A hypothetical electrochemical cell is shown below
`ThetaPt(s)|A_(2)(1 atm)|A^(-)(xM)||B^(+)(yM)|B(s)^(o+)`
The emf measured is 0.30V The cell reaction is :-

Options

A. `A_(2)+2B^(+)to2A^(-)+2B`
B. `A_(2)+2e^(-)to2A^(-),2B^(+)+2e^(-)to2B`
C. The cell reaction cannot predicted
D. `2A^(-)+2B_(+)toA_(2)+2B`

Answer

To determine the cell reaction for the given electrochemical cell, we can follow these steps:

### Step 1: Identify the components of the cell
The cell is represented as:
`ThetaPt(s)|A_(2)(1 atm)|A^(-)(xM)||B^(+)(yM)|B(s)^(o+)`

From this representation:
- The left side (anode) contains A2 and A^(-).
- The right side (cathode) contains B^+ and B.

### Step 2: Determine the reactions at the anode and cathode
In an electrochemical cell:
- At the anode, oxidation occurs.
- At the cathode, reduction occurs.

**Anode Reaction:**
A2 is oxidized to A^(-):
\[ \text{A}_2 \rightarrow 2 \text{A}^- + 2 \text{e}^- \]

**Cathode Reaction:**
B^+ is reduced to B:
\[ \text{B}^+ + \text{e}^- \rightarrow \text{B} \]

### Step 3: Combine the half-reactions
To find the overall cell reaction, we need to combine the half-reactions. We have:
- From the anode: A2 gives 2 A^- and releases 2 electrons.
- From the cathode: B^+ gains 1 electron to form B.

To balance the electrons, we can multiply the cathode reaction by 2:
\[ 2 \text{B}^+ + 2 \text{e}^- \rightarrow 2 \text{B} \]

### Step 4: Write the overall cell reaction
Now, we can combine the balanced half-reactions:
\[ \text{A}_2 + 2 \text{B}^+ \rightarrow 2 \text{A}^- + 2 \text{B} \]

### Final Cell Reaction
The overall cell reaction is:
\[ \text{A}_2 + 2 \text{B}^+ \rightarrow 2 \text{A}^- + 2 \text{B} \]