In the ionic equation $BiO_{3}^{-} + 6H^{+} + x e^{-} \longrightarrow Bi^{3+} + 3H_{2}O$,the value of $x$ is:

In acidic medium,$K_2Cr_2O_7$ shows oxidising action as represented in the half reaction:
$Cr_2O_7^{2-} + XH^+ + Ye^- \rightarrow 2A + ZH_2O$
$X, Y, Z$ and $A$ are respectively:

The heating of $(NH_4)_2Cr_2O_7$ produces another chromium compound along with $N_2$ gas. The change of the oxidation state of $Cr$ in the reaction is

Aqueous solutions of $Na_2S_2O_3$ on reaction with $Cl_2$ gives

How many moles of electrons are required to reduce $2.5 \ mol$ of $Cr_2O_7^{2-}$ to $Cr^{3+}$?

Balance the following equations in basic medium by ion-electron method and oxidation number methods and identify the oxidising agent and the reducing agent.
$(a)$ $P_{4(s)} + OH_{(aq)}^{-} \to PH_{3(g)} + HPO_{2(aq)}^{-}$
$(b)$ $N_{2}H_{4(l)} + ClO_{3(aq)}^{-} \to NO_{(g)} + Cl_{(g)}^{-}$
$(c)$ $Cl_{2}O_{7(g)} + H_{2}O_{2(aq)} \to ClO_{2(aq)}^{-} + O_{2(g)} + H^{+}$