Cr2O72-I- I2-Cr3-Eocell-0-79- EoCr2O72-1-33 V- Eot2

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

Standard Electrode Potentials

Cr2O72-+I-→ I...

Question

$C r_{2} O_{7}^{2 -} + I^{-} \to I_{2} + C r^{3 +} E_{c e l l}^{o} = 0.79; E_{C r_{2} O_{7}^{2 -}}^{o} = 1.33 V, E_{t_{2}}^{o}$

- 0.10 V

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

+ 0.18

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

- 0.54 V

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

0.54 V

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

Open in App

Solution

The correct option is D $0.54 V$
$I^{-}$ get oxidized to $I_{2}$ hence will form anode and get reduced to $C r^{3 +}$ hence will form cathode.
$E_{c e l l}^{o} = E_{c a t h o d e}^{o} - E_{a n o d e}^{o}$
, $E_{c e l l}^{o} = E_{C r_{2} O_{7}^{- 2}} - E_{I_{2}}^{o}$ .
0.79=1.33− $E_{I_{2}}^{o}$ ,
$E_{I_{2}}^{o}$ =1.33−0.79,
$E_{I_{2}}^{o}$ =0.54V.

Suggest Corrections

Similar questions

Q. Given

E_{C l_{2} / C l^{-}}^{o} = 1.36 V, E_{C r^{3 +} / C r}^{\circ} = - 0.74 V

E_{C r_{2} O_{7}^{2 -} / C r^{3 +}}^{o} = 1.33 V, E_{M n O_{4}^{-} / M n^{2 +}}^{o} = 1.51 V

.
Among the following, the strongest reducing agent is :

Q. Calculate the equilibrium constant for the following reaction:

3 S n + 2 C r_{2} O_{7}^{2 -} + 28 H^{+} \to 3 S n^{4 +} + 4 C r^{3 -} + 14 H_{2} O

E_{S n / S n^{4 +}}^{o} = 0.009 V

E_{C r_{2} O_{7}^{2 -} / C r^{3 -}}^{o} = 1.33 V

Given that : $I_{2} + 2 e^{-} \to 2 I^{-}; E_{I_{2} / I^{-}}^{o}$ = 0.54V; $B r_{2} + 2 e^{-} \to 2 B r^{-}; E_{B r_{2} / B r^{-}}^{o}$ = 1.69V. Predict which of the following is true:

Q. Balance the following equations by ion electron method.

{C r_{2} O_{7}}^{2 -} + I^{-} + H^{+} \to C r^{3 +} + I_{2} + H_{2} O

Q. Balance the following equations by oxidation number method.

{C r_{2} O_{7}}^{2 -} + I^{-} + H^{+} \to C r^{3 +} + I_{2} + H_{2} O

Join BYJU'S Learning Program

Cr2O2−7+I−→I2+Cr3+Eocell=0.79;EoCr2O2−7=1.33V,Eot2

The correct option is D 0.54VI− get oxidized to I2 hence will form anode and get reduced to Cr3+ hence will form cathode.Eocell=Eocathode−Eoanode,Eocell=ECr2O−27−EoI2.0.79=1.33−EoI2,EoI2=1.33−0.79,EoI2=0.54V.

$C r_{2} O_{7}^{2 -} + I^{-} \to I_{2} + C r^{3 +} E_{c e l l}^{o} = 0.79; E_{C r_{2} O_{7}^{2 -}}^{o} = 1.33 V, E_{t_{2}}^{o}$