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Step-by-Step Solution
Step 1: List the known quantities
• Current, I = 2\\,\text{A}
• Time, t = 5\\,\text{hours} = 5 \times 3600 = 18000\\,\text{seconds}
• Mass of metal deposited, W = 22.2\\,\text{g}
• Atomic mass of metal, M = 177
Step 2: Use Faraday’s First Law of Electrolysis
According to Faraday’s first law of electrolysis, the mass of a substance ( W ) deposited is given by:
W = \frac{I\,t\,E}{96500}
Here, E is the equivalent mass of the metal and 96500 C/mol is approximately the value of the Faraday constant.
Step 3: Calculate the equivalent mass E
From W = \frac{I\,t\,E}{96500} , rearrange to solve for E :
E = \frac{W \times 96500}{I \times t}
Substitute the numerical values:
E = \frac{22.2 \times 96500}{2 \times 18000} \approx 59.5\\,\text{g/equivalent}
Step 4: Determine the oxidation state n
For a metal of atomic mass M deposited in oxidation state n , the equivalent mass E is:
E = \frac{M}{n}
So,
59.5 = \frac{177}{n}
Solving for n :
n = \frac{177}{59.5} \approx 2.97
Step 5: Round to the nearest whole number
Rounding 2.97 to the nearest whole number gives 3 .
Therefore, the oxidation state of the metal in its salt is:
\boxed{3}
