Balance the following redox reaction- Cr- - Sn4- Cr3- - Sn2-

The correct option is B 3Cr^+ + Sn^4+→ Cr^3+ + 2Sn^2+ Cr^++Sn^4+⟶ Cr^3+ + Sn^2+ Oxidation reduction method : Cr^+ ⟶ Cr^3++2e^ Sn^4++2e^ ⟶ ...

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

Ion Electron Method

Balance the f...

Question

Balance the following redox reaction:
$C r^{+} + S n^{4 +} \to C r^{3 +} + S n^{2 +}$

C r^{+} + S n^{4 +} \to C r^{3 +} + S n^{2 +}

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

3 C r^{+} + S n^{4 +} \to C r^{3 +} + 2 S n^{2 +}

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

C r^{+} + 4 S n^{4 +} \to C r^{3 +} + 2 S n^{2 +}

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

C r^{+} + 4 S n^{4 +} \to 3 C r^{3 +} + 2 S n^{2 +}

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

Open in App

Solution

Suggest Corrections

Similar questions

C r^{+} + S n^{4 +} \to C r^{3 +} + S n^{2 +}

2 S n^{2 +} \to S n^{4 +} + S n

S n^{4 +} | S n^{2 +}

+ 0.15 V

C r^{3 +} | C r

- 0.74 V

E_{C r^{3 +} / C r}^{\circ} = - 0.72 V, E_{F e^{2 +} / F e}^{\circ} = - 0.42 V

C r | C r^{3 +} (0.1 M) | | F e^{2 +} (0.01 M) | F e

E_{C r^{3 +} / C r}^{\circ} = - 0.72, E_{F e^{2 +} / F e}^{\circ} = - 0.42 V

C r | C r^{3 +} (0.1 M) | | F e^{2 +} (0.01 M) | F e

Join BYJU'S Learning Program