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100 ml solution of each 0.1 M AuCl and 0...

100 ml solution of each 0.1 M AuCl and 0.1 AgCl is electrolysed by passing 1 amp current for 15 minutes then which of the following will be deposited?
given `Au^+ + e^- rarr Au E^@ = 1.69V`
`Ag^+ + e^- rarr Ag E^@ = 0.80V`

A

only Au

B

Only Ag

C

Both Au & Ag

D

None of Au & Ag

Text Solution

AI Generated Solution

The correct Answer is:
To solve the problem, we need to determine which metal, gold (Au) or silver (Ag), will be deposited during the electrolysis of their respective solutions when a current is passed. ### Step-by-Step Solution: 1. **Determine the Charge Passed (Q)**: - Current (I) = 1 A - Time (t) = 15 minutes = 15 × 60 seconds = 900 seconds - Charge (Q) = I × t = 1 A × 900 s = 900 C 2. **Calculate the Moles of Electrons (n)**: - Faraday's constant (F) = 96500 C/mol - Moles of electrons (n) = Q / F = 900 C / 96500 C/mol ≈ 0.00932 mol 3. **Determine the Moles of Each Metal Ion in Solution**: - For both AuCl and AgCl, the concentration is 0.1 M in a 100 ml solution. - Moles of AuCl = 0.1 mol/L × 0.1 L = 0.01 mol - Moles of AgCl = 0.1 mol/L × 0.1 L = 0.01 mol 4. **Identify the Standard Reduction Potentials**: - For Au: Au⁺ + e⁻ → Au, E° = 1.69 V - For Ag: Ag⁺ + e⁻ → Ag, E° = 0.80 V 5. **Determine Which Metal Will Be Deposited**: - The metal with the higher standard reduction potential will be reduced first. In this case, Au (1.69 V) has a higher reduction potential than Ag (0.80 V). - Therefore, Au will be deposited first. 6. **Check if There is Sufficient Charge to Deposit Gold**: - The moles of electrons available (0.00932 mol) is less than the moles of Au available (0.01 mol). - Since the charge available is sufficient to deposit gold before the current runs out, gold will be deposited. ### Conclusion: - Only gold (Au) will be deposited during the electrolysis.
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