Question

If $E_{A{u}^{+}/Au}^{\circ }$ is 1.69 V & $E_{A{u}^{3+}/Au}^{\circ }$ is 1.40 V, then $E_{A{u}^{3+}/A{u}^{+}}^{\circ }$ will be:

• A. 0.19 V
• B. 2.945 V
• C. 1.255 V
• D. None

Answer 1

The correct option is C 1.255 V

Given,
$A{u}^{+}+e^{-}⟶Au\left(s\right)E^{\circ }=1.69V-\left(1\right)\left[\Delta G_1^{\circ }\right]$
$A{u}^{3+}+3e^{-}⟶Au\left(s\right)E^{\circ }=1.40V-\left(2\right)\left[\Delta G_2^{\circ }\right]$ From (2) - (1), we get
$A{u}^{3+}+2e^{-}⟶A{u}^{+}$ ---(3) $\left[\Delta G_3^{\circ }\right]$
So,
$\Delta G_3^{\circ }=\Delta G_2^{\circ }-\Delta G_1^{\circ }$
$-2\times F\times E^{\circ }=-3\times F\times 1.40+1\times 1.69\times F$
$E^{\circ }=1.255V$