The number of electrons required to reduce chromium completely in $Cr_2O_7^{2-}$ to $Cr^{3+}$ in acidic medium is:

For the redox reaction $MnO_4^- + C_2O_4^{2-} + H^{+} \to Mn^{2+} + CO_2 + H_2O$,the correct coefficients of the reactants for the balanced reaction are:
$MnO_4^-$ : $C_2O_4^{2-}$ : $H^{+}$

The reaction,$K_2Cr_2O_7 + m \, FeSO_4 + n \, H_2SO_4 \longrightarrow Cr_2(SO_4)_3 + p \, Fe_2(SO_4)_3 + K_2SO_4 + q \, H_2O$ when balanced,$m, n, p$ and $q$ are,respectively:

For the balanced redox reaction $a C_2O_4^{2-} + b MnO_4^- + c H^+ \rightarrow x Mn^{2+} + y H_2O + z CO_2$, the values of $a$ and $x$ respectively are:

$Cr _2 O _7^{2-} \rightarrow Cr ^{3+}_{aq}$
How many moles of $Sn^{2+}$ will be oxidized by $1 \ mol \ Cr_2O_7^{2-}$ into $Sn^{4+}$?

The coefficients $x, y, p, q,$ and $r$ in the following balanced equation are respectively:
$xMnO_4^{2-}{_{\text{(aq)}}} + yH_2O_{\text{(l)}} \rightarrow pMnO_2{_{\text{(s)}}} + qMnO_4^{-}{_{\text{(aq)}}} + rOH^{-}{_{\text{(aq)}}}$