The $ EMF $ of a galvanic cell by coupling two electrodes $ M_{1} \left|M_{1}^{2+}\left(0.1 \, M\right)\right|\left|M_{2}^{2+}\left(0.01\,M\right)\right|M_{2} $ is $ +1.47\,V $ . If the $ E^{\circ} $ value (reduction potential) of $ M_{2} $ electrode is $ 0.9\, V, E^{\circ} $ (reduction potential) value of $ M_{1} $ electrode in volts would be
$ \left[Assume \frac{2.303RT\left(T=298\,k\right)}{F}=0.06\right] $