The emf `(E_(cell)^(@))` of the cell reaction, `3Sn^(4+)+2Crto3Sn^(2+)+2Cr^(3+)` is 0.89V.
Calculate `DeltaG^(@)` for the reaction `(F=96,500" C "mol^(-1) and VC-=J)`.

Explanation

To calculate the standard Gibbs free energy change (ΔG°) for the given cell reaction, we can use the relationship between ΔG°, the number of moles of electrons transferred (n), the Faraday constant (F), and the standard cell potential (E°cell). The formula is:

\[ \Delta G° = -nFE°_{cell} \]

### Step-by-Step Solution:

**Step 1: Identify the values needed for the calculation.**
- We have the standard cell potential (E°cell) = 0.89 V.
- The Faraday constant (F) = 96,500 C/mol.

**Step 2: Determine the number of moles of electrons transferred (n).**
From the balanced cell reaction:
\[ 3 \text{Sn}^{4+} + 2 \text{Cr} \rightarrow 3 \text{Sn}^{2+} + 2 \text{Cr}^{3+} \]
Each Sn ion goes from +4 to +2, which means each Sn ion gains 2 electrons. Therefore, for 3 Sn ions:
\[ 3 \text{Sn}^{4+} \text{ gains } 3 \times 2 = 6 \text{ electrons} \]
The chromium (Cr) is oxidized from 0 to +3, which means each Cr loses 3 electrons. For 2 Cr atoms:
\[ 2 \text{Cr} \text{ loses } 2 \times 3 = 6 \text{ electrons} \]
Thus, the total number of moles of electrons transferred (n) = 6.

**Step 3: Substitute the values into the ΔG° equation.**
Now we can substitute the values into the equation:
\[ \Delta G° = -nFE°_{cell} \]
\[ \Delta G° = -6 \times 96500 \, \text{C/mol} \times 0.89 \, \text{V} \]

**Step 4: Calculate ΔG°.**
Calculating the above expression:
\[ \Delta G° = -6 \times 96500 \times 0.89 \]
\[ \Delta G° = -6 \times 86535 \]
\[ \Delta G° = -519210 \, \text{J/mol} \]

**Step 5: Convert to kJ/mol (if necessary).**
To convert Joules to kilojoules:
\[ \Delta G° = -519.21 \, \text{kJ/mol} \]

### Final Answer:
\[ \Delta G° \approx -519.21 \, \text{kJ/mol} \]