If triangle G-smallcirc- the half cell MnO-4-MnO-2 in an acid solution is xF then the value of x-Given- E-smallcirc-MnO-4-Mn-2- - 1-5V- space E-smallcirc-MnO-2-Mn-2- - 1-25V- -0-75 -0-95 None of these -0-85

Question

If triangle G-smallcirc- the half cell MnO-4-MnO-2 in an acid solution is xF then the value of x-Given- E-smallcirc-MnO-4-Mn-2- - 1-5V- space E-smallcirc-MnO-2-Mn-2- - 1-25V- -0-75  -0-95 None of these -0-85

Toppr Admin · Accepted Answer

The expression for the standard free energy change is $Δ G^{0} = - n F E^{0}$ But it is equal to xF.
Hence, $x = \frac{- n F E^{0}}{F} = - n E^{0}$
But for the given half reaction, $n = 3$ and $E^{0} = E_{M n O_{4}^{-} | M n^{2 +}}^{0} - E_{M n O_{2} | M n^{2 +}}^{0} 1.5 - 1.25 = 0.25 V$
Hence, $x = - n E^{0} = - 3 \times 0.25 = - 0.75$

If △G∘ for the half cell MnO−4|MnO2 in an acid solution is xF then find the value of x.(Given: E∘MnO−4|Mn2+=1.5V; E∘MnO2|Mn2+=1.25V)−0.75−0.95None of these−0.85

The expression for the standard free energy change is ΔG0=−nFE0 But it is equal to xF.Hence, x=−nFE0F=−nE0But for the given half reaction, n=3 and E0=E0MnO−4|Mn2+−E0MnO2|Mn2+1.5−1.25=0.25VHence, x=−nE0=−3×0.25=−0.75

If $△ G^{\circ}$ for the half cell $M n O_{4}^{-} | M n O_{2}$ in an acid solution is $x F$ then find the value of $x$ .
(Given: $E_{M n O_{4}^{-} | M n^{2 +}}^{\circ} = 1.5 V; E_{M n O_{2} | M n^{2 +}}^{\circ} = 1.25 V$ )
$- 0.75$
$- 0.95$
None of these
$- 0.85$