A solution of $N i (N O_{3})_{2}$ is electrolysis between plantinum electrodes using a current a current of $5$ amperes for $20 m i n u t e s$ . What mass of $N i$ is deposited at the cathode?

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Solution

Current passed = 5A

Time = 20 × 60 = 1200 s
Charge = current × time
$= 5 \times 1200 = 6000 C$
According to the reaction, $N i^{2 +} + 2 e^{-} \to N i$

Nickel deposited by $2 \times 96487 C = 58.71 g$

Therefore, nickel deposited by $6000 C = \frac{58.71 \times 6000}{2 X 96487} g$
= 1.825 g
Hence, 1.825 g of nickel will be deposited at the cathode.

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A solution of Ni(NO3)2 is electrolysis between plantinum electrodes using a current a current of 5 amperes for 20 minutes. What mass of Ni is deposited at the cathode?

Current passed = 5A Time = 20 × 60 = 1200 s Charge = current × time =5×1200=6000C According to the reaction, Ni2++2e−→Ni Nickel deposited by 2×96487C=58.71g Therefore, nickel deposited by 6000C=58.71×60002X96487g = 1.825 g Hence, 1.825 g of nickel will be deposited at the cathode.

A solution of $N i (N O_{3})_{2}$ is electrolysis between plantinum electrodes using a current a current of $5$ amperes for $20 m i n u t e s$ . What mass of $N i$ is deposited at the cathode?