Standard electrode potential for `Sn^(4+)//Sn^(2+)` couple is `0.15 V` and that for the `Cr^(3+)//Cr` couple is `-0.74 V`. These two couples in their standard state are connected to make a cell. The cell potential will be `+0.89 V`, `+0.18 V`, `+1.83 V`, +1.199 V`

Options

A. `+0.89 V`
B. `+0.18 V`
C. `+1.83 V`
D. `+1.199 V`

Explanation

To solve the problem of finding the cell potential for the `Sn^(4+)//Sn^(2+)` and `Cr^(3+)//Cr` electrochemical couples, we will follow these steps:

### Step 1: Identify the Standard Electrode Potentials
We are given the following standard electrode potentials:
- For the `Sn^(4+)//Sn^(2+)` couple: \( E^\circ_{Sn} = +0.15 \, V \)
- For the `Cr^(3+)//Cr` couple: \( E^\circ_{Cr} = -0.74 \, V \)

### Step 2: Determine the Anode and Cathode
The electrode with the higher standard electrode potential acts as the cathode (where reduction occurs), while the one with the lower standard electrode potential acts as the anode (where oxidation occurs).

- Since \( E^\circ_{Sn} = +0.15 \, V \) is greater than \( E^\circ_{Cr} = -0.74 \, V \), we identify:
- **Cathode**: `Sn^(4+)//Sn^(2+)` (reduction occurs)
- **Anode**: `Cr^(3+)//Cr` (oxidation occurs)

### Step 3: Write the Half-Reactions
- **At the Cathode** (reduction):
\[ Sn^{4+} + 2e^- \rightarrow Sn^{2+} \quad (E^\circ = +0.15 \, V) \]

- **At the Anode** (oxidation):
\[ Cr \rightarrow Cr^{3+} + 3e^- \quad (E^\circ = -0.74 \, V) \]

### Step 4: Calculate the Cell Potential
The overall cell potential \( E^\circ_{cell} \) is calculated using the formula:
\[ E^\circ_{cell} = E^\circ_{cathode} - E^\circ_{anode} \]

Substituting the values:
\[ E^\circ_{cell} = E^\circ_{Sn} - E^\circ_{Cr} \]
\[ E^\circ_{cell} = 0.15 \, V - (-0.74 \, V) \]
\[ E^\circ_{cell} = 0.15 \, V + 0.74 \, V \]
\[ E^\circ_{cell} = 0.89 \, V \]

### Conclusion
The cell potential for the electrochemical cell constructed from the `Sn^(4+)//Sn^(2+)` and `Cr^(3+)//Cr` couples is \( 0.89 \, V \).

### Final Answer
The correct answer is **+0.89 V**.

---