In acidic medium Cr2O7-2 is an oxidising agent Cr2O7-2 - 14 H- - 6e- - 2Cr-3 -7 H2O - E- Cr 2 O 7 -2 -Cr -3 -1-33V -The electrode potential of half cell at pH - 1 keeping concentration of other species unity is-

You visited us 1 times! Enjoying our articles? Unlock Full Access!

Equivalent, Molar Conductivity and Cell Constant

In acidic med...

Question

In acidic medium $C r_{2} O_{7}^{- 2}$ is an oxidising agent
$C r_{2} O_{7}^{- 2} + 14 H^{+} + 6 e^{-} \to 2 C r^{+ 3} + 7 H_{2} O$
$[E^{\circ} {C r}_{2} O_{7}^{- 2} / C r^{+ 3} = 1.33 V]$
The electrode potential of half cell at pH = 1 keeping concentration of other species unity is:

2.1574 V

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

0.5026 V

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

1.1921 V

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

1.4679 V

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

Open in App

Solution

Suggest Corrections

Similar questions

C r_{2} O_{7}^{- 2}

C r_{2} O_{7}^{- 2} + 14 H^{+} + 6 e^{-} \leftarrow 2 C r^{+ 3} + 7 H_{2} O

[E_{C r_{2} O_{7}^{- 2} / C r^{+ 3}}^{^{\circ}} = 1.33 V]

p H = 1

{C r}_{2} O_{7}^{- 2} + 14 H^{+} + 6 e^{-} \to 2 {C r}^{+ 3} + 7 H_{2} O; E^{o} = 1.33 V

[{C r}_{2} O_{7}^{- 2}] = 4.5

[{C r}^{+ 3}] = 15

E

1.067 V

C r_{2} O_{7}^{- 2} + F e^{+ 2} \to C r^{+ 3} + H_{2} O

C r_{2} O_{7}^{2 -} (a q, 1 M) + 14 H^{+} (a q) + 6 e^{-} \to 2 C r^{3 +} (a q, 1 M) + 7 H_{2} O

p H = 2.0

25^{\circ} C

C r^{+ 3}, C r_{2} O_{7}^{2 -}

p O H = 11

0.01 M

C r^{3 +}

C r_{2} O_{7}^{2 -}

C r_{2} O_{7}^{2 -} + 14 H^{+} + 6 e^{-} \to 2 C r^{3 +} + 7 H_{2} O

E^{\circ} = 1.33 V

Join BYJU'S Learning Program