#current calculation

Variation of Conductivity with Concentration and Measurement

What is the time (in seconds) required to deposit 0.355 g of cobalt from a CoSO₄ solution using a current of 0.2 A? (Ato

Co²⁺ + 2e⁻ → Co . 1 mol Co (59 g) requires 2F. Moles = (0.355/59) = 0.006017 mol , Charge = 0.006017 × 2 × 96500 = 1161.24 C . t = (Q/I) = (1161.24/0.2) = 5806.2 s .

Ref: NCERT Class 12 Chemistry > Chapter 2: Electrochemistry > Topic: Variation of Conductivity with Concentration and Measurement

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Electrolytic Cells and Electrolysis and Faraday's Laws

A dry cell delivers 0.25 A for 9650 s. How many grams of zinc are oxidized at the anode? (Atomic mass of Zn = 65 g/mol,

Charge = 0.25 × 9650 = 2412.5 C . Zn → Zn²⁺ + 2e⁻ , 1 mol Zn (65 g) requires 2F. Faradays = (2412.5/96500) = 0.025 F , Moles = (0.025/2) = 0.0125 mol , Mass = 0.0125 × 65 = 0.8125 g .

Ref: NCERT Class 12 Chemistry > Chapter 2: Electrochemistry > Topic: Electrolytic Cells and Electrolysis and Faraday's Laws

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Electrochemical Cells - Galvanic Cells and Electrode Potential

What is the time (in seconds) required to deposit 0.585 g of chromium from a Cr₂(SO₄)₃ solution using a current of 0.5 A

Cr³⁺ + 3e⁻ → Cr . 1 mol Cr (52 g) requires 3F. Moles = (0.585/52) = 0.01125 mol , Charge = 0.01125 × 3 × 96500 = 3256.875 C . t = (Q/I) = (3256.875/0.5) = 6513.75 s .

Ref: NCERT Class 12 Chemistry > Chapter 2: Electrochemistry > Topic: Electrochemical Cells - Galvanic Cells and Electrode Potential

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Electrochemical Cells - Galvanic Cells and Electrode Potential

What is the time (in seconds) required to deposit 0.585 g of chromium from a Cr₂(SO₄)₃ solution using a current of 0.5 A

Cr³⁺ + 3e⁻ → Cr . 1 mol Cr (52 g) requires 3F. Moles = (0.585/52) = 0.01125 mol , Charge = 0.01125 × 3 × 96500 = 3256.875 C . t = (Q/I) = (3256.875/0.5) = 6513.75 s .

Ref: NCERT Class 12 Chemistry > Chapter 2: Electrochemistry > Topic: Electrochemical Cells - Galvanic Cells and Electrode Potential

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Electrochemical Cells - Galvanic Cells and Electrode Potential

A current of 1 A deposits 0.635 g of Cu from CuSO₄ in 1930 s. What is the time required to deposit 0.54 g of Al from Al₂

Cu: Cu²⁺ + 2e⁻ → Cu , Moles = (0.635/63.5) = 0.01 mol , Charge = 0.01 × 2 × 96500 = 1930 C , matches 1 × 1930 . Al: Al³⁺ + 3e⁻ → Al , Moles = (0.54/27) = 0.02 mol , Charge = 0.02 × 3 × 96500 = 5790 C . t = (5790/1) = 5790 s .

Ref: NCERT Class 12 Chemistry > Chapter 2: Electrochemistry > Topic: Electrochemical Cells - Galvanic Cells and Electrode Potential

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Electrochemical Cells - Galvanic Cells and Electrode Potential

A current of 1 A deposits 0.635 g of Cu from CuSO₄ in 1930 s. What is the time required to deposit 0.54 g of Al from Al₂

Cu: Cu²⁺ + 2e⁻ → Cu , Moles = (0.635/63.5) = 0.01 mol , Charge = 0.01 × 2 × 96500 = 1930 C , matches 1 × 1930 . Al: Al³⁺ + 3e⁻ → Al , Moles = (0.54/27) = 0.02 mol , Charge = 0.02 × 3 × 96500 = 5790 C . t = (5790/1) = 5790 s .

Ref: NCERT Class 12 Chemistry > Chapter 2: Electrochemistry > Topic: Electrochemical Cells - Galvanic Cells and Electrode Potential

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Corrosion and Applications of Electrochemistry

A current of 0.2 A deposits 0.127 g of Cu from CuSO₄ in 9650 s. What is the current required to deposit 0.27 g of Al fro

Cu: Charge = 0.2 × 9650 = 1930 C , Moles = (0.127/63.5) = 0.002 mol , Charge = 0.002 × 2 × 96500 = 1930 C , matches. Al: Al³⁺ + 3e⁻ → Al , Moles = (0.27/27) = 0.01 mol , Charge = 0.01 × 3 × 96500 = 2895 C . I = (2895/9650) = 0.3 A .

Ref: NCERT Class 12 Chemistry > Chapter 2: Electrochemistry > Topic: Corrosion and Applications of Electrochemistry

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Corrosion and Applications of Electrochemistry

A current of 0.2 A deposits 0.127 g of Cu from CuSO₄ in 9650 s. What is the current required to deposit 0.27 g of Al fro

Cu: Charge = 0.2 × 9650 = 1930 C , Moles = (0.127/63.5) = 0.002 mol , Charge = 0.002 × 2 × 96500 = 1930 C , matches. Al: Al³⁺ + 3e⁻ → Al , Moles = (0.27/27) = 0.01 mol , Charge = 0.01 × 3 × 96500 = 2895 C . I = (2895/9650) = 0.3 A .

Ref: NCERT Class 12 Chemistry > Chapter 2: Electrochemistry > Topic: Corrosion and Applications of Electrochemistry

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