Calculate the equilibrium concentration of all ions in an ideal solution prepared by mixing `25.00mL` of `0.100M TI^(+)` with `25.00 mL` of `0.200M Co^(3+)`
`E^(@) (TI^(+)//TI^(3+)) =- 1.25 V, E^(@) (Co^(3+)//Co^(2+)) = 1.84V`

Explanation

To solve the problem of calculating the equilibrium concentration of all ions in the solution prepared by mixing `25.00 mL` of `0.100 M Ti^(+)` with `25.00 mL` of `0.200 M Co^(3+)`, we will follow these steps:

### Step 1: Calculate Initial Moles of Each Ion
1. **Calculate moles of Ti^(+)**:
\[ \text{Moles of Ti}^{+} = \text{Molarity} \times \text{Volume} = 0.100 \, \text{M} \times 0.025 \, \text{L} = 0.0025 \, \text{moles} \]

2. **Calculate moles of Co^(3+)**:
\[ \text{Moles of Co}^{3+} = \text{Molarity} \times \text{Volume} = 0.200 \, \text{M} \times 0.025 \, \text{L} = 0.0050 \, \text{moles} \]

### Step 2: Determine the Cell Potential (E°cell)
1. **Use the standard reduction potentials**:
\[ E^{\circ}_{\text{cell}} = E^{\circ}_{\text{cathode}} - E^{\circ}_{\text{anode}} = 1.84 \, \text{V} - (-1.25 \, \text{V}) = 1.84 + 1.25 = 3.09 \, \text{V} \]

### Step 3: Write the Half-Reactions
1. **Anode Reaction (Oxidation)**:
\[ \text{Ti}^{+} \rightarrow \text{Ti}^{3+} + 2e^{-} \]

2. **Cathode Reaction (Reduction)**:
\[ \text{Co}^{3+} + e^{-} \rightarrow \text{Co}^{2+} \]

### Step 4: Balance the Electrons
1. **Multiply the cathode reaction by 2** to balance the electrons:
\[ 2\text{Co}^{3+} + 2e^{-} \rightarrow 2\text{Co}^{2+} \]

### Step 5: Write the Overall Reaction
1. **Combine the half-reactions**:
\[ \text{Ti}^{+} + 2\text{Co}^{3+} \rightarrow \text{Ti}^{3+} + 2\text{Co}^{2+} \]

### Step 6: Calculate the Change in Moles
1. **Initial Moles**:
- Ti^(+): 0.0025 moles
- Co^(3+): 0.0050 moles
- Co^(2+): 0 moles
- Ti^(3+): 0 moles

2. **Change in Moles**:
- Ti^(+): 0.0025 - 0.0025 = 0
- Co^(3+): 0.0050 - 0.0050 = 0
- Co^(2+): 0 + 0.0050 = 0.0050
- Ti^(3+): 0 + 0.0025 = 0.0025

### Step 7: Calculate Final Concentrations
1. **Total Volume**:
\[ V_{\text{total}} = 25.00 \, \text{mL} + 25.00 \, \text{mL} = 50.00 \, \text{mL} = 0.050 \, \text{L} \]

2. **Concentrations**:
- \([Ti^{3+}] = \frac{0.0025 \, \text{moles}}{0.050 \, \text{L}} = 0.050 \, \text{M}\)
- \([Co^{2+}] = \frac{0.0050 \, \text{moles}}{0.050 \, \text{L}} = 0.100 \, \text{M}\)
- \([Co^{3+}] = 0 \, \text{M}\)
- \([Ti^{+}] = 0 \, \text{M}\)

### Final Answer
- \([Ti^{+}] = 0 \, \text{M}\)
- \([Ti^{3+}] = 0.050 \, \text{M}\)
- \([Co^{3+}] = 0 \, \text{M}\)
- \([Co^{2+}] = 0.100 \, \text{M}\)