Question

$6\times {10}^{-3}$ moles of $K_{2}C{r}_2{O}_7$ react completely with $9\times {10}^{-3}$ moles of $X^{n+}$ to give $X{O}_3^{-}$ and $C{r}^{3+}$. The value of $n$ is :

• A. 1
• B. 2
• C. 3
• D. None of these

Answer 1

The correct option is A 1

When we write the oxidation and reduction halves we find that
Cr is getting reduced from +6 to +3 and since there are two Cr ions so the n factor for Cr will be 6.
Similarly X will get oxidized from n to +5, so the n factor for X will be 5-n
We know that:
The number of equivalents of $K_{2}C{r}_2{O}_7$ = the number of equivalents of $X^{n+}$
So, moles of $K_{2}C{r}_2{O}_7$ × n factor of $K_{2}C{r}_2{O}_7$ = moles of $X^{n+}$ × n factor of $X^{n+}$
$6\times {10}^{-3}\times 6=9\times {10}^{-3}(5-n)$
$4=5-n$
$n=1$