QCAA Chemistry Oxidation and reduction

+4

Vanadium in $V_2O_5(s)$ is acting as an oxidising agent in reaction 1 as the oxidation state decreased from +5 to +4 in $V_2O_4(s)$.

Reaction 1: $V_2O_5(s)$ is converted to $V_2O_4(s)$ to allow $SO_2$ to be converted to $SO_3$.
Reaction 2: $V_2O_4(s)$ is converted back to $V_2O_5(s)$ by reacting with $O_2$.
Therefore, $V_2O_5(s)$ acts as a catalyst because it undergoes a temporary chemical change during the reaction but remains chemically unchanged at the end.

Z

$Q^{2+}$

$Z(NO_3)_2(aq)$ and A(s)

$\text{Z}^{2+} + 2e^- \rightleftharpoons \text{Z}$ $\text{A}^{2+} + 2e^- \rightleftharpoons \text{A}$
$\text{X}^{2+} + 2e^- \rightleftharpoons \text{X}$
$\text{Q}^{2+} + 2e^- \rightleftharpoons \text{Q}$
Z can reduce $\text{X}^{2+}$, $\text{Q}^{2+}$ and $\text{A}^{2+}$, therefore it is the strongest reducing agent.
X can only reduce $\text{Q}^{2+}$ therefore it is the second weakest reducing agent.
Q cannot reduce any metal ions, therefore it is the weakest reducing agent.

$3 \times (+1) + As + 3 \times (-2) = 0$
$3 + As - 6 = 0$
$As - 3 = 0$
$As = +3$

Balanced oxidation half-equation:
$H_3AsO_3 + H_2O \rightarrow H_3AsO_4 + 2H^+ + 2e^-$

Balanced reduction half-equation:
$NO_3^- + 2H^+ + e^- \rightarrow NO_2 + H_2O$

Multiply by 2:
$2NO_3^- + 4H^+ + 2e^- \rightarrow 2NO_2 + 2H_2O$

Balanced redox equation:
$H_3AsO_3 + 2NO_3^- + 2H^+ \rightarrow H_3AsO_4 + 2NO_2 + H_2O$

$NO_3^-$