Pt(s)|H$$_2$$(g)(1 bar)|H$$^+$$(aq)(1M)||M$$^{3+}$$(aq), M$$^+$$(aq)|Pt(s)
The E$$_{cell}$$ for the given cell is 0.1115 V at 298 K
When $$\frac{[M^+(aq)]}{[M^{3+}(aq)]} = 10^a$$
The value of a is _____
Given: E$$^0_{M^{3+}/M^+}$$ = 0.2 V
$$\frac{2.303RT}{F}$$ = 0.059 V

The given electrochemical cell is

$$Pt(s)|H_2(g,1\ \text{bar})|H^+(aq,1\ M)||M^{3+}(aq),M^+(aq)|Pt(s)$$

The corresponding half-cell reactions are:

• Anode (oxidation):

$$H_2(g)\rightarrow 2H^+(aq)+2e^-$$

• Cathode (reduction):

$$M^{3+}(aq)+2e^-\rightarrow M^+(aq)$$

Hence, the overall cell reaction is

$$H_2(g)+M^{3+}(aq)\rightarrow 2H^+(aq)+M^+(aq)$$

The standard cell potential is

$$E^\circ_{\text{cell}}=E^\circ_{\text{cathode}}-E^\circ_{\text{anode}}$$

$$E^\circ_{\text{cell}}=0.20-0=0.20\ \text{V}$$

Using the Nernst equation at $$298\ \text{K}$$,

$$E_{\text{cell}}=E^\circ_{\text{cell}}-\frac{0.059}{n}\log Q$$

where $$n=2$$ and

$$Q=\frac{[H^+]^2[M^+]}{[H_2][M^{3+}]}$$

Since $$[H^+]=1\ \text{M}$$ and $$P_{H_2}=1\ \text{bar}$$,

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

Substituting the given values,

$$0.1115=0.20-\frac{0.059}{2}\log\left(\frac{[M^+]}{[M^{3+}]}\right)$$

$$0.1115=0.20-0.0295\log\left(\frac{[M^+]}{[M^{3+}]}\right)$$

Rearranging,

$$0.0295\log\left(\frac{[M^+]}{[M^{3+}]}\right)=0.20-0.1115$$

$$0.0295\log\left(\frac{[M^+]}{[M^{3+}]}\right)=0.0885$$

$$\log\left(\frac{[M^+]}{[M^{3+}]}\right)=\frac{0.0885}{0.0295}=3$$

Since

$$\frac{[M^+]}{[M^{3+}]}=10^a$$

we have

$$\log(10^a)=a=3$$

Conclusion

The required value is

$$\boxed{a=3}$$

Hence, the correct answer is 3.