If E∘M+∣M=1.7V and E∘M3+∣M=1.4V then E∘M3+,M+∣Ptwill be about - Sri Chaitanya Infinity Learn Best Online Courses for NCERT Solutions, CBSE, ICSE, JEE, NEET, Olympiad and Class 6 to 12

We have

$(i) M^{+} + e^{-} \to M; E^{\circ} = 1.7 V; Δ G^{\circ} = - F (1.7 V)$

$(ii) M^{3 +} + 3 e^{-} \to M; E^{\circ} = 1.4 V; Δ G^{\circ} = - 3 F (1.4 V)$

Thus $M^{3 +} + 2 e^{-} \to M^{+}$ ; $Δ G^{\circ} = - 2 F (E^{\circ} M^{3 +}, M^{+} ∣ Pt)$

Equation (iii) is obtained by subtracting Eq. (i) from Eq. (ii). Hence

$- 2 F (E_{M^{3 +} \cdot M^{+} ∣ Pt}^{\circ}) = - 3 F (1.4 V) + F (1.7 V)$ or $E_{M^{3 +}, M^{+} {Pt}^{\circ}} = 1.25 V$

If $E^{\circ} (M^{+} ∣ M) = 1.7 V$ and $E^{\circ} (M^{3 +} ∣ M) = 1.4 V$ then $E^{\circ} (M^{3 +}, M^{+} ∣ Pt)$ will be about