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 ...

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In the calibration of an ammeter using Faraday's laws of electrolysis, the amount of metal deposited at the electrode is directly proportional to the amount of electric charge passed through the cell. Faraday's first law states that the amount of substance produced by the passage of current through an electrolyte is directly proportional to the quantity of electricity passed through it. We can use this law to calculate the correction to be applied to the ammeter. First, we need to calculate the quantity of electricity that passed through the cell. We can use the formula: Q = It where Q is the quantity of electricity, I is the current, and t is the time. In this case, I = 1.2 A and t = 40 minutes = 2400 seconds. Q = (1.2 A) x (2400 s) = 2880 C Next, we need to calculate the amount of copper deposited at the electrode. We can use Faraday's first law, which states that the amount of substance produced is directly proportional to the quantity of electricity passed through it. The constant of proportionality is called the electrochemical equivalent, which is the amount of substance produced by the passage of one coulomb of electricity. For copper, the electrochemical equivalent is 0.000329 g/C. The amount of copper deposited can be calculated using the formula: m = ZQ where m is the mass of copper deposited, Z is the electrochemical equivalent of copper, and Q is the quantity of electricity passed through the cell. In this case, Z = 0.000329 g/C and Q = 2880 C. m = (0.000329 g/C) x (2880 C) = 0.948 g The expected amount of copper deposited is 0.990 g, but we obtained 0.948 g. This means that there was an error in the measurement of the current. To calculate the correction to be applied to the ammeter, we can use the formula: % error = [(expected value - measured value) / expected value] x 100% % error = [(0.990 g - 0.948 g) / 0.990 g] x 100% = 4.24% The % error is positive, which means that the measured value is less than the expected value. To correct for this error, we need to increase the current by the same percentage. The ammeter reading was kept constant at 1.2 A, so we need to calculate 4.24% of 1.2 A. 0.0424 x 1.2 A = 0.051 A Therefore, the correction to be applied to the ammeter is 0.05 A (option C).