.015 \mathrm{~A}$ for. Tags and topics: Electrochemistry,Faraday's Laws | Conductance of Electrolytes | Kohlrausch Law,JEE MAIN,JEE Advanced,JEE Mains,NEET, Class 11 Chemistry,Class 12 Chemistry,JEE Main Chemistry.">
.015 \mathrm{~A}$ for. Tags and topics: Electrochemistry,Faraday's Laws | Conductance of Electrolytes | Kohlrausch Law,JEE MAIN,JEE Advanced,JEE Mains,NEET, Class 11 Chemistry,Class 12 Chemistry,JEE Main Chemistry.">
.015 \mathrm{~A}$ for. Tags and topics: Electrochemistry,Faraday's Laws | Conductance of Electrolytes | Kohlrausch Law,JEE MAIN,JEE Advanced,JEE Mains,NEET, Class 11 Chemistry,Class 12 Chemistry,JEE Main Chemistry.">
The mass of zinc produced by the electrolysis of zine sulphate solution with a steady current of $0.015 \mathrm{~A}$ for 15 minutes is _________ $\times 10^{-4} \mathrm{~g}$.
(Atomic mass of zinc $=65.4 \mathrm{~amu}$)
Solution
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Step-by-Step Solution
Step 1: Identify the Known Quantities
• Current, I = 0.015\,\mathrm{A}
• Time, t = 15\,\text{minutes} = 15 \times 60 = 900\,\mathrm{s}
• Atomic mass of zinc, M = 65.4\,\mathrm{g/mol}
• Charge on zinc during electrolysis, n = 2 (since \mathrm{Zn} is deposited from \mathrm{Zn}^{2+} )
• Faraday's constant, F = 96500\,\mathrm{C/mol} (approximately)
Step 2: Write the Formula for the Mass Deposited
The general formula for the mass m of a substance deposited during electrolysis is:
m = \frac{M \times I \times t}{n \times F},
where
• m = mass of metal deposited (in grams),
• M = molar mass of the metal (in g/mol),
• I = current (in amperes),
• t = time (in seconds),
• n = number of electrons transferred per ion (valency), and
• F = Faraday's constant (in C/mol).
Step 3: Substitute the Values
Substitute M = 65.4\,\mathrm{g/mol} , I = 0.015\,\mathrm{A} , t = 900\,\mathrm{s} , n=2, and F = 96500\,\mathrm{C/mol} into the formula:
m = \frac{65.4 \times 0.015 \times 900}{2 \times 96500}.
Step 4: Perform the Calculation
1. Multiply current by time:
0.015 \times 900 = 13.5.
2. Multiply by molar mass of Zn:
65.4 \times 13.5 = 882.9.
3. Multiply n \times F :
2 \times 96500 = 193000.
4. Divide to find m :
m = \frac{882.9}{193000} \approx 0.00458\,\mathrm{g}.
This can be written as 4.58 \times 10^{-3}\,\mathrm{g} .
Step 5: Express the Final Answer
The question specifically asks for the mass in terms of 10^{-4}\,\mathrm{g} . Converting 4.58 \times 10^{-3}\,\mathrm{g} :
4.58 \times 10^{-3}\,\mathrm{g} = 45.8 \times 10^{-4}\,\mathrm{g}.
Therefore, the mass of zinc produced is
\boxed{45.8 \times 10^{-4}\,\mathrm{g}}.