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

Options

A. `+1.19" V "`
B. `+0.89" V "`
C. `+0.18" V "`
D. `+1.83" V "`

Explanation

To find the cell potential for the electrochemical cell formed by the `Sn^(4+)//Sn^(2+)` and `Cr^(3+)//Cr` couples, we can follow these steps:

### Step 1: Identify the half-reactions and their standard electrode potentials.
- For the `Sn^(4+)//Sn^(2+)` couple, the standard electrode potential (E°) is given as +0.15 V.
- For the `Cr^(3+)//Cr` couple, the standard electrode potential (E°) is given as -0.74 V.

### Step 2: Determine which half-reaction will act as the cathode and which as the anode.
- The half-reaction with the higher (more positive) standard electrode potential will act as the cathode, where reduction occurs.
- The half-reaction with the lower (more negative) standard electrode potential will act as the anode, where oxidation occurs.

In this case:
- `Sn^(4+) + 2e^- → Sn^(2+)` (E° = +0.15 V) will act as the cathode.
- `Cr^(3+) + 3e^- → Cr` (E° = -0.74 V) will act as the anode.

### Step 3: Write the overall cell reaction.
The overall cell reaction can be constructed by combining the half-reactions. However, we need to balance the number of electrons transferred in both half-reactions.

1. Multiply the tin half-reaction by 3 to balance the electrons:
- `3(Sn^(4+) + 2e^- → Sn^(2+))` gives `3Sn^(4+) + 6e^- → 3Sn^(2+)`

2. Multiply the chromium half-reaction by 2:
- `2(Cr^(3+) + 3e^- → Cr)` gives `2Cr^(3+) + 6e^- → 2Cr`

Now we can combine these balanced half-reactions:
- `3Sn^(4+) + 2Cr → 3Sn^(2+) + 2Cr^(3+)`

### Step 4: Calculate the cell potential (E°cell).
The cell potential can be calculated using the formula:
\[ E°_{cell} = E°_{cathode} - E°_{anode} \]

Substituting the values:
\[ E°_{cell} = E°_{Sn^{4+}/Sn^{2+}} - E°_{Cr^{3+}/Cr} \]
\[ E°_{cell} = 0.15 V - (-0.74 V) \]
\[ E°_{cell} = 0.15 V + 0.74 V \]
\[ E°_{cell} = 0.89 V \]

### Final Answer:
The cell potential for the electrochemical cell formed by the `Sn^(4+)//Sn^(2+)` and `Cr^(3+)//Cr` couples is **0.89 V**.

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