How many Faradays are required to reduce 1 mol of $ Cr_2O_7^{2-} $ to $ Cr^{3+} $ in acidic medium?

Question

Consider following reaction :
$Fe(OH)_2 + 2e^- \rightarrow Fe(s) + 2OH^-$ , $E^{\circ} = -0.88 \text{ V}$
$AgBr(s) + e^- \rightarrow Ag(s) + Br^-(aq)$ , $E^{\circ} = 0.07 \text{ V}$
Select the correct statement.
(1) $E^{\circ}_{cell} = -0.95 \text{V}$
(2) $E^{\circ}_{cell}$ is an extensive property
(3) $Fe$ is getting reduced
(4) Net cell reaction: $Fe(s) + 2OH^-(aq) + 2AgBr(s) \rightarrow Fe(OH)_2 + 2Ag(s) + 2Br^-(aq)$

Answer 1

Step 1: Conceptual Understanding:
The number of Faradays required for a reduction process is determined by the total number of moles of electrons transferred for each mole of substance being reduced. This value is directly related to the change in oxidation state of the substance.
Step 2: Detailed Explanation:
Consider the dichromate ion $ Cr_2O_7^{2-} $, where the oxidation state of chromium ($ Cr $) is $ +6 $.
When reduced to $ Cr^{3+} $, the oxidation state of chromium decreases to $ +3 $.
The change in oxidation state for each $ Cr $ atom is $ 6 - 3 = 3 $. Since there are two chromium atoms in each mole of $ Cr_2O_7^{2-} $, the total number of moles of electrons required for the reduction is $ 2 \times 3 = 6 $ moles of electrons.
Since the charge of 1 mole of electrons is equivalent to 1 Faraday (F), a total of 6 Faradays are required to reduce 1 mole of $ Cr_2O_7^{2-} $.
Step 3: Conclusion:
Therefore, a total of 6 Faradays are required to reduce 1 mole of $ Cr_2O_7^{2-} $ to $ Cr^{3+} $.

How many Faradays are required to reduce 1 mol of \( Cr_2O_7^{2-} \) to \( Cr^{3+} \) in acidic medium?

Show Hint

The Correct Option is C

Solution and Explanation

Top Questions on Electrochemistry

Questions Asked in CBSE Class XII exam