8.23 ×10-3 mol of InCl (s) is placed in 1.00 of 0.010 M HCl (aq) at 25C. The InC

Question

8.23 ×10-3 mol of InCl (s) is placed in 1.00 of 0.010 M HCl (aq) at 25C. The InCl (s) dissolves quite quickly. and then the following reaction occurs:

3In+ (aq) 2 In (s) + In3+ (aq)
As this disproportionation proceeds, the solution is analyzed at intervals to determine the concentration on In+ (aq) that remains.

What will be the standard cell potential for this reaction?

Time (s)                                               [In+] (mol. L-1)

0                                                            8.23 × 10-3

240                                                        6.41 × 10-3

480                                                        5.00 × 10-3

720                                                        3.89 × 10-3

1000                                                      3.03 × 10-3

1200                                                     3.03 × 10-3

10000                                                    3.03 × 10-3

Explanation / Answer

For the given disproportionation reaction,

Standard cell potential of the cell,

Eo = Ecathode - Eanode

     = -0.14 - (-0.443)

     = 0.303 V

Related Questions

Navigate

• Browse (All)
• Browse 8
• Subjects

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.