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: BPlease 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.
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
My answer: B
Please tell me is my logic correct? If yes, then is the answer also correct?
Thanks!!!
Thanks!!!
Devansh Rastogi , 6 Years ago
Grade 12
