Arrange these species by their ability to act as an oxidizing agent. Best oxidizing agent...

Step 1
Given ionic species:
${\displaystyle N{i}^{2+},M{g}^{2+},Z{n}^{2+},A{u}^{3+}}$
Step 2
From the standard reduction potential data
${\displaystyle N{i}^{2+}+2e^{-}\Rightarrow Ni}$
${\displaystyle E^{\circ }=-0.26v}$
${\displaystyle M{g}^{2+}+2e^{-}\Rightarrow Mg}$
${\displaystyle E^{\circ }=-2.37v}$
${\displaystyle Z{n}^{2+}+2e^{-}\Rightarrow Zn}$
${\displaystyle E^{-}=-0.76v}$
${\displaystyle A{u}^{3+}+3e^{-}\Rightarrow Au}$
${\displaystyle E^{\circ }=1.40v}$
Reduction potential more act as strong oxidining agent means it
Can take electrons from less reducetion potentid species
Based on this: best oxiding agent is ${\displaystyle A{u}^{3+}}$
Poor oxiding agent: ${\displaystyle M{g}^{2+}}$
Oxidizing agent Capacity increasing order ${\displaystyle M{g}^{2+}<Z{n}^{2+}<N{i}^{2+}<A{u}^{3+}}$
Best oxidizing agent
${\displaystyle A{u}^{3+}}$
${\displaystyle N{i}^{2+}}$
${\displaystyle Z{n}^{2+}}$ Poor oxigizing agent ${\displaystyle M{g}^{2+}}$

Solution:
${\displaystyle N{i}^{2+}+2e^{-}\Rightarrow Ni{E}^0=-0.26v}$
${\displaystyle Z{n}^{2+}+2e^{-}\Rightarrow Zn{E}^0=-0.76v}$
${\displaystyle M{g}^{2+}+2e^{-}\Rightarrow Mg{E}^0=-2.37v}$
${\displaystyle A{u}^{3+}+3e^{-}\Rightarrow Au{E}^0=1.40v}$
Oxidizing agent capacity increasing order:
${\displaystyle M{g}^{2+}<Z{n}^{2+}<N{i}^{2+}<A{u}^{3+}}$

Solution from given: then $(1x+5)=(4x+1)$ $\{\text{the qualioy of parallelegram}\}$ $-1x=-4$ $x=2$ then $PQ=2x+5=2\times 2+5=9cm$