Question 12:How much charge is required for the following reductions: (i) 1 mol of Al3+ to Al. (ii) 1 mol of Cu2+ to Cu. (iii) 1 mol of MnO4– to Mn2+. Chapter 3: Electrochemistry Chemistry Class 12 solutions are developed for assisting understudies with working on their score and increase knowledge of the subjects. How much charge is required for the following reductions: (i) 1 mol of Al3+ to Al. (ii) 1 mol of Cu2+ to Cu. (iii) 1 mol of MnO4? to Mn2+. is solved by our expert teachers. You can get ncert solutions and notes for class 12 chapter 3 absolutely free. NCERT Solutions for class 12 Chemistry Chapter 3: Electrochemistry is very essencial for getting good marks in CBSE Board examinations
Question 12:How much charge is required for the following reductions:
(i) 1 mol of Al3+ to Al.
(ii) 1 mol of Cu2+ to Cu.
(iii) 1 mol of MnO4– to Mn2+.
Answer:(i)
Formula required charge n × F
n = difference of charge on ions
F is constant and equal to 96487 Coulombs
Here n = 3
Hence required charge = 3 × 96487 Coulombs
= 289461 Coulombs
= 2.89 ×10 –5 Coulombs
(ii)
n = 2
plug the value in formula we get
Required charge= 2 × 96487 Coulombs
= 192974 Coulombs
= 1.93 × 105 Coulombs
(iii)
Charge on Mn in MnO4–
Charge on Oxygen is – 2
Mn + 4O = – 1
Mn +4(–2) = – 1
Mn = +7
So our reaction is
MN7+ → Mn2+
n = 7– 2 = 5
Required charge will = 5 × 96487 Coulombs
= 482435 Coulombs
= 4.82 × 105 Coulombs
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