Question

If $E_{A{u}^{+}/Au}^0$ is $1.69V$ and $E_{A{u}^{3+}/Au}^0$ is $1.40V$, then $E_{A{u}^{+}/A{u}^{3+}}^0$ will be :

• A. $0.19V$
• B. $2.945V$
• C. $1.255V$
• D. $noneofthese$

Answer 1

Answer: (C)

$1.255V$

$A{u}^{+}+e^{-}\to Au\left(s\right)$ $E^{\circ }=1.69V$ .............(1) $\Delta G_1^{\circ }$
$A{u}^{3+}+3e^{-}\to Au\left(s\right)$ $E^{\circ }=1.40V$ ..................(2) $\Delta G_2^{\circ }$
From (2) - (1)
$A{u}^{3+}+2e^{-}\to A{u}^{+}$ ;
$\Delta G_3^{\circ }=\Delta G_2^{\circ }-\Delta G_1^{\circ }$
$-2\times F\times F^{\circ }=-3\times F\times 1.40+1\times 1.69\times F;E^{\circ }=1.255V$

$\therefore$ $E_{A{u}^{+}|A{u}^{3+}}^{\circ }=-1.255V$

Option (C) is correct.