A solution of copper sulphate is electrolysed for 10 minutes with a current of 1.5 amperes. The mass of copper…

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Solution & Explanation

Related Formula

W = Z \cdot I \cdot t

Z = (E)/(F) = Molar Massn \cdot F

Core Logic

Cathode Reaction: $Cu²⁺ + 2e⁻ arrow Cu(s)$ $Cu^{2+} + 2e^{-} \rightarrow Cu(s)$ Here, $n = 2$ $n = 2$. Molar mass of Cu = $63 ~g mol⁻¹$ $63 \mathrm{~g \, mol}^{-1}$. Equivalent mass of Cu ( $E$ $E$) $= (63)/(2) = 31.5 ~g eq⁻¹$ .

Given variables: Current ( $I$ $I$) = $1.5 A$ $1.5 \text{ A}$. Time ( $t$ $t$) = $10 minutes = 10 \times 60 = 600 s$ . Faraday ( $F$ $F$) approx = $96500 ~C$ $96500 \mathrm{~C}$.

Step 1: Calculate Mass

W = (E \cdot I \cdot t)/(F)

W = (63 \times 1.5 \times 600)/(2 \times 96500)

W = (56700)/(193000)

W = 0.29378 g \approx 0.2938 g

Pattern Recognition

Use $W = (M \cdot I \cdot t)/(nF)$ $W = \frac{M \cdot I \cdot t}{nF}$. Never forget to convert time from minutes to seconds, and remember that Copper in Copper Sulphate ( $CuSO₄$ $CuSO_4$) is in the +2 state ( $n=2$ $n=2$).

Chapter Mix

Class 12 Chemistry: Electrochemistry

More Electrochemistry Previous-Year Questions

Q55 neet_2026_03_may_morning Nernst Equation and Electrode Potential

Calculate emf of the half cell given below: Pt (s) |

H₂

(g, 2 atm) | HCl (aq, 0.02 M)

E_H₂ / H⁺° = 0

(Given : 2.303 RTF = 0.059, 2 = 0.3010)

A. 0.109 V
B. 0.035 V
C. -0.035 V
D. -0.109 V

Solution

Related Formula

E = E° - 2.303RTnF Q

Core Logic

Reaction for the oxidation half-cell:

H₂(g) arrow 2H⁺(aq) + 2e⁻

Here, $n = 2$ $n = 2$.

Nernst equation:

E = E° - (0.059)/(n) [H^+]²PH₂

Given parameters: $[H^+] = 0.02 ~M = 2 \times 10⁻² ~M$ $PH₂ = 2 ~atm$ $P_{\mathrm{H}_2} = 2 \mathrm{~atm}$ $E° = 0 ~V$ $E^{\circ} = 0 \mathrm{~V}$

Step 1: Calculate E

E = 0 - (0.059)/(2) ((0.02)²)/(2)

E = -0.0295 ( (0.0004)/(2) )

E = -0.0295 (2 \times 10⁻⁴)

E = -0.0295 ( 2 + 10⁻⁴)

E = -0.0295 \times (0.3010 - 4)

E = -0.0295 \times (-3.699)

E = +0.1091 ~V \approx 0.109 ~V

Pattern Recognition

Since standard potential of SHE is 0V, when oxidizing hydrogen gas into protons, lower proton concentration ( $<1M$ $<1M$) and higher gas pressure ( $>1atm$ $>1\text{atm}$) makes the cell more spontaneous (positive $E$ $E$).

Chapter Mix

Class 12 Chemistry: Electrochemistry

A solution of copper sulphate is electrolysed for 10 minutes with a current of 1.5 amperes. The mass of copper deposited at cathode is: (Given: Molar mass of Cu = $63 ~g mol⁻¹$ $63 \mathrm{~g \, mol}^{-1}$ ; 1 F = $96487 ~C mol⁻¹$ $96487 \mathrm{~C \, mol}^{-1}$)

Solution & Explanation

Related Formula

Core Logic

Step 1: Calculate Mass

Pattern Recognition

Chapter Mix

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Solution

Related Formula

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Step 1: Calculate E

Pattern Recognition

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A solution of copper sulphate is electrolysed for 10 minutes with a current of 1.5 amperes. The mass of copper deposited at cathode is: (Given: Molar mass of Cu = 63 ~g mol⁻¹$63 \mathrm{~g \, mol}^{-1}$ ; 1 F = 96487 ~C mol⁻¹$96487 \mathrm{~C \, mol}^{-1}$)

Solution & Explanation

Related Formula

Core Logic

Step 1: Calculate Mass

Pattern Recognition

Chapter Mix

Reference Study Guides

More Electrochemistry Previous-Year Questions

Solution

Related Formula

Core Logic

Step 1: Calculate E

Pattern Recognition

Chapter Mix

More Electrochemistry Questions — neet_2026_03_may_morning

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A solution of copper sulphate is electrolysed for 10 minutes with a current of 1.5 amperes. The mass of copper deposited at cathode is: (Given: Molar mass of Cu = $63 ~g mol⁻¹$ $63 \mathrm{~g \, mol}^{-1}$ ; 1 F = $96487 ~C mol⁻¹$ $96487 \mathrm{~C \, mol}^{-1}$)