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Hello, as already mentioned in the question, both H2 and Ni are being liberated and deposited respectively, at the cathode. i.e., At Cathode: Ni+2e → Ni(s) 2H20 + 2e → H2 + 4OH- From this, we say that Ni:H2 = 1:1 i.e., half of the current is used to liberate H2 and half of it used to deposit Ni. This after calculating mass of Ni deposited (from the default formula, m=(M.I.t)nF, which comes out to be 7.883g). This value should be divided by 2 to get actual weight of Ni deposited. Given answer: A My answer: B Please tell me is my logic correct? If yes, then is the answer also correct? Thanks!!!

Hello, as already mentioned in the question, both H2 and Ni are being liberated and deposited respectively, at the cathode.
i.e.,
At Cathode:
Ni+2e → Ni(s)
2H20 + 2e → H2 + 4OH-
From this, we say that Ni:H2 = 1:1
i.e., half of the current is used to liberate H2 and half of it used to deposit Ni.
This after calculating mass of Ni deposited (from the default formula, m=(M.I.t)nF, which comes out to be 7.883g). This value should be divided by 2 to get actual weight of Ni deposited.
Given answer: A
My answer: B
Please tell me is my logic correct? If yes, then is the answer also correct?
Thanks!!!

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Grade:12

2 Answers

Shankhadip Bhattacharjee
13 Points
4 years ago
We know, for each i.e, amount of current  necessary are equal. But from the formula that is already mentioned in question we are calculating
Only amount of current needed to liberate Ni .The total current will be double. 
Vikas TU
14149 Points
4 years ago
Dear student 
Your logic is correct...................................................
Good Luck 
Cheers 

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