In the calibration of an ammeter using faraday's laws of electrolysis, the ammeter reading is kept constant at 1.2A. If 0.990g of copper is deposited in 40 minutes , the correction to be applied to the ammeter is
0.03A
0.04A
0.05A
0.06A
Explanation
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Discussions (28)
Given the electrochemical equivalent of copper (Z) is approximately 3.29 × 10^-4 g/C, the current (I) is 1.2 A, and the time (t) is 40 minutes or 2400 seconds (40 × 60):
Given mass = Z × I × t
= 3.29 × 10^-4 g/C × 1.2 A × 2400 s
= 3.29 × 10^-4 × 2880
= 0.94752 g
≈ 0.948 g
Given the actual mass deposited is 0.990 g, we can find the actual current using the formula m = ZIt,
Making I_actual the subject of formula,
I_actual = m / (Zt).
I_actual = 0.990 g / (3.29 × 10^-4 g/C × 2400 s)
= 0.990 / 0.7896
≈ 1.253 A
Correction = Actual current - Measured current
= 1.253 A - 1.2 A
= 0.053 A
≈ 0.05 A
The correct answer is C.
N/B: Z = 63.5/(2 × 96500)
= 3.290155 × 10^-4
≈ 3.29 × 10^-4
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Chai!!! I used 64 g as the mass Of 1mOle Of Cu, instead Of 63.5g which makes me get 0.043A instead Of 0.05A
I really don't understand they said calculate the correction which unit is given as Ampere and they gave us current already
