`Pt(s) H_(2)(g)(1bar)|H^(+)(aq)(1M)||M^(+3)(aq),M^(+)(aq)|Pt(s)` The `E_cell` for the given cell is `0.1115V` at `298K` when `([M^(+)(aq)]/[M^(+3)(aq)])=10^(a)` The value of a is Given: `E_(M^(+3)//M^(+))^@=0.2V`
`(2.303RT)/(F)=0.059V`

Answer

To solve the given electrochemical cell problem, we will follow these steps:

### Step 1: Write the Nernst Equation
The Nernst equation relates the cell potential (E_cell) to the standard cell potential (E°) and the concentrations of the reactants and products. The equation is given by:

\[ E_{cell} = E^{\circ} - \frac{RT}{nF} \ln Q \]

Where:
- \(E_{cell}\) = cell potential (0.1115 V)
- \(E^{\circ}\) = standard cell potential
- \(R\) = universal gas constant (8.314 J/(mol·K))
- \(T\) = temperature in Kelvin (298 K)
- \(n\) = number of moles of electrons transferred
- \(F\) = Faraday's constant (96485 C/mol)
- \(Q\) = reaction quotient

### Step 2: Identify the Standard Cell Potential (E°)
The standard reduction potential for the half-reaction \(M^{+3} + 3e^- \rightarrow M^{+}\) is given as \(E^{\circ}_{M^{+3}/M^{+}} = 0.2 V\). The standard potential for the hydrogen half-reaction is \(E^{\circ}_{H^{+}/H_{2}} = 0 V\).

The overall standard cell potential can be calculated as:

\[ E^{\circ}_{cell} = E^{\circ}_{cathode} - E^{\circ}_{anode} = 0.2 V - 0 V = 0.2 V \]

### Step 3: Calculate the Reaction Quotient (Q)
The reaction quotient \(Q\) for the given cell can be expressed as:

\[ Q = \frac{[M^{+}]}{[M^{+3}]} \]

Given that \(\frac{[M^{+}]}{[M^{+3}]} = 10^{a}\), we can substitute this into the equation.

### Step 4: Substitute Values into the Nernst Equation
Now we can substitute the known values into the Nernst equation:

\[ 0.1115 V = 0.2 V - \frac{(2.303)(0.059 V)}{n} \log(10^{a}) \]

Here, \(n = 3\) (since 3 electrons are transferred in the reduction of \(M^{+3}\) to \(M^{+}\)).

### Step 5: Simplify the Equation
Substituting \(n = 3\) into the equation gives:

\[ 0.1115 V = 0.2 V - \frac{(2.303)(0.059 V)}{3} \cdot a \]

Calculating the term \(\frac{(2.303)(0.059 V)}{3}\):

\[ \frac{(2.303)(0.059)}{3} \approx 0.0452 V \]

Now substituting this back into the equation:

\[ 0.1115 V = 0.2 V - 0.0452 V \cdot a \]

### Step 6: Solve for \(a\)
Rearranging the equation to isolate \(a\):

\[ 0.0452 V \cdot a = 0.2 V - 0.1115 V \]

Calculating the right side:

\[ 0.0452 V \cdot a = 0.0885 V \]

Now, solving for \(a\):

\[ a = \frac{0.0885 V}{0.0452 V} \approx 1.96 \]

### Final Answer
The value of \(a\) is approximately **2**.