)-2E_(2)^(0)`
D. `3E_(1)^(0)+2E_(2)^(0)`

Answer

To find the standard cell potential (E°cell) for the galvanic cell formed from the given half-cells, we will follow these steps:

### Step 1: Identify the half-reactions and their standard potentials
We have two half-reactions:
1. \( A^{2+} + 2e^- \leftrightarrow A \) with \( E^{0}_{1} = 0.8 \, V \) (this is the reduction half-reaction).
2. \( B \leftrightarrow B^{3+} + 3e^- \) with \( E^{0}_{2} = -0.3 \, V \) (this is the oxidation half-reaction).

### Step 2: Determine the direction of the reactions
- The half-reaction for A is a reduction (gaining electrons).
- The half-reaction for B is an oxidation (losing electrons).

### Step 3: Balance the number of electrons in the half-reactions
To combine the half-reactions, we need to ensure that the number of electrons lost in the oxidation reaction equals the number of electrons gained in the reduction reaction.

- The reduction half-reaction involves 2 electrons.
- The oxidation half-reaction involves 3 electrons.

To balance the electrons, we can multiply the reduction half-reaction by 3 and the oxidation half-reaction by 2:

- Reduction: \( 3(A^{2+} + 2e^- \leftrightarrow A) \) becomes \( 3A^{2+} + 6e^- \leftrightarrow 3A \)
- Oxidation: \( 2(B \leftrightarrow B^{3+} + 3e^-) \) becomes \( 2B \leftrightarrow 2B^{3+} + 6e^- \)

### Step 4: Write the overall balanced cell reaction
Combining the balanced half-reactions, we get:
\[ 3A^{2+} + 2B \leftrightarrow 3A + 2B^{3+} + 6e^- \]
The electrons cancel out, leaving us with the overall balanced reaction.

### Step 5: Calculate the standard cell potential (E°cell)
The standard cell potential can be calculated using the formula:
\[ E^{0}_{cell} = E^{0}_{reduction} - E^{0}_{oxidation} \]

From our half-reactions:
- \( E^{0}_{reduction} = 0.8 \, V \) (for A)
- \( E^{0}_{oxidation} = -(-0.3) \, V = 0.3 \, V \) (for B, since we take the negative of the oxidation potential)

Now substituting in the values:
\[ E^{0}_{cell} = 0.8 \, V - (-0.3 \, V) \]
\[ E^{0}_{cell} = 0.8 \, V + 0.3 \, V \]
\[ E^{0}_{cell} = 1.1 \, V \]

### Final Answer:
The standard cell potential \( E^{0}_{cell} \) is \( 1.1 \, V \).

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