$2 {K M n O}_{4} + 5 H_{2} S + {6 H}^{+} \to {2 M n}^{2 +} + {2 K}^{+} + 5 S + {8 H}_{2} O$ . In the above reaction, how many electrons would be involved in the oxidation of $1$ mole of reductant?

Two

Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses

Five

No worries! We‘ve got your back. Try BYJU‘S free classes today!

Ten

No worries! We‘ve got your back. Try BYJU‘S free classes today!

One

No worries! We‘ve got your back. Try BYJU‘S free classes today!

Open in App

Solution

The correct option is B Two
$2 K M n O_{4} + 5 H_{2} S + {6 H}^{+} ⟶ 2 {M n}^{2 +} + {2 K}^{+} + 5 S + 8 H_{2} O$
The oxidation state of $S$ in $H_{2} S$ is $- 2$ & in $S$ in $0$
$∴ H_{2} S$ is a reductant.
$∴$ No of electrons involved in oxidation of $1$ mole of reductant $= 2$

Suggest Corrections

Similar questions

2 M n O_{4}^{-} + 5 H_{2} C_{2} O_{4} + 6 H^{+} \to 2 M n^{2 +} + 10 C O_{2} + 8 H_{2} O

O^{18}

2 M n O_{4}^{-} + 5 H_{2} O_{2}^{18} + 6 H^{+} \to 2 M n^{2 +} + 8 H_{2} O + 5 O_{2}

{2 m n O}_{4}^{-} + 6 H^{+} + {5 H}_{2} C_{2} O_{4} ⇌ {2 M N}^{2 +} + 8 H_{2} O + {10 C O}_{2}

E_{{M n O}_{4} M n^{2 +},}^{0} = 1.51 V

E_{{C O}_{3} C_{2} O_{4}^{2 -}}^{0} = - 0.49 V

298 K

2 M n O_{4}^{-} + 6 H^{+} + 5 H_{2} C_{2} O_{4} \to 2 M n^{2 +} + 8 H_{2} O + 10 C O_{2} (E_{c e l l}^{\circ} = 2.0 V)

2 M n O_{4}^{-} + 5 C_{2} O_{4}^{2 -} + 16 H^{+} ⟶ 10 {C O}_{2} + 2 {M n}^{2 +} + 8 H_{2} O

Join BYJU'S Learning Program