Question

For the oxidation-reduction reaction;
$K_{2}C{r}_2{O}_7+X{H}_{2}S{O}_4+YS{O}_2\to K_{2}S{O}_4+C{r}_2(S{O}_4{)}_3+Z{H}_{2}O$
The values X, Y and Z are:

• A. 1, 3, 1
• B. 4, 1, 4
• C. 3, 2, 3
• D. 2, 1, 2

Answer 1

The correct option is A 1, 3, 1

From the reaction, $C{r}_2{O}_7$ is reduced

to$C{r}^{3+}andS{O}_2$ is oxidized to

$S{O}_4^2$

$C{r}_2{O}_7+14H++6e\to C{r}^{3+}+7H_{2}O$ (Reduction) (i)

$3S{O}_2+6H_{2}O\to 3S{O}_4^{2-}+12H^{+}+6e$ (Oxidation) (ii)

Addition of (i) and (ii) gives

$C{r}_2{O}_7+2H++3S{O}_2\to 3S{O}_4^{2-}+2C{r}_{3+}+H_{2}O$

$K_{2}C{r}_2{O}_7+H_{2}S{O}_4+3S{O}_2\to 3K_{2}S{O}_4+C{r}_2(S{O}_4{)}_3+H_{2}O$

$HenceX=1,Y=3andZ=1$